Electromagnetic Waves

NEET Ug . NEET Physics . Physics . Electromagnetic Waves

Chapter 23: Electromagnetic Waves

Ch 23

NEET Physics

NEET Ug

Introduction

Key Points

Maxwell identified inconsistency in Ampere’s circuital law for magnetic field outside a capacitor with time-varying current.
He introduced displacement current to resolve this inconsistency.
Maxwell’s equations unify electric and magnetic fields with their sources (charge and current densities).
Hertz experimentally confirmed electromagnetic waves in 1885; Marconi used them for communication revolution.

Solved Examples

Ex 1: Explain why displacement current is needed in Ampere’s law.
- Charging capacitor shows magnetic field discontinuity if only conduction current considered.
- Displacement current term added to maintain continuity and satisfy Kirchhoff’s law.

Practice Set

Q1: What did Maxwell add to Ampere’s law?
- A) Magnetic flux
- B) Displacement current
- C) Electric charge
- D) Electric flux
Q2: Who experimentally demonstrated electromagnetic waves?
- A) Maxwell
- B) Marconi
- C) Hertz
- D) Faraday
Q3: Maxwell’s equations relate which of the following?
- A) Only electric fields
- B) Only magnetic fields
- C) Electric and magnetic fields with charges and currents
- D) Only charges and currents

Answer Key

Q1: B - Displacement current added to fix Ampere’s law inconsistency.
Q2: C - Hertz demonstrated electromagnetic waves experimentally.
Q3: C - Maxwell’s equations unify electric and magnetic fields with sources.

Memory Aids

Displacement current = changing electric flux × permittivity.
Maxwell’s equations unify electricity and magnetism.
Hertz = electromagnetic waves proof.

Displacement Current

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current I_D = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 1 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current I_D = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 2 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current (I_D) = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 3 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current I_D = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 4 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\) acts like current in capacitor gap.
Electric field between capacitor plates: \(E = \frac{Q}{\varepsilon_0 A}\).
Electric flux: \(\Phi_E = EA\).
No real charge flows through capacitor gap, but changing \(E\) creates displacement current.

Diagram 5 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if \(\frac{d\Phi_E}{dt} = 5 \times 10^{-6} \text{ Wb/s} \) and \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \).
- \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt} = 8.85 \times 10^{-12} \times 5 \times 10^{-6} = 4.425 \times 10^{-17} \text{ A}\)

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) \(E = \frac{Q}{\varepsilon_0 A}\)
- B) \(E = \frac{Q}{A}\)
- C) \(E = \varepsilon_0 Q A\)
- D) \(E = \frac{A}{Q}\)
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field \(E = \frac{Q}{\varepsilon_0 A}\).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\).
Electric flux \(\Phi_E = EA\).
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current I_D = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 6 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current (I_D) = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 7 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current I_D = ε₀ dΦ_E/dt acts like current in capacitor gap.
Electric field between capacitor plates: E = Q/(ε₀A).
Electric flux: Φ_E = EA.
No real charge flows through capacitor gap, but changing E creates displacement current.

Diagram 8 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if dΦ_E/dt = 5 × 10^-6 Wb/s and ε₀ = 8.85 × 10^-12 F/m .
- I_D = ε₀ dΦ_E/dt = 8.85 × 10^-12 × 5 × 10^-6 = 4.425 × 10^-17 A

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) E = Q/(ε₀A)
- B) E = Q/A
- C) E = ε₀QA
- D) E = A/Q
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field E = Q/(ε₀A).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current I_D = ε₀ dΦ_E/dt.
Electric flux Φ_E = EA.
No real charge crosses capacitor gap.

Concepts

Changing electric field produces magnetic field (Maxwell’s insight).
Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\) acts like current in capacitor gap.
Electric field between capacitor plates: \(E = \frac{Q}{\varepsilon_0 A}\).
Electric flux: \(\Phi_E = EA\).
No real charge flows through capacitor gap, but changing \(E\) creates displacement current.

Diagram 9 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate displacement current if \(\frac{d\Phi_E}{dt} = 5 \times 10^{-6} \text{ Wb/s} \) and \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \).
- \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt} = 8.85 \times 10^{-12} \times 5 \times 10^{-6} = 4.425 \times 10^{-17} \text{ A}\)

Practice Set

Q1: Displacement current arises due to?
- A) Constant electric field
- B) Changing electric field
- C) Magnetic field
- D) Static charge
Q2: Electric field between capacitor plates is given by?
- A) \(E = \frac{Q}{\varepsilon_0 A}\)
- B) \(E = \frac{Q}{A}\)
- C) \(E = \varepsilon_0 Q A\)
- D) \(E = \frac{A}{Q}\)
Q3: Displacement current is?
- A) Actual charge flow
- B) Magnetic flux
- C) Equivalent current due to changing electric field
- D) Zero

Answer Key

Q1: B - Displacement current arises from changing electric field.
Q2: A - Electric field \(E = \frac{Q}{\varepsilon_0 A}\).
Q3: C - Displacement current behaves like current due to changing electric field.

Memory Aids

Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\).
Electric flux \(\Phi_E = EA\).
No real charge crosses capacitor gap.

Inconsistency of Ampere’s Circuital Law

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 10 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 11 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 12 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current \(I\) enters plate \(P_1\) but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops \(C_1\) (around wire) and \(C_2\) (between plates) give different magnetic field results using Ampere’s law.
Loop \(C_1\): \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I \neq 0\).
Loop \(C_2\): \(\oint \vec{B} \cdot d\vec{l} = 0\) as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 13 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop \(C_1\) encloses conduction current \(I\), magnetic field non-zero.
- Loop \(C_2\) encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop \(C_2\) in capacitor gap shows magnetic field?
- A) Equal to \(\mu_0 I\)
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop \(C_2\) has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 14 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 15 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current I enters plate P₁ but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops C₁ (around wire) and C₂ (between plates) give different magnetic field results using Ampere’s law.
Loop C₁: ∮ B · dl = μ₀I ≠ 0.
Loop C₂: ∮ B · dl = 0 as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 16 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop C₁ encloses conduction current I, magnetic field non-zero.
- Loop C₂ encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop C₂ in capacitor gap shows magnetic field?
- A) Equal to μ₀I
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop C₂ has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Key Points

Charging capacitor shows current \(I\) enters plate \(P_1\) but no conduction current beyond it.
Kirchhoff’s first law seems violated as current appears to stop.
Two loops \(C_1\) (around wire) and \(C_2\) (between plates) give different magnetic field results using Ampere’s law.
Loop \(C_1\): \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I \neq 0\).
Loop \(C_2\): \(\oint \vec{B} \cdot d\vec{l} = 0\) as conduction current is zero.
This discontinuity contradicts continuous magnetic field observed physically.
Maxwell proposed modifying Ampere’s law to include displacement current.

Diagram 17 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Explain why magnetic field is discontinuous using Ampere’s law.
- Loop \(C_1\) encloses conduction current \(I\), magnetic field non-zero.
- Loop \(C_2\) encloses no conduction current, magnetic field zero.
- Physical magnetic field is continuous, so Ampere’s law incomplete.

Practice Set

Q1: What causes inconsistency in Ampere’s law?
- A) Magnetic field discontinuity
- B) Electric field discontinuity
- C) Charge accumulation
- D) Current continuity
Q2: Loop \(C_2\) in capacitor gap shows magnetic field?
- A) Equal to \(\mu_0 I\)
- B) Zero
- C) Infinite
- D) Negative
Q3: Why is Kirchhoff’s law apparently violated?
- A) Current stops at capacitor plate
- B) No displacement current considered
- C) Magnetic field is zero
- D) Electric field is constant

Answer Key

Q1: A - Magnetic field discontinuity causes inconsistency.
Q2: B - Loop \(C_2\) has zero conduction current, so magnetic field zero.
Q3: B - Displacement current missing causes Kirchhoff’s law violation.

Memory Aids

Ampere’s law incomplete without displacement current.
Magnetic field must be continuous.
Displacement current fixes discontinuity.

Modified Ampere’s Law by Maxwell

Key Points

Maxwell modified Ampere’s law to include displacement current I_D.
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current (I_D).
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current I_D.
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current (I_D).
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current I_D.
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current (I_D).
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current I_D.
Generalized law: ∮ B · dl = μ₀ [I + I_D].
Displacement current: I_D = ε₀ dΦ_E/dt, where Φ_E = EA.
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- ∮ B · dl = μ₀ [I + ε₀ dΦ_E/dt]

Practice Set

Q1: Displacement current formula is?
- A) I_D = ε₀ dΦ_E/dt
- B) I_D = μ₀I
- C) I_D = Q/A
- D) I_D = 0
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula I_D = ε₀ dΦ_E/dt.
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: ∮ B · dl = μ₀ [I + I_D].
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Key Points

Maxwell modified Ampere’s law to include displacement current \(I_D\).
Generalized law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 [I + I_D]\).
Displacement current: \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\), where \(\Phi_E = EA\).
This modification ensures magnetic field continuity and Kirchhoff’s law validity.

Solved Examples

Ex 1: Write generalized Ampere’s law.
- \(\oint \vec{B} \cdot d\vec{l} = \mu_0 [I + \varepsilon_0 \frac{d\Phi_E}{dt}]\)

Practice Set

Q1: Displacement current formula is?
- A) \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\)
- B) \(I_D = \mu_0 I\)
- C) \(I_D = \frac{Q}{A}\)
- D) \(I_D = 0\)
Q2: Generalized Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither
Q3: Displacement current exists where?
- A) Outside capacitor plates
- B) Inside capacitor gap with changing electric flux
- C) In resistors
- D) In open air only

Answer Key

Q1: A - Displacement current formula \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\).
Q2: C - Generalized law includes both currents.
Q3: B - Displacement current exists in capacitor gap with changing flux.

Memory Aids

Generalized Ampere’s law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 [I + I_D]\).
Displacement current equals conduction current in magnitude.
Ensures magnetic field continuity.

Consistency of Modified Ampere’s Law

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop \(C_1\): \(I_D = 0\), magnetic field due to conduction current only.
Loop \(C_2\): \(I = 0\), magnetic field due to displacement current \(I_D\).
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current \(I_D = I\).
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show \(I_D = I\) using electric flux.
- \(\Phi_E = EA = \frac{q}{\varepsilon_0 A} \times A = \frac{q}{\varepsilon_0}\)
- \(\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dq}{dt} = \frac{I}{\varepsilon_0}\)
- \(I_D = \varepsilon_0 \times \frac{I}{\varepsilon_0} = I\)

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current \(I_D = I\).
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop C₁: I_D = 0, magnetic field due to conduction current only.
Loop C₂: I = 0, magnetic field due to displacement current I_D.
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current I_D = I.
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show I_D = I using electric flux.
- Φ_E = EA = q/(ε₀A) × A = q/ε₀
- dΦ_E/dt = 1/ε₀ dq/dt = I/ε₀
- I_D = ε₀ × I/ε₀ = I

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current I_D = I.
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Key Points

Loop \(C_1\): \(I_D = 0\), magnetic field due to conduction current only.
Loop \(C_2\): \(I = 0\), magnetic field due to displacement current \(I_D\).
Magnetic field is continuous inside and outside capacitor plates.
Displacement current magnitude equals conduction current \(I_D = I\).
Kirchhoff’s first law holds with displacement current included.

Solved Examples

Ex 1: Show \(I_D = I\) using electric flux.
- \(\Phi_E = EA = \frac{q}{\varepsilon_0 A} \times A = \frac{q}{\varepsilon_0}\)
- \(\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dq}{dt} = \frac{I}{\varepsilon_0}\)
- \(I_D = \varepsilon_0 \times \frac{I}{\varepsilon_0} = I\)

Practice Set

Q1: Displacement current magnitude compared to conduction current is?
- A) Zero
- B) Half
- C) Equal
- D) Double
Q2: Kirchhoff’s first law applies when?
- A) Only conduction current present
- B) Only displacement current present
- C) Both conduction and displacement currents present
- D) No current present
Q3: Magnetic field continuity ensured by?
- A) Conduction current only
- B) Displacement current only
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: C - Displacement current equals conduction current.
Q2: C - Kirchhoff’s law valid with both currents.
Q3: C - Magnetic field continuity needs both currents.

Memory Aids

Displacement current \(I_D = I\).
Kirchhoff’s law valid with displacement current.
Magnetic field continuous everywhere.

Maxwell’s Equations

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · dS = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · dS = 0.
3. Faraday’s law: ∮_l E · dl = - d/dt ∮_S B · dS.
4. Modified Ampere’s law: ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · dS = q/ε₀
- 2) ∮_S B · dS = 0
- 3) ∮_l E · dl = - d/dt ∮_S B · dS
- 4) ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?

A) Only conduction current
B) Only displacement current
C) Both conduction and displacement currents
D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · dS = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · dS = 0.
3. Faraday’s law: ∮_l E · dl = - d/dt ∮_S B · dS.
4. Modified Ampere’s law: ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · dS = q/ε₀
- 2) ∮_S B · dS = 0
- 3) ∮_l E · dl = - d/dt ∮_S B · dS
- 4) ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?

A) Only conduction current
B) Only displacement current
C) Both conduction and displacement currents
D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · d_S = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · d_S = 0.
3. Faraday’s law: ∮_l E · d_l = - d/dt ∮_S B · d_S.
4. Modified Ampere’s law: ∮_l B · d_l = μ₀ [I + ε₀ d/dt ∮_S E · d_S].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · d_S = q/ε₀
- 2) ∮_S B · d_S = 0
- 3) ∮_l E · d_l = - d/dt ∮_S B · d_S
- 4) ∮_l B · d_l = μ₀ [I + ε₀ d/dt ∮_S E · d_S]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?

A) Only conduction current
B) Only displacement current
C) Both conduction and displacement currents
D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · dS = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · dS = 0.
3. Faraday’s law: ∮_l E · dl = - d/dt ∮_S B · dS.
4. Modified Ampere’s law: ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · dS = q/ε₀
- 2) ∮_S B · dS = 0
- 3) ∮_l E · dl = - d/dt ∮_S B · dS
- 4) ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · dS = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · dS = 0.
3. Faraday’s law: ∮_l E · dl = - d/dt ∮_S B · dS.
4. Modified Ampere’s law: ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · dS = q/ε₀
- 2) ∮_S B · dS = 0
- 3) ∮_l E · dl = - d/dt ∮_S B · dS
- 4) ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?

A) Only conduction current
B) Only displacement current
C) Both conduction and displacement currents
D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · d_S = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · d_S = 0.
3. Faraday’s law: ∮_l E · d_l = - d/dt ∮_S B · d_S.
4. Modified Ampere’s law: ∮_l B · d_l = μ₀ [I + ε₀ d/dt ∮_S E · d_S].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · d_S = q/ε₀
- 2) ∮_S B · d_S = 0
- 3) ∮_l E · d_l = - d/dt ∮_S B · d_S
- 4) ∮_l B · d_l = μ₀ [I + ε₀ d/dt ∮_S E · d_S]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?

A) Only conduction current
B) Only displacement current
C) Both conduction and displacement currents
D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: ∮_S E · dS = q_inside/ε₀.
2. Gauss’s law for magnetism: ∮_S B · dS = 0.
3. Faraday’s law: ∮_l E · dl = - d/dt ∮_S B · dS.
4. Modified Ampere’s law: ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS].
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) ∮_S E · dS = q/ε₀
- 2) ∮_S B · dS = 0
- 3) ∮_l E · dl = - d/dt ∮_S B · dS
- 4) ∮_l B · dl = μ₀ [I + ε₀ d/dt ∮_S E · dS]

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by ε₀
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/ε₀.
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = μ₀(conduction + displacement current).

Key Points

Four fundamental laws of electricity and magnetism.
Explain production of electromagnetic waves.
1. Gauss’s law for electricity: \(\oint_S \vec{E} \cdot d\vec{S} = \frac{q_{inside}}{\varepsilon_0}\).
2. Gauss’s law for magnetism: \(\oint_S \vec{B} \cdot d\vec{S} = 0\).
3. Faraday’s law: \(\oint_l \vec{E} \cdot d\vec{l} = - \frac{d}{dt} \oint_S \vec{B} \cdot d\vec{S}\).
4. Modified Ampere’s law: \(\oint_l \vec{B} \cdot d\vec{l} = \mu_0 \left[I + \varepsilon_0 \frac{d}{dt} \oint_S \vec{E} \cdot d\vec{S} \right]\).
Changing electric field produces magnetic field and vice versa.

Solved Examples

Ex 1: Write Maxwell’s four equations.
- 1) \(\oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0}\)
- 2) \(\oint_S \vec{B} \cdot d\vec{S} = 0\)
- 3) \(\oint_l \vec{E} \cdot d\vec{l} = - \frac{d}{dt} \oint_S \vec{B} \cdot d\vec{S}\)
- 4) \(\oint_l \vec{B} \cdot d\vec{l} = \mu_0 \left[I + \varepsilon_0 \frac{d}{dt} \oint_S \vec{E} \cdot d\vec{S} \right]\)

Practice Set

Q1: Gauss’s law for electricity states?
- A) Magnetic flux through closed surface is zero
- B) Electric flux through closed surface equals charge inside divided by \(\varepsilon_0\)
- C) Induced emf equals rate of change of magnetic flux
- D) Magnetic field line integral equals current
Q2: Faraday’s law relates induced emf to?
- A) Electric flux change
- B) Magnetic flux change
- C) Charge change
- D) Current change
Q3: Modified Ampere’s law includes?
- A) Only conduction current
- B) Only displacement current
- C) Both conduction and displacement currents
- D) Neither

Answer Key

Q1: B - Electric flux equals charge inside divided by permittivity.
Q2: B - Induced emf proportional to rate of change of magnetic flux.
Q3: C - Includes both conduction and displacement currents.

Memory Aids

Gauss’s law (electric): flux = charge/\(\varepsilon_0\).
Gauss’s law (magnetic): flux = 0.
Faraday: emf = -d(magnetic flux)/dt.
Ampere-Maxwell: magnetic field line integral = \(\mu_0\)(conduction + displacement current).

Electromagnetic Waves

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 18 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 19 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 20 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 21 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light \( c = 3 \times 10^{8} \) m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 22 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 23 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 24 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light c = 3 × 10⁸ m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 25 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Definition and Properties

Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
Produced by alternating electric and magnetic fields as per Maxwell's equations.
Frequency of the wave equals frequency of oscillating fields or accelerated charges.
Carry energy and momentum through space at speed of light \( c = 3 \times 10^{8} \) m/s in vacuum.

Production of Electromagnetic Waves

Accelerated charges emit electromagnetic waves.
Oscillating charges produce oscillating electric and magnetic fields regenerating each other.
Energy of wave comes from energy of accelerating charge.

Diagram 26 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: An antenna oscillates charges at frequency 1 MHz. What is the frequency of emitted EM wave?
Solution: Frequency of EM wave = frequency of oscillation = 1 MHz.

Practice Set

Q1: What causes electromagnetic waves?
A. Stationary charges
B. Uniform motion charges
C. Accelerated charges
D. None of these
Q2: The electric and magnetic fields in an EM wave are:
A. Parallel
B. Perpendicular to each other and to wave direction
C. Perpendicular only to each other
D. Randomly oriented
Q3: If an EM wave frequency is doubled, its wavelength:
A. Doubles
B. Halves
C. Remains same
D. Quadruples

Answer Key

Q1: C - Accelerated charges produce EM waves.
Q2: B - Fields are perpendicular to each other and direction of propagation.
Q3: B - Wavelength inversely proportional to frequency.

Memory Aids

EM waves = Oscillating E and B fields at right angles.
Accelerated charges = EM wave source.
Frequency of wave = frequency of oscillation.

Hertz Experiment

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 27 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 28 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 29 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 30 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 31 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 32 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 33 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 34 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 35 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 36 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 37 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 38 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 39 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 40 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 41 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 42 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 43 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 44 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 45 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 46 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 47 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 48 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 49 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 50 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 51 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 52 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 53 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 54 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 55 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 56 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 57 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 58 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 59 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 60 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 61 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 62 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 63 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 64 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 65 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency ν = 1/(2π√LC).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Setup and Working

Two metal plates connected to spheres and induction coil produce high-frequency oscillations.
High potential ionizes air gap, causing sparks and oscillating electric field.
Oscillating magnetic field perpendicular to electric field is produced.
EM waves generated propagate through space and detected by a loop with gap producing sparks.

Diagram 66 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: What causes the spark in Hertz's detector?
Solution: EM waves induce emf in detector loop causing spark.

Practice Set

Q1: What ionizes air in Hertz's experiment?
A. Low potential difference
B. High potential difference
C. Magnetic field
D. Electric current
Q2: The resonant frequency in Hertz's setup depends on:
A. Inductance and capacitance
B. Resistance only
C. Voltage only
D. Current only
Q3: What is the role of induction coil?
A. Produce sparks
B. Cause oscillations in spheres
C. Detect EM waves
D. None of these

Answer Key

Q1: B - High potential difference ionizes air.
Q2: A - Resonant frequency \( \nu = \frac{1}{2\pi\sqrt{LC}} \).
Q3: B - Induction coil causes oscillations in spheres.

Memory Aids

High voltage = air ionization = sparks.
Resonant frequency depends on L and C.
Induction coil drives oscillations.

Nature of Electromagnetic Waves

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 67 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 68 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 69 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 70 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 71 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 72 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field E_x = E₀ sin(kz - ωt) = E₀ sin[2π(z/λ - νt)].
Magnetic field B_y = B₀ sin(kz - ωt) = B₀ sin[2π(z/λ - νt)].
Relation: E/B = c or E₀/B₀ = c.
Speed in vacuum: c = 1/√(μ₀ε₀).
Speed in medium: v = 1/√(μϵ), where μ and ϵ are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum p = U/c where U is energy transferred.

Diagram 73 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: u_E = 1/2 ε₀E².
Energy density of magnetic field: u_B = 1/(2μ₀) B².
Total energy density: u = u_E + u_B = 1/2 ε₀E²_rms + 1/(2μ₀) B²_rms.
Using c = E_rms/B_rms and c = 1/√(μ₀ε₀), u = ε₀E²_rms = B²_rms/μ₀.

Solved Examples

Ex 1: Calculate energy density if E_rms = 100 V/m.
Solution: u = ε₀E²_rms = 8.85 × 10^-12 × (100)² = 8.85 × 10^-8 J/m³.

Practice Set

Q1: The ratio E/B in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. 1/2 ε₀E²
B. 1/(2μ₀) B²
C. ε₀E²
D. B²/ε₀

Answer Key

Q1: B - E/B = c, speed of light.
Q2: A - c = 1/√(μ₀ε₀).
Q3: B - Magnetic energy density = 1/(2μ₀) B².

Memory Aids

EM wave speed c = 1/√(μ₀ε₀).
Energy density u = ε₀E²_rms = B²_rms/μ₀.
Fields perpendicular and transverse wave.

Wave Properties and Equations

Electric field \( E_x = E_0 \sin(kz - \omega t) = E_0 \sin[2\pi(z/\lambda - \nu t)] \).
Magnetic field \( B_y = B_0 \sin(kz - \omega t) = B_0 \sin[2\pi(z/\lambda - \nu t)] \).
Relation: \( \frac{E}{B} = c \) or \( \frac{E_0}{B_0} = c \).
Speed in vacuum: \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \).
Speed in medium: \( v = \frac{1}{\sqrt{\mu \epsilon}} \), where \( \mu \) and \( \epsilon \) are permeability and permittivity.
EM waves are transverse; E and B fields perpendicular to each other and propagation direction.
Carry energy and momentum; momentum \( p = \frac{U}{c} \) where \( U \) is energy transferred.

Diagram 74 – NEET Physics Electromagnetic Waves

Energy Density

Energy density of electric field: \( u_E = \frac{1}{2} \epsilon_0 E^2 \).
Energy density of magnetic field: \( u_B = \frac{1}{2 \mu_0} B^2 \).
Total energy density: \( u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2_{rms} + \frac{1}{2 \mu_0} B^2_{rms} \).
Using \( c = \frac{E_{rms}}{B_{rms}} \) and \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \), \( u = \epsilon_0 E^2_{rms} = \frac{B^2_{rms}}{\mu_0} \).

Solved Examples

Ex 1: Calculate energy density if \( E_{rms} = 100 \) V/m.
Solution: \( u = \epsilon_0 E^2_{rms} = 8.85 \times 10^{-12} \times (100)^2 = 8.85 \times 10^{-8} \) J/m³.

Practice Set

Q1: The ratio \( \frac{E}{B} \) in EM waves equals:
A. Speed of sound
B. Speed of light
C. Frequency
D. Wavelength
Q2: The speed of EM wave in vacuum depends on:
A. Permeability and permittivity
B. Frequency
C. Amplitude
D. Medium temperature
Q3: Energy density of magnetic field is:
A. \( \frac{1}{2} \epsilon_0 E^2 \)
B. \( \frac{1}{2 \mu_0} B^2 \)
C. \( \epsilon_0 E^2 \)
D. \( \frac{B^2}{\epsilon_0} \)

Answer Key

Q1: B - \( E/B = c \), speed of light.
Q2: A - \( c = 1/\sqrt{\mu_0 \epsilon_0} \).
Q3: B - Magnetic energy density \( = \frac{1}{2 \mu_0} B^2 \).

Memory Aids

EM wave speed \( c = 1/\sqrt{\mu_0 \epsilon_0} \).
Energy density \( u = \epsilon_0 E^2_{rms} = B^2_{rms}/\mu_0 \).
Fields perpendicular and transverse wave.

Electromagnetic Spectrum

Overview

Electromagnetic spectrum: orderly distribution of waves by frequency or wavelength.
Wavelength range: gamma rays (shortest) to radio waves (longest).
Frequency and energy increase as wavelength decreases.
Visible light: small part of spectrum, wavelength 400–700 nm, frequency 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Energy relation: E = hν (energy proportional to frequency).

Diagram 75 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of visible light with wavelength 500 nm.
- Use ν = c/λ, c = 3 × 10⁸ m/s, λ = 500 × 10^-9 m.
- ν = 3 × 10⁸/(500 × 10^-9) = 6 × 10¹⁴ Hz.

Practice Set

Q1: Which wave has the highest frequency?
- A. Radio waves
- B. Microwaves
- C. X-rays
- D. Gamma rays
Q2: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q3: Energy of a photon is directly proportional to:
- A. Wavelength
- B. Frequency
- C. Amplitude
- D. Speed

Answer Key

Q1: D. Gamma rays have the highest frequency and shortest wavelength.
Q2: A. Visible light ranges from 400 to 700 nm wavelength.
Q3: B. Energy E = hν is proportional to frequency.

Memory Aids

Frequency ↑ → Wavelength ↓ → Energy ↑
Visible light colors: VIBGYOR (Violet shortest wavelength)
Energy formula: E = hν

Overview

Electromagnetic spectrum: orderly distribution of waves by frequency or wavelength.
Wavelength range: gamma rays (shortest) to radio waves (longest).
Frequency and energy increase as wavelength decreases.
Visible light: small part of spectrum, wavelength 400–700 nm, frequency 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Energy relation: E = hν (energy proportional to frequency).

Diagram 76 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of visible light with wavelength 500 nm.
- Use ν = c/λ, c = 3 × 10⁸ m/s, λ = 500 × 10^-9 m.
- ν = 3 × 10⁸/(500 × 10^-9) = 6 × 10¹⁴ Hz.

Practice Set

Q1: Which wave has the highest frequency?
- A. Radio waves
- B. Microwaves
- C. X-rays
- D. Gamma rays
Q2: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q3: Energy of a photon is directly proportional to:
- A. Wavelength
- B. Frequency
- C. Amplitude
- D. Speed

Answer Key

Q1: D. Gamma rays have the highest frequency and shortest wavelength.
Q2: A. Visible light ranges from 400 to 700 nm wavelength.
Q3: B. Energy E = hν is proportional to frequency.

Memory Aids

Frequency ↑ → Wavelength ↓ → Energy ↑
Visible light colors: VIBGYOR (Violet shortest wavelength)
Energy formula: E = hν

Overview

Electromagnetic spectrum: orderly distribution of waves by frequency or wavelength.
Wavelength range: gamma rays (shortest) to radio waves (longest).
Frequency and energy increase as wavelength decreases.
Visible light: small part of spectrum, wavelength 400–700 nm, frequency 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Energy relation: E = hν (energy proportional to frequency).

Diagram 77 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of visible light with wavelength 500 nm.
- Use ν = c/λ, c = 3 × 10⁸ m/s, λ = 500 × 10^-9 m.
- ν = 3 × 10⁸/(500 × 10^-9) = 6 × 10¹⁴ Hz.

Practice Set

Q1: Which wave has the highest frequency?
- A. Radio waves
- B. Microwaves
- C. X-rays
- D. Gamma rays
Q2: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q3: Energy of a photon is directly proportional to:
- A. Wavelength
- B. Frequency
- C. Amplitude
- D. Speed

Answer Key

Q1: D. Gamma rays have the highest frequency and shortest wavelength.
Q2: A. Visible light ranges from 400 to 700 nm wavelength.
Q3: B. Energy E = hν is proportional to frequency.

Memory Aids

Frequency ↑ → Wavelength ↓ → Energy ↑
Visible light colors: VIBGYOR (Violet shortest wavelength)
Energy formula: E = hν

Overview

Electromagnetic spectrum: orderly distribution of waves by frequency or wavelength.
Wavelength range: gamma rays (shortest) to radio waves (longest).
Frequency and energy increase as wavelength decreases.
Visible light: small part of spectrum, wavelength 400–700 nm, frequency 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Energy relation: \( E = h \nu \) (energy proportional to frequency).

Diagram 78 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of visible light with wavelength 500 nm.
- Use \( \nu = \frac{c}{\lambda} \), \( c = 3 \times 10^8 \) m/s, \( \lambda = 500 \times 10^{-9} \) m.
- \( \nu = \frac{3 \times 10^8}{500 \times 10^{-9}} = 6 \times 10^{14} \) Hz.

Practice Set

Q1: Which wave has the highest frequency?
- A. Radio waves
- B. Microwaves
- C. X-rays
- D. Gamma rays
Q2: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q3: Energy of a photon is directly proportional to:
- A. Wavelength
- B. Frequency
- C. Amplitude
- D. Speed

Answer Key

Q1: D. Gamma rays have the highest frequency and shortest wavelength.
Q2: A. Visible light ranges from 400 to 700 nm wavelength.
Q3: B. Energy \( E = h \nu \) is proportional to frequency.

Memory Aids

Frequency \( \uparrow \) → Wavelength \( \downarrow \) → Energy \( \uparrow \)
Visible light colors: VIBGYOR (Violet shortest wavelength)
Energy formula: \( E = h \nu \)

Radio Waves

Properties and Uses

Frequency range: 500 kHz to 1000 MHz.
Used in communication: radio, TV, cellular phones.

Solved Examples

Ex 1: Find wavelength of radio wave with frequency 1 MHz.
- λ = c/ν = 3 × 10⁸/(1 × 10⁶) = 300 m.

Practice Set

Q1: Radio waves frequency range is:
- A. 500 kHz to 1000 MHz
- B. 1 MHz to 10 GHz
- C. 100 MHz to 1 GHz
- D. 10 kHz to 500 kHz
Q2: Radio waves are mainly used for:
- A. Cooking
- B. Communication
- C. Medical imaging
- D. Sterilization
Q3: Wavelength of a 500 kHz radio wave is:
- A. 600 m
- B. 300 m
- C. 150 m
- D. 1000 m

Answer Key

Q1: A. Radio waves frequency range is 500 kHz to 1000 MHz.
Q2: B. Radio waves are used in communication systems.
Q3: A. λ = 3 × 10⁸/(5 × 10⁵) = 600 m.

Memory Aids

Radio waves = Communication waves
Frequency range: 500 kHz to 1000 MHz
Wavelength λ = c/ν

Properties and Uses

Frequency range: 500 kHz to 1000 MHz.
Used in communication: radio, TV, cellular phones.

Solved Examples

Ex 1: Find wavelength of radio wave with frequency 1 MHz.
- λ = c/ν = 3 × 10⁸/(1 × 10⁶) = 300 m.

Practice Set

Q1: Radio waves frequency range is:
- A. 500 kHz to 1000 MHz
- B. 1 MHz to 10 GHz
- C. 100 MHz to 1 GHz
- D. 10 kHz to 500 kHz
Q2: Radio waves are mainly used for:
- A. Cooking
- B. Communication
- C. Medical imaging
- D. Sterilization
Q3: Wavelength of a 500 kHz radio wave is:
- A. 600 m
- B. 300 m
- C. 150 m
- D. 1000 m

Answer Key

Q1: A. Radio waves frequency range is 500 kHz to 1000 MHz.
Q2: B. Radio waves are used in communication systems.
Q3: A. λ = 3 × 10⁸/(5 × 10⁵) = 600 m.

Memory Aids

Radio waves = Communication waves
Frequency range: 500 kHz to 1000 MHz
Wavelength λ = c/ν

Properties and Uses

Frequency range: 500 kHz to 1000 MHz.
Used in communication: radio, TV, cellular phones.

Solved Examples

Ex 1: Find wavelength of radio wave with frequency 1 MHz.
- λ = c/ν = 3 × 10⁸/(1 × 10⁶) = 300 m.

Practice Set

Q1: Radio waves frequency range is:
- A. 500 kHz to 1000 MHz
- B. 1 MHz to 10 GHz
- C. 100 MHz to 1 GHz
- D. 10 kHz to 500 kHz
Q2: Radio waves are mainly used for:
- A. Cooking
- B. Communication
- C. Medical imaging
- D. Sterilization
Q3: Wavelength of a 500 kHz radio wave is:
- A. 600 m
- B. 300 m
- C. 150 m
- D. 1000 m

Answer Key

Q1: A. Radio waves frequency range is 500 kHz to 1000 MHz.
Q2: B. Radio waves are used in communication systems.
Q3: A. λ = 3 × 10⁸/(5 × 10⁵) = 600 m.

Memory Aids

Radio waves = Communication waves
Frequency range: 500 kHz to 1000 MHz
Wavelength λ = c/ν

Properties and Uses

Frequency range: 500 kHz to 1000 MHz.
Used in communication: radio, TV, cellular phones.

Solved Examples

Ex 1: Find wavelength of radio wave with frequency 1 MHz.
- \( \lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{1 \times 10^6} = 300 \) m.

Practice Set

Q1: Radio waves frequency range is:
- A. 500 kHz to 1000 MHz
- B. 1 MHz to 10 GHz
- C. 100 MHz to 1 GHz
- D. 10 kHz to 500 kHz
Q2: Radio waves are mainly used for:
- A. Cooking
- B. Communication
- C. Medical imaging
- D. Sterilization
Q3: Wavelength of a 500 kHz radio wave is:
- A. 600 m
- B. 300 m
- C. 150 m
- D. 1000 m

Answer Key

Q1: A. Radio waves frequency range is 500 kHz to 1000 MHz.
Q2: B. Radio waves are used in communication systems.
Q3: A. \( \lambda = \frac{3 \times 10^8}{5 \times 10^5} = 600 \) m.

Memory Aids

Radio waves = Communication waves
Frequency range: 500 kHz to 1000 MHz
Wavelength \( \lambda = \frac{c}{\nu} \)

Microwaves

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 79 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 80 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 81 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 82 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 83 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 84 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 85 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 86 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- λ = 3 × 10⁸/(10 × 10⁹) = 0.03 m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Properties and Uses

Shortest wavelength radio waves, smaller than other radio waves but longer than visible light.
Used in radar for aircraft navigation and speed detection of moving objects.

Diagram 87 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
- \( \lambda = \frac{3 \times 10^8}{10 \times 10^9} = 0.03 \) m (3 cm).

Practice Set

Q1: Microwaves are used in:
- A. Communication only
- B. Radar systems
- C. X-ray imaging
- D. Sterilization
Q2: Microwave wavelength is generally:
- A. Longer than radio waves
- B. Shorter than visible light
- C. Between radio waves and visible light
- D. Same as infrared waves
Q3: Radar using microwaves can detect speed of:
- A. Stationary objects
- B. Moving balls and automobiles
- C. Only airplanes
- D. Only ships

Answer Key

Q1: B. Microwaves are used in radar systems.
Q2: C. Microwaves have wavelengths between radio waves and visible light.
Q3: B. Radar detects speed of moving balls and automobiles.

Memory Aids

Microwaves = Radar + short radio waves
Wavelength between radio and visible light
Used for speed detection

Infrared Waves

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- λ = 3 × 10⁸/(1 × 10¹³) = 3 × 10^-5 m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Properties and Uses

Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Used in infrared lamps for physical therapy.
Used in infrared detectors on Earth satellites.
Used in LEDs for remote controls (TV, AC, CD players).
Used in solar water heaters and cookers.
Used in greenhouses to maintain temperature (greenhouse effect).
Infrared radiation trapped by greenhouse gases maintains Earth's temperature.

Solved Examples

Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
- \( \lambda = \frac{3 \times 10^8}{1 \times 10^{13}} = 3 \times 10^{-5} \) m (30 µm).

Practice Set

Q1: Infrared waves frequency range is:
- A. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- B. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz
Q2: Infrared waves are used in:
- A. Physical therapy lamps
- B. Water sterilization
- C. X-ray imaging
- D. Cancer therapy
Q3: Greenhouse effect is caused by:
- A. Visible light trapped by atmosphere
- B. Infrared radiation trapped by greenhouse gases
- C. Ultraviolet rays absorbed by ozone
- D. Gamma rays heating Earth

Answer Key

Q1: A. Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
Q2: A. Infrared lamps are used in physical therapy.
Q3: B. Infrared radiation trapped by CO₂ and water vapor causes greenhouse effect.

Memory Aids

Infrared = Heat waves
Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz
Greenhouse effect traps infrared radiation

Visible Rays

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Ultraviolet Rays

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- ν = 3 × 10⁸/(700 × 10^-9) = 4.29 × 10¹⁴ Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Properties and Uses

Wavelength: 400 to 700 nm.
Frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
Visible to human eye; colors from violet (shortest wavelength) to red (longest).
Used in electric bulbs, tube lights, lasers, and cinema screens.

Solved Examples

Ex 1: Find frequency of red light with wavelength 700 nm.
- \( \nu = \frac{3 \times 10^8}{700 \times 10^{-9}} = 4.29 \times 10^{14} \) Hz.

Practice Set

Q1: Visible light wavelength range is:
- A. 400–700 nm
- B. 700–1000 nm
- C. 100–400 nm
- D. 1–10 nm
Q2: Which color has the shortest wavelength?
- A. Red
- B. Green
- C. Violet
- D. Yellow
Q3: Visible light frequency range is:
- A. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. Visible light wavelength is 400–700 nm.
Q2: C. Violet has the shortest wavelength in visible spectrum.
Q3: A. Frequency range is 4 × 10¹⁴ to 7 × 10¹⁴ Hz.

Memory Aids

VIBGYOR for colors
Wavelength increases from violet to red
Frequency decreases from violet to red

Ultraviolet Rays

Properties and Uses

Wavelength range: 0.6 to 400 nm.
Frequency range: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
Used for sterilizing surgical instruments and killing germs in water purifiers.
Used in LASIK eye surgery and mineral fluorescence detection.
Used in burglar alarms.

Solved Examples

Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
- λ = 3 × 10⁸/(1 × 10¹⁵) = 3 × 10^-7 m (300 nm).

Practice Set

Q1: Ultraviolet rays wavelength range is:
- A. 0.6–400 nm
- B. 400–700 nm
- C. 700–1000 nm
- D. 1–10 nm
Q2: UV rays are used for:
- A. Cooking
- B. Sterilization
- C. Communication
- D. Heating
Q3: UV rays frequency range is:
- A. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. UV wavelength range is 0.6–400 nm.
Q2: B. UV rays sterilize instruments and kill germs.
Q3: A. Frequency range is 8 × 10¹⁴ to 5 × 10¹⁷ Hz.

Memory Aids

UV = Sterilization and fluorescence
Wavelength shorter than visible light
Frequency higher than visible light

Properties and Uses

Wavelength range: 0.6 to 400 nm.
Frequency range: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
Used for sterilizing surgical instruments and killing germs in water purifiers.
Used in LASIK eye surgery and mineral fluorescence detection.
Used in burglar alarms.

Solved Examples

Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
- λ = 3 × 10⁸/(1 × 10¹⁵) = 3 × 10^-7 m (300 nm).

Practice Set

Q1: Ultraviolet rays wavelength range is:
- A. 0.6–400 nm
- B. 400–700 nm
- C. 700–1000 nm
- D. 1–10 nm
Q2: UV rays are used for:
- A. Cooking
- B. Sterilization
- C. Communication
- D. Heating
Q3: UV rays frequency range is:
- A. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. UV wavelength range is 0.6–400 nm.
Q2: B. UV rays sterilize instruments and kill germs.
Q3: A. Frequency range is 8 × 10¹⁴ to 5 × 10¹⁷ Hz.

Memory Aids

UV = Sterilization and fluorescence
Wavelength shorter than visible light
Frequency higher than visible light

Properties and Uses

Wavelength range: 0.6 to 400 nm.
Frequency range: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
Used for sterilizing surgical instruments and killing germs in water purifiers.
Used in LASIK eye surgery and mineral fluorescence detection.
Used in burglar alarms.

Solved Examples

Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
- λ = 3 × 10⁸/(1 × 10¹⁵) = 3 × 10^-7 m (300 nm).

Practice Set

Q1: Ultraviolet rays wavelength range is:
- A. 0.6–400 nm
- B. 400–700 nm
- C. 700–1000 nm
- D. 1–10 nm
Q2: UV rays are used for:
- A. Cooking
- B. Sterilization
- C. Communication
- D. Heating
Q3: UV rays frequency range is:
- A. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. UV wavelength range is 0.6–400 nm.
Q2: B. UV rays sterilize instruments and kill germs.
Q3: A. Frequency range is 8 × 10¹⁴ to 5 × 10¹⁷ Hz.

Memory Aids

UV = Sterilization and fluorescence
Wavelength shorter than visible light
Frequency higher than visible light

Properties and Uses

Wavelength range: 0.6 to 400 nm.
Frequency range: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
Used for sterilizing surgical instruments and killing germs in water purifiers.
Used in LASIK eye surgery and mineral fluorescence detection.
Used in burglar alarms.

Solved Examples

Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
- λ = 3 × 10⁸/(1 × 10¹⁵) = 3 × 10^-7 m (300 nm).

Practice Set

Q1: Ultraviolet rays wavelength range is:
- A. 0.6–400 nm
- B. 400–700 nm
- C. 700–1000 nm
- D. 1–10 nm
Q2: UV rays are used for:
- A. Cooking
- B. Sterilization
- C. Communication
- D. Heating
Q3: UV rays frequency range is:
- A. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. UV wavelength range is 0.6–400 nm.
Q2: B. UV rays sterilize instruments and kill germs.
Q3: A. Frequency range is 8 × 10¹⁴ to 5 × 10¹⁷ Hz.

Memory Aids

UV = Sterilization and fluorescence
Wavelength shorter than visible light
Frequency higher than visible light

Properties and Uses

Wavelength range: 0.6 to 400 nm.
Frequency range: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
Used for sterilizing surgical instruments and killing germs in water purifiers.
Used in LASIK eye surgery and mineral fluorescence detection.
Used in burglar alarms.

Solved Examples

Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
- \( \lambda = \frac{3 \times 10^8}{1 \times 10^{15}} = 3 \times 10^{-7} \) m (300 nm).

Practice Set

Q1: Ultraviolet rays wavelength range is:
- A. 0.6–400 nm
- B. 400–700 nm
- C. 700–1000 nm
- D. 1–10 nm
Q2: UV rays are used for:
- A. Cooking
- B. Sterilization
- C. Communication
- D. Heating
Q3: UV rays frequency range is:
- A. 8 × 10¹⁴ to 5 × 10¹⁷ Hz
- B. 3 × 10¹¹ to 4 × 10¹⁴ Hz
- C. 4 × 10¹⁴ to 7 × 10¹⁴ Hz
- D. 3 × 10¹⁸ to 3 × 10²² Hz

Answer Key

Q1: A. UV wavelength range is 0.6–400 nm.
Q2: B. UV rays sterilize instruments and kill germs.
Q3: A. Frequency range is 8 × 10¹⁴ to 5 × 10¹⁷ Hz.

Memory Aids

UV = Sterilization and fluorescence
Wavelength shorter than visible light
Frequency higher than visible light

X-rays

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- ν = 3 × 10⁸/(0.1 × 10^-9) = 3 × 10¹⁸ Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Properties and Uses

Wavelength range: 10^-4 nm to 10 nm.
Used for bone fracture detection and metal flaw detection.
Used in crystal structure investigation and detecting foreign bodies in humans.

Solved Examples

Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
- \( \nu = \frac{3 \times 10^8}{0.1 \times 10^{-9}} = 3 \times 10^{18} \) Hz.

Practice Set

Q1: X-rays wavelength range is:
- A. 10^-4 nm to 10 nm
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: X-rays are used for:
- A. Detecting fractures
- B. Communication
- C. Cooking
- D. Sterilization
Q3: X-rays help detect:
- A. Mineral fluorescence
- B. Foreign bodies in human body
- C. Speed of moving objects
- D. Greenhouse gases

Answer Key

Q1: A. X-rays wavelength range is 10^-4 nm to 10 nm.
Q2: A. Used for detecting fractures in bones.
Q3: B. Detect foreign bodies like stones and bullets.

Memory Aids

X-rays = Medical imaging and metal inspection
Wavelength very short, frequency very high
Used to detect fractures and flaws

Gamma Rays

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 88 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 89 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 90 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 91 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 92 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 93 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 94 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 95 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- ν = 3 × 10⁸/(10^-12) = 3 × 10²⁰ Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

Properties and Uses

Wavelength range: 10^-14 m to 10^-10 m.
Frequency range: 3 × 10¹⁸ Hz to 3 × 10²² Hz.
Used in cancer therapy to destroy cancer cells.
Used to preserve food by killing microorganisms.
Used to study atomic nucleus structure.

Diagram 96 – NEET Physics Electromagnetic Waves

Solved Examples

Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
- \( \nu = \frac{3 \times 10^8}{10^{-12}} = 3 \times 10^{20} \) Hz.

Practice Set

Q1: Gamma rays wavelength range is:
- A. 10^-14 m to 10^-10 m
- B. 400–700 nm
- C. 0.6–400 nm
- D. 500 kHz to 1000 MHz
Q2: Gamma rays are used for:
- A. Cancer therapy
- B. Communication
- C. Cooking
- D. Sterilization
Q3: Gamma rays help in:
- A. Studying atomic nucleus
- B. Detecting fractures
- C. Speed detection
- D. Greenhouse effect

Answer Key

Q1: A. Gamma rays have wavelength 10^-14 m to 10^-10 m.
Q2: A. Used in cancer therapy to destroy cells.
Q3: A. Used to study atomic nucleus structure.

Memory Aids

Gamma rays = Highest frequency, shortest wavelength
Used in cancer therapy and food preservation
Frequency range: 3 × 10¹⁸ to 3 × 10²² Hz

NEET Physics — ALL CHAPTERS

Basic Mathematics

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Units and Measurement

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Motion in a Straight Line

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Motion in a Plane

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Laws of Motion

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Work, Energy and Power

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System of Particles and Rotational Motion

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Gravitation

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Mechanical Properties of Solids

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Mechanical Properties of Fluids

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Thermal Properties of Matter

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Thermodynamics

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Kinetic Theory

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Oscillations

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Waves

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Electric Charges and Fields

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Electrostatic Potential and Capacitance

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Current Electricity

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Moving Charges and Magnetism

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Magnetism and Matter

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Electromagnetic Induction

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Alternating Current

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Electromagnetic Waves

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Ray Optics and Optical Instruments

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Wave Optics

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Dual Nature of Radiation and Matter

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Atoms

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Nuclei

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Semiconductor Electronics: Materials, Devices and Simple Circuits

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