Chapter 21: Electromagnetic Induction

Introduction

Key Points

• Moving electric charges produce magnetic fields (Oersted, Ampere).
• Electric current is induced in a closed coil by varying magnetic fields (Faraday, Henry).
• Electromagnetic induction: generation of electric current by changing magnetic fields.

Solved Examples

Ex 1

• Given a coil and a moving magnet, determine if current is induced.
• When magnet moves relative to coil, current is induced due to changing magnetic flux.
• Use concept: Current induced only when magnetic field changes.

Practice Set

• Q1: Who discovered that moving electric charges produce magnetic fields?
  • A. Faraday
  • B. Oersted
  • C. Henry
  • D. Ampere
• Q2: What causes electric current in a coil during electromagnetic induction?
  • A. Constant magnetic field
  • B. Varying magnetic field
  • C. Static electric field
  • D. No magnetic field
• Q3: Electromagnetic induction is the phenomenon of
  • A. Generating magnetic field by electric current
  • B. Generating electric current by varying magnetic field
  • C. Generating electric current by static magnetic field
  • D. Generating electric current by constant magnetic field

Answer Key

• Q1: B - Oersted discovered moving charges produce magnetic fields.
• Q2: B - Electric current is induced by varying magnetic fields.
• Q3: B - Electromagnetic induction is current generation by changing magnetic fields.

Memory Aids

• Moving charges → Magnetic field (Oersted).
• Changing magnetic field → Induced current (Faraday's law).
• Static field → No induction.

Key Points

• Moving electric charges produce magnetic fields (Oersted, Ampere).
• Electric current is induced in a closed coil by varying magnetic fields (Faraday, Henry).
• Electromagnetic induction: generation of electric current by changing magnetic fields.

Solved Examples

Ex 1

• Given a coil and a moving magnet, determine if current is induced.
• When magnet moves relative to coil, current is induced due to changing magnetic flux.
• Use concept: Current induced only when magnetic field changes.

Practice Set

• Q1: Who discovered that moving electric charges produce magnetic fields?
  • A. Faraday
  • B. Oersted
  • C. Henry
  • D. Ampere
• Q2: What causes electric current in a coil during electromagnetic induction?
  • A. Constant magnetic field
  • B. Varying magnetic field
  • C. Static electric field
  • D. No magnetic field
• Q3: Electromagnetic induction is the phenomenon of
  • A. Generating magnetic field by electric current
  • B. Generating electric current by varying magnetic field
  • C. Generating electric current by static magnetic field
  • D. Generating electric current by constant magnetic field

Answer Key

• Q1: B - Oersted discovered moving charges produce magnetic fields.
• Q2: B - Electric current is induced by varying magnetic fields.
• Q3: B - Electromagnetic induction is current generation by changing magnetic fields.

Memory Aids

• Moving charges → Magnetic field (Oersted).
• Changing magnetic field → Induced current (Faraday's law).
• Static field → No induction.

Faraday and Henry Experiments

Experiment 1: Moving Magnet and Coil

• Moving bar magnet towards coil induces current (galvanometer deflects).
• Current flows only while magnet moves (changing magnetic flux).
• Direction of current reverses if magnet moves away.

Experiment 2: Current Induced by Moving Current-Carrying Coil

• Current in coil C1 induced by motion of current-carrying coil C2.
• Changing magnetic field from moving coil induces current in nearby coil.

Experiment 3: Two Coils with Battery and Galvanometer

• Coil C2 connected to battery and key K; coil C1 connected to galvanometer G.
• Closing key K causes current in C2, changing magnetic field around C1.
• Changing magnetic field induces current in C1, detected by galvanometer deflection.
• Deflection is momentary when key is pressed or released; no deflection when key is held closed.

Solved Examples

Ex 1

• Calculate induced current direction when magnet moves towards coil.
• Use Lenz's law: induced current opposes change in magnetic flux.

Practice Set

• Q1: What happens to galvanometer when magnet moves towards coil?
  • A. No deflection
  • B. Deflection indicating induced current
  • C. Deflection indicating no current
  • D. Constant deflection
• Q2: When key K is kept closed, galvanometer shows
  • A. Continuous deflection
  • B. No deflection
  • C. Momentary deflection
  • D. Increasing deflection
• Q3: Induced current in coil C1 is due to
  • A. Constant magnetic field from C2
  • B. Changing magnetic field from C2
  • C. Static electric field
  • D. No magnetic field

Answer Key

• Q1: B - Moving magnet changes magnetic flux, inducing current.
• Q2: B - No deflection when key is held closed (steady current).
• Q3: B - Induced current caused by changing magnetic field.

Memory Aids

• Magnet motion → Induced current (galvanometer deflects).
• Steady current → No induction.
• Changing magnetic field → Induction.

Experiment 1: Moving Magnet and Coil

• Moving bar magnet towards coil induces current (galvanometer deflects).
• Current flows only while magnet moves (changing magnetic flux).
• Direction of current reverses if magnet moves away.

Experiment 2: Current Induced by Moving Current-Carrying Coil

• Current in coil C1 induced by motion of current-carrying coil C2.
• Changing magnetic field from moving coil induces current in nearby coil.

Experiment 3: Two Coils with Battery and Galvanometer

• Coil C2 connected to battery and key K; coil C1 connected to galvanometer G.
• Closing key K causes current in C2, changing magnetic field around C1.
• Changing magnetic field induces current in C1, detected by galvanometer deflection.
• Deflection is momentary when key is pressed or released; no deflection when key is held closed.

Solved Examples

Ex 1

• Calculate induced current direction when magnet moves towards coil.
• Use Lenz's law: induced current opposes change in magnetic flux.

Practice Set

• Q1: What happens to galvanometer when magnet moves towards coil?
  • A. No deflection
  • B. Deflection indicating induced current
  • C. Deflection indicating no current
  • D. Constant deflection
• Q2: When key K is kept closed, galvanometer shows
  • A. Continuous deflection
  • B. No deflection
  • C. Momentary deflection
  • D. Increasing deflection
• Q3: Induced current in coil C1 is due to
  • A. Constant magnetic field from C2
  • B. Changing magnetic field from C2
  • C. Static electric field
  • D. No magnetic field

Answer Key

• Q1: B - Moving magnet changes magnetic flux, inducing current.
• Q2: B - No deflection when key is held closed (steady current).
• Q3: B - Induced current caused by changing magnetic field.

Memory Aids

• Magnet motion → Induced current (galvanometer deflects).
• Steady current → No induction.
• Changing magnetic field → Induction.

Magnetic Flux

Definition and Formula

• Magnetic flux (ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (θ = 90°).

Definition and Formula

• Magnetic flux (ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (θ = 90°).

Definition and Formula

• Magnetic flux (ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (θ = 90°).

Definition and Formula

• Magnetic flux (ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (θ = 90°).

Definition and Formula

• Magnetic flux (ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (θ = 90°).

Definition and Formula

• Magnetic flux ( ϕ) through a surface of area vector vector A in a uniform magnetic field vector B is ϕ = vector B · vector A.
• Formula: ϕ = BA cos θ, where:
• B = magnetic field strength (Tesla, T)
• A = area of the surface (m²)
• θ = angle between vector B and vector A
• Area vector vector A is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if B = 0.5 T, A = 2 m², and θ = 60°.
  • Use ϕ = BA cos θ = 0.5 × 2 × cos 60° = 0.5 × 2 × 0.5 = 0.5 Wb.

Practice Set

• Q1: Magnetic flux is maximum when θ is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field B = 1 T, area A = 3 m², and θ = 90°, magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area 4 m² is placed in a magnetic field of 0.2 T at an angle of 60°. Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux ϕ = BA cos θ is maximum when cos θ = 1, i.e., θ = 0°.
• Q2: B) 0 Wb — At 90°, cos 90° = 0, so flux is zero.
• Q3: B) 0.8 Wb — ϕ = 0.2 × 4 × cos 60° = 0.8 Wb.

Memory Aids

• Magnetic flux ϕ = BA cos θ — remember cos 0° = 1 for max flux.
• Area vector vector A is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector ( θ = 90°).

Definition and Formula

• Magnetic flux (\( \phi \)) through a surface of area vector \( \vec{A} \) in a uniform magnetic field \( \vec{B} \) is \( \phi = \vec{B} \cdot \vec{A} \).
• Formula: \( \phi = BA \cos \theta \), where:
• \( B \) = magnetic field strength (Tesla, T)
• \( A \) = area of the surface (m\(^2\))
• \( \theta \) = angle between \( \vec{B} \) and \( \vec{A} \)
• Area vector \( \vec{A} \) is perpendicular to the surface.
• Magnetic flux measures the amount of magnetic field passing through the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if \( B = 0.5\,T \), \( A = 2\,m^2 \), and \( \theta = 60^\circ \).
  • Use \( \phi = BA \cos \theta = 0.5 \times 2 \times \cos 60^\circ = 0.5 \times 2 \times 0.5 = 0.5\,Wb \).

Practice Set

• Q1: Magnetic flux is maximum when \( \theta \) is:
  • A) 0°
  • B) 90°
  • C) 45°
  • D) 180°
• Q2: If magnetic field \( B = 1\,T \), area \( A = 3\,m^2 \), and \( \theta = 90^\circ \), magnetic flux is:
  • A) 3 Wb
  • B) 0 Wb
  • C) 1 Wb
  • D) 90 Wb
• Q3: A surface of area \( 4\,m^2 \) is placed in a magnetic field of \( 0.2\,T \) at an angle of \( 60^\circ \). Find magnetic flux:
  • A) 0.4 Wb
  • B) 0.8 Wb
  • C) 1.6 Wb
  • D) 2.0 Wb

Answer Key

• Q1: A) 0° — Magnetic flux \( \phi = BA \cos \theta \) is maximum when \( \cos \theta = 1 \), i.e., \( \theta = 0^\circ \).
• Q2: B) 0 Wb — At \( 90^\circ \), \( \cos 90^\circ = 0 \), so flux is zero.
• Q3: B) 0.8 Wb — \( \phi = 0.2 \times 4 \times \cos 60^\circ = 0.8\,Wb \).

Memory Aids

• Magnetic flux \( \phi = BA \cos \theta \) — remember \( \cos 0^\circ = 1 \) for max flux.
• Area vector \( \vec{A} \) is always perpendicular to surface.
• Flux zero when field is perpendicular to area vector (\( \theta = 90^\circ \)).

Faraday's Law

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: ε = -dϕ/dt.
• For N turns, ε = -N dϕ/dt.
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: ε = -N dϕ/dt = -50 × 0.1/0.02 = -250 V.

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - ε = -dϕ/dt = -(0.1-0.5)/0.1 = 4 V.
• Q3: A - ε = -N dϕ/dt = -100 × 0.02 = 2 V.

Memory Aids

• "Change in flux → emf"
• Formula: ε = -N dϕ/dt
• Lenz's law gives polarity (opposes change)

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: ε = -dϕ/dt.
• For N turns, ε = -N dϕ/dt.
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: ε = -N dϕ/dt = -50 × 0.1/0.02 = -250 V.

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - ε = -dϕ/dt = -(0.1-0.5)/0.1 = 4 V.
• Q3: A - ε = -N dϕ/dt = -100 × 0.02 = 2 V.

Memory Aids

• "Change in flux → emf"
• Formula: ε = -N dϕ/dt
• Lenz's law gives polarity (opposes change)

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: ε = -dϕ/dt.
• For N turns, ε = -N dϕ/dt.
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: ε = -N dϕ/dt = -50 × 0.1/0.02 = -250 V.

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - ε = -dϕ/dt = -(0.1-0.5)/0.1 = 4 V.
• Q3: A - ε = -N dϕ/dt = -100 × 0.02 = 2 V.

Memory Aids

• "Change in flux → emf"
• Formula: ε = -N dϕ/dt
• Lenz's law gives polarity (opposes change)

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: ε = -dϕ/dt.
• For N turns, ε = -N dϕ/dt.
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: ε = -N dϕ/dt = -50 × 0.1/0.02 = -250 V.

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - ε = -dϕ/dt = -(0.1-0.5)/0.1 = 4 V.
• Q3: A - ε = -N dϕ/dt = -100 × 0.02 = 2 V.

Memory Aids

• "Change in flux → emf"
• Formula: ε = -N dϕ/dt
• Lenz's law gives polarity (opposes change)

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: ε = -dϕ/dt.
• For N turns, ε = -N dϕ/dt.
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: ε = -N dϕ/dt = -50 × 0.1/0.02 = -250 V.

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - ε = -dϕ/dt = -(0.1-0.5)/0.1 = 4 V.
• Q3: A - ε = -N dϕ/dt = -100 × 0.02 = 2 V.

Memory Aids

• "Change in flux → emf"
• Formula: ε = -N dϕ/dt
• Lenz's law gives polarity (opposes change)

Explanation

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: \( \varepsilon = -\frac{d\phi}{dt} \).
• For N turns, \( \varepsilon = -N \frac{d\phi}{dt} \).
• Polarity given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil with 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  Solution: \( \varepsilon = -N \frac{d\phi}{dt} = -50 \times \frac{0.1}{0.02} = -250 \text{ V} \).

Practice Set

• Q1: What causes emf in a coil?
  A) Constant magnetic flux
  B) Change in magnetic flux
  C) Electric current
  D) Resistance
  
• Q2: If magnetic flux changes from 0.5 Wb to 0.1 Wb in 0.1 s in a single turn coil, emf induced is:
  A) 4 V
  B) 0.4 V
  C) 5 V
  D) 0.05 V
  
• Q3: A coil of 100 turns has flux changing at 0.02 Wb/s. Induced emf is:
  A) 2 V
  B) 0.5 V
  C) 5 V
  D) 0.2 V
  

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: A - \( \varepsilon = -\frac{d\phi}{dt} = -\frac{0.1-0.5}{0.1} = 4 \text{ V} \).
• Q3: A - \( \varepsilon = -N \frac{d\phi}{dt} = -100 \times 0.02 = 2 \text{ V} \).

Memory Aids

• "Change in flux → emf"
• Formula: \( \varepsilon = -N \frac{d\phi}{dt} \)
• Lenz's law gives polarity (opposes change)

Lenz's Law

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, ε = - dϕ/dt, where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) ε = dϕ/dt
  • B) ε = - dϕ/dt
  • C) ε = φ × t
  • D) ε = 0
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in ε = - dϕ/dt shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: ε = - dϕ/dt
• Induced current always fights the cause.

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, ε = - dϕ/dt, where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) ε = dϕ/dt
  • B) ε = - dϕ/dt
  • C) ε = φ × t
  • D) ε = 0
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in ε = - dϕ/dt shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: ε = - dϕ/dt
• Induced current always fights the cause.

Lenz's Law

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, ε = - dϕ/dt, where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) ε = dϕ/dt
  • B) ε = - dϕ/dt
  • C) ε = φ × t
  • D) ε = 0
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in ε = - dϕ/dt shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: ε = - dϕ/dt
• Induced current always fights the cause.

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, ε = - dϕ/dt, where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) ε = dϕ/dt
  • B) ε = - dϕ/dt
  • C) ε = φ × t
  • D) ε = 0
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in ε = - dϕ/dt shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: ε = - dϕ/dt
• Induced current always fights the cause.

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, ε = - dϕ/dt, where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) ε = dϕ/dt
  • B) ε = - dϕ/dt
  • C) ε = φ × t
  • D) ε = 0
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in ε = - dϕ/dt shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: ε = - dϕ/dt
• Induced current always fights the cause.

Explanation

• Induced current opposes the cause of flux change.
• When magnet's North pole approaches coil, coil acts as North pole (repels).
• When magnet's North pole moves away, coil acts as South pole (attracts).
• Direction of induced current shown by galvanometer.

Solved Examples

• Ex 1: Magnet's North pole moves toward coil, galvanometer deflects right. What is coil's pole?
  Solution: Coil acts as North pole (repels magnet).
• Ex 2: Magnet's North pole moves away, galvanometer deflects left. Coil's pole?
  Solution: Coil acts as South pole (attracts magnet).

Practice Set

• Q1: Lenz's law states induced current:
  A) Supports flux change
  B) Opposes flux change
  C) Is independent
  D) Is zero
  
• Q2: When magnet approaches coil, coil behaves like:
  A) South pole
  B) North pole
  C) Neutral
  D) No pole
  
• Q3: Galvanometer shows current direction. If magnet moves away, coil's induced pole is:
  A) North
  B) South
  C) Neutral
  D) None
  

Answer Key

• Q1: B - Induced current opposes flux change.
• Q2: B - Coil acts as North pole to repel approaching magnet.
• Q3: B - Coil acts as South pole to attract receding magnet.

Memory Aids

• "Induced current fights change"
• Approach → coil repels (same pole)
• Recede → coil attracts (opposite pole)

Explanation

• Induced current opposes the cause of flux change.
• When magnet's North pole approaches coil, coil acts as North pole (repels).
• When magnet's North pole moves away, coil acts as South pole (attracts).
• Direction of induced current shown by galvanometer.

Solved Examples

• Ex 1: Magnet's North pole moves toward coil, galvanometer deflects right. What is coil's pole?
  Solution: Coil acts as North pole (repels magnet).
• Ex 2: Magnet's North pole moves away, galvanometer deflects left. Coil's pole?
  Solution: Coil acts as South pole (attracts magnet).

Practice Set

• Q1: Lenz's law states induced current:
  A) Supports flux change
  B) Opposes flux change
  C) Is independent
  D) Is zero
  
• Q2: When magnet approaches coil, coil behaves like:
  A) South pole
  B) North pole
  C) Neutral
  D) No pole
  
• Q3: Galvanometer shows current direction. If magnet moves away, coil's induced pole is:
  A) North
  B) South
  C) Neutral
  D) None
  

Answer Key

• Q1: B - Induced current opposes flux change.
• Q2: B - Coil acts as North pole to repel approaching magnet.
• Q3: B - Coil acts as South pole to attract receding magnet.

Memory Aids

• "Induced current fights change"
• Approach → coil repels (same pole)
• Recede → coil attracts (opposite pole)

Lenz's Law

Explanation

• Lenz's law states that the induced current in a circuit flows to oppose the change in magnetic flux that produces it.
• Mathematically, \( \varepsilon = - \frac{d\phi}{dt} \), where the negative sign shows opposition.
• This law aligns with the conservation of energy: work must be done against the induced emf to change the flux.
• Energy spent by an external source appears as electrical energy in the circuit.
• Example: Moving a magnet towards a loop induces current that creates a magnetic field opposing the magnet's motion.

Solved Examples

• Ex 1: A magnet is moved towards a coil. Direction of induced current?
  • Answer: Anticlockwise to create a north pole opposing the magnet's approach.
• Ex 2: Magnet moved away from coil. Induced current direction?
  • Answer: Clockwise to create a south pole attracting the magnet, opposing its motion.

Practice Set

• Q1: Lenz's law states that induced current flows to:
  • A) Increase magnetic flux
  • B) Oppose change in magnetic flux
  • C) Align with magnetic field
  • D) Flow randomly
• Q2: The formula for induced emf according to Lenz's law is:
  • A) \( \varepsilon = \frac{d\phi}{dt} \)
  • B) \( \varepsilon = - \frac{d\phi}{dt} \)
  • C) \( \varepsilon = \phi \times t \)
  • D) \( \varepsilon = 0 \)
• Q3: When a magnet is moved towards a coil, the induced current:
  • A) Attracts the magnet
  • B) Has no effect
  • C) Repels the magnet
  • D) Depends on coil resistance

Answer Key

• Q1: B - Induced current opposes the change in magnetic flux as per Lenz's law.
• Q2: B - Negative sign in \( \varepsilon = - \frac{d\phi}{dt} \) shows opposition.
• Q3: C - Induced current creates a magnetic field repelling the approaching magnet.

Memory Aids

• "Lenz's law = Oppose change"
• Formula: \( \varepsilon = - \frac{d\phi}{dt} \)
• Induced current always fights the cause.

Direction of Induced EMF

Explanation

• Magnetic flux ϕ = B A cos θ, depends on magnetic field B, area A, and angle θ.
• Induced emf depends on the rate of change of magnetic flux.
• Flux increases if B increases and θ = 0° (field perpendicular to loop).
• Flux decreases if B decreases and θ = 0°.
• Flux is zero or negligible if θ = 90° (field parallel to loop), regardless of B change.
• Induced emf magnitude is proportional to the relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field B increases perpendicular to coil. Induced emf direction?
  • Answer: Opposes increase, direction given by Lenz's law.
• Ex 2: Magnetic field decreases at 0° angle. Induced emf direction?
  • Answer: Induced emf tries to increase flux, opposite to decrease.

Practice Set

• Q1: Magnetic flux is zero when θ =:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: Induced emf depends on:
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Temperature
• Q3: If magnetic field B is constant but angle θ changes from 0° to 90°, induced emf is:
  • A) Maximum
  • B) Zero
  • C) Half
  • D) Double

Answer Key

• Q1: C - Flux is zero when field is parallel to loop (θ = 90°).
• Q2: B - Induced emf depends on rate of change of magnetic flux, not coil resistance.
• Q3: B - Flux change is zero if B constant and angle changes to 90°, so emf is zero.

Memory Aids

• Flux ϕ = B A cos θ
• Emf ∝ rate of flux change
• Zero flux at 90° angle

Explanation

• Magnetic flux ϕ = B A cos θ, depends on magnetic field B, area A, and angle θ.
• Induced emf depends on the rate of change of magnetic flux.
• Flux increases if B increases and θ = 0° (field perpendicular to loop).
• Flux decreases if B decreases and θ = 0°.
• Flux is zero or negligible if θ = 90° (field parallel to loop), regardless of B change.
• Induced emf magnitude is proportional to the relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field B increases perpendicular to coil. Induced emf direction?
  • Answer: Opposes increase, direction given by Lenz's law.
• Ex 2: Magnetic field decreases at 0° angle. Induced emf direction?
  • Answer: Induced emf tries to increase flux, opposite to decrease.

Practice Set

• Q1: Magnetic flux is zero when θ =:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: Induced emf depends on:
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Temperature
• Q3: If magnetic field B is constant but angle θ changes from 0° to 90°, induced emf is:
  • A) Maximum
  • B) Zero
  • C) Half
  • D) Double

Answer Key

• Q1: C - Flux is zero when field is parallel to loop (θ = 90°).
• Q2: B - Induced emf depends on rate of change of magnetic flux, not coil resistance.
• Q3: B - Flux change is zero if B constant and angle changes to 90°, so emf is zero.

Memory Aids

• Flux ϕ = B A cos θ
• Emf ∝ rate of flux change
• Zero flux at 90° angle

Explanation

• Magnetic flux ϕ = B A cos θ, depends on magnetic field B, area A, and angle θ.
• Induced emf depends on the rate of change of magnetic flux.
• Flux increases if B increases and θ = 0° (field perpendicular to loop).
• Flux decreases if B decreases and θ = 0°.
• Flux is zero or negligible if θ = 90° (field parallel to loop), regardless of B change.
• Induced emf magnitude is proportional to the relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field B increases perpendicular to coil. Induced emf direction?
  • Answer: Opposes increase, direction given by Lenz's law.
• Ex 2: Magnetic field decreases at 0° angle. Induced emf direction?
  • Answer: Induced emf tries to increase flux, opposite to decrease.

Practice Set

• Q1: Magnetic flux is zero when θ =:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: Induced emf depends on:
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Temperature
• Q3: If magnetic field B is constant but angle θ changes from 0° to 90°, induced emf is:
  • A) Maximum
  • B) Zero
  • C) Half
  • D) Double

Answer Key

• Q1: C - Flux is zero when field is parallel to loop (θ = 90°).
• Q2: B - Induced emf depends on rate of change of magnetic flux, not coil resistance.
• Q3: B - Flux change is zero if B constant and angle changes to 90°, so emf is zero.

Memory Aids

• Flux ϕ = B A cos θ
• Emf ∝ rate of flux change
• Zero flux at 90° angle

Explanation

• Magnetic flux ϕ = B A cos θ, depends on magnetic field B, area A, and angle θ.
• Induced emf depends on the rate of change of magnetic flux.
• Flux increases if B increases and θ = 0° (field perpendicular to loop).
• Flux decreases if B decreases and θ = 0°.
• Flux is zero or negligible if θ = 90° (field parallel to loop), regardless of B change.
• Induced emf magnitude is proportional to the relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field B increases perpendicular to coil. Induced emf direction?
  • Answer: Opposes increase, direction given by Lenz's law.
• Ex 2: Magnetic field decreases at 0° angle. Induced emf direction?
  • Answer: Induced emf tries to increase flux, opposite to decrease.

Practice Set

• Q1: Magnetic flux is zero when θ =:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: Induced emf depends on:
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Temperature
• Q3: If magnetic field B is constant but angle θ changes from 0° to 90°, induced emf is:
  • A) Maximum
  • B) Zero
  • C) Half
  • D) Double

Answer Key

• Q1: C - Flux is zero when field is parallel to loop ( θ = 90°).
• Q2: B - Induced emf depends on rate of change of magnetic flux, not coil resistance.
• Q3: B - Flux change is zero if B constant and angle changes to 90°, so emf is zero.

Memory Aids

• Flux ϕ = B A cos θ
• Emf ∝ rate of flux change
• Zero flux at 90° angle

Explanation

• Magnetic flux \( \phi = B A \cos \theta \), depends on magnetic field \( B \), area \( A \), and angle \( \theta \).
• Induced emf depends on the rate of change of magnetic flux.
• Flux increases if \( B \) increases and \( \theta = 0^\circ \) (field perpendicular to loop).
• Flux decreases if \( B \) decreases and \( \theta = 0^\circ \).
• Flux is zero or negligible if \( \theta = 90^\circ \) (field parallel to loop), regardless of \( B \) change.
• Induced emf magnitude is proportional to the relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field \( B \) increases perpendicular to coil. Induced emf direction?
  • Answer: Opposes increase, direction given by Lenz's law.
• Ex 2: Magnetic field decreases at \( 0^\circ \) angle. Induced emf direction?
  • Answer: Induced emf tries to increase flux, opposite to decrease.

Practice Set

• Q1: Magnetic flux is zero when \( \theta = \):
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: Induced emf depends on:
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Temperature
• Q3: If magnetic field \( B \) is constant but angle \( \theta \) changes from 0° to 90°, induced emf is:
  • A) Maximum
  • B) Zero
  • C) Half
  • D) Double

Answer Key

• Q1: C - Flux is zero when field is parallel to loop (\( \theta = 90^\circ \)).
• Q2: B - Induced emf depends on rate of change of magnetic flux, not coil resistance.
• Q3: B - Flux change is zero if \( B \) constant and angle changes to 90°, so emf is zero.

Memory Aids

• Flux \( \phi = B A \cos \theta \)
• Emf \( \propto \) rate of flux change
• Zero flux at \( 90^\circ \) angle

Motional Electromotive Force

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment dvector l: dε = (vector v × vector B) · dvector l
• Total emf across the rod: ε = ∫ (vector v × vector B) · dvector l = v B l sin θ
• Induced electric field in the rod: vector E_in = - (vector v × vector B) = (vector B × vector v)
• If any two of vector v, vector B, and dvector l are parallel or antiparallel, emf ε = 0.

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: ε = v B l = 4 × 0.3 × 0.5 = 0.6 V

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity vector v and magnetic field vector B are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle 30° to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — ε = v B l = 5 × 0.2 × 1 = 1 V
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — ε = v B l sin 30° = 3 × 0.5 × 0.4 × 0.5 = 0.3 V

Memory Aids

• Motional emf ∝ v, B, l, sin θ
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment dvector l: dε = (vector v × vector B) · dvector l
• Total emf across the rod: ε = ∫ (vector v × vector B) · dvector l = v B l sin θ
• Induced electric field in the rod: vector E_in = - (vector v × vector B) = (vector B × vector v)
• If any two of vector v, vector B, and dvector l are parallel or antiparallel, emf ε = 0.

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: ε = v B l = 4 × 0.3 × 0.5 = 0.6 V

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity vector v and magnetic field vector B are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle 30° to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — ε = v B l = 5 × 0.2 × 1 = 1 V
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — ε = v B l sin 30° = 3 × 0.5 × 0.4 × 0.5 = 0.3 V

Memory Aids

• Motional emf ∝ v, B, l, sin θ
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment dvector l: dε = (vector v × vector B) · dvector l
• Total emf across the rod: ε = ∫ (vector v × vector B) · dvector l = v B l sin θ
• Induced electric field in the rod: vector E_in = - (vector v × vector B) = (vector B × vector v)
• If any two of vector v, vector B, and dvector l are parallel or antiparallel, emf ε = 0.

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: ε = v B l = 4 × 0.3 × 0.5 = 0.6 V

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity vector v and magnetic field vector B are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle 30° to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — ε = v B l = 5 × 0.2 × 1 = 1 V
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — ε = v B l sin 30° = 3 × 0.5 × 0.4 × 0.5 = 0.3 V

Memory Aids

• Motional emf ∝ v, B, l, sin θ
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment dvector l: dε = (vector v × vector B) · dvector l
• Total emf across the rod: ε = ∫ (vector v × vector B) · dvector l = v B l sin θ
• Induced electric field in the rod: vector E_in = - (vector v × vector B) = (vector B × vector v)
• If any two of vector v, vector B, and dvector l are parallel or antiparallel, emf ε = 0.

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: ε = v B l = 4 × 0.3 × 0.5 = 0.6 V

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity vector v and magnetic field vector B are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle 30° to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — ε = v B l = 5 × 0.2 × 1 = 1 V
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — ε = v B l sin 30° = 3 × 0.5 × 0.4 × 0.5 = 0.3 V

Memory Aids

• Motional emf ∝ v, B, l, sin θ
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment dvector l: dε = (vector v × vector B) · dvector l
• Total emf across the rod: ε = ∫ (vector v × vector B) · dvector l = v B l sin θ
• Induced electric field in the rod: vector E_in = - (vector v × vector B) = (vector B × vector v)
• If any two of vector v, vector B, and dvector l are parallel or antiparallel, emf ε = 0.

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: ε = v B l = 4 × 0.3 × 0.5 = 0.6 V

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity vector v and magnetic field vector B are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle 30° to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — ε = v B l = 5 × 0.2 × 1 = 1 V
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — ε = v B l sin 30° = 3 × 0.5 × 0.4 × 0.5 = 0.3 V

Memory Aids

• Motional emf ∝ v, B, l, sin θ
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Concept and Formulas

• When a conducting rod moves in a magnetic field cutting the field lines, free electrons shift from one end to another, inducing an emf called motional emf.
• Emf induced in a small segment \( d\vec{l} \): \( d\varepsilon = (\vec{v} \times \vec{B}) \cdot d\vec{l} \)
• Total emf across the rod: \( \varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l} = v B l \sin \theta \)
• Induced electric field in the rod: \( \vec{E}_{in} = - (\vec{v} \times \vec{B}) = (\vec{B} \times \vec{v}) \)
• If any two of \( \vec{v} \), \( \vec{B} \), and \( d\vec{l} \) are parallel or antiparallel, emf \( \varepsilon = 0 \).

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find the emf induced.
  Solution: \( \varepsilon = v B l = 4 \times 0.3 \times 0.5 = 0.6 \text{ V} \)

Practice Set

• Q1: A rod of length 1 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. What is the induced emf?
  A) 1 V
  B) 0.5 V
  C) 2 V
  D) 0.1 V
• Q2: If velocity \( \vec{v} \) and magnetic field \( \vec{B} \) are parallel, the induced emf is:
  A) Maximum
  B) Zero
  C) Depends on length
  D) Depends on angle
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle \( 30^\circ \) to a magnetic field of 0.5 T. Calculate emf.
  A) 0.3 V
  B) 0.6 V
  C) 0.9 V
  D) 0.15 V

Answer Key

• Q1: A) 1 V — \( \varepsilon = v B l = 5 \times 0.2 \times 1 = 1 \text{ V} \)
• Q2: B) Zero — Parallel vectors produce zero emf.
• Q3: A) 0.3 V — \( \varepsilon = v B l \sin 30^\circ = 3 \times 0.5 \times 0.4 \times 0.5 = 0.3 \text{ V} \)

Memory Aids

• Motional emf \( \propto v, B, l, \sin \theta \)
• Zero emf if velocity and magnetic field are parallel
• Use right-hand rule for direction of induced field

Methods of Producing Induced emf

By Changing Magnetic Field

• Changing magnetic field intensity B in a closed circuit changes flux ϕ, inducing emf.
• Formula: ε = - dϕ/dt = - d/dt (B A cos θ)
• Or ε = - A cos θ dB/dt, where dB/dt is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area A inside magnetic field changes flux, inducing emf.
• Formula: ε = - dϕ/dt = - B cos θ dA/dt
• For rectangular loop moving out of field with velocity v: ε = B l v cos θ

By Changing Orientation of Coil or Angle θ

• Coil of area A, turns N, rotating with angular velocity ω in magnetic field B.
• Flux linked: ϕ = N B A cos (ω t)
• Induced emf: ε = - dϕ/dt = N B A ω sin (ω t) = ε0 sin (ω t)
• ε0 = N B A ω is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length l rotates with angular velocity ω in perpendicular magnetic field B.
• Time period T = 2π/ω.
• Induced emf: ε = - B dA/dt = - B π l2/T = - B l2 ω/2 = - π B l2 ν, where ν = ω/2π is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: ε0 = N B A ω = 100 × 0.2 × 0.01 × 50 = 10 V
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: ε = - π B l2 ν = - 3.14 × 0.5 × 0.3^2 × 60 = -8.5 V

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf ε = 15 sin 200 t. If N=100, A=0.01 m^2, B=0.3 T, find angular velocity ω.
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — ε0 = N B A ω = 50 × 0.3 × 0.02 × 40 = 12 V
• Q2: B) 10 V — ε = π B l2 ν = 3.14 × 0.4 × 0.4^2 × 100 = 10 V
• Q3: C) 200 rad/s — ε0 = N B A ω = 15 ⇒ ω = 15/(100 × 0.01 × 0.3) = 200

Memory Aids

• Peak emf ε0 = N B A ω
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf ∝ B l2 ν

By Changing Magnetic Field

• Changing magnetic field intensity B in a closed circuit changes flux ϕ, inducing emf.
• Formula: ε = - dϕ/dt = - d/dt (B A cos θ)
• Or ε = - A cos θ dB/dt, where dB/dt is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area A inside magnetic field changes flux, inducing emf.
• Formula: ε = - dϕ/dt = - B cos θ dA/dt
• For rectangular loop moving out of field with velocity v: ε = B l v cos θ

By Changing Orientation of Coil or Angle θ

• Coil of area A, turns N, rotating with angular velocity ω in magnetic field B.
• Flux linked: ϕ = N B A cos (ω t)
• Induced emf: ε = - dϕ/dt = N B A ω sin (ω t) = ε0 sin (ω t)
• ε0 = N B A ω is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length l rotates with angular velocity ω in perpendicular magnetic field B.
• Time period T = 2π/ω.
• Induced emf: ε = - B dA/dt = - B π l2/T = - B l2 ω/2 = - π B l2 ν, where ν = ω/2π is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: ε0 = N B A ω = 100 × 0.2 × 0.01 × 50 = 10 V
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: ε = - π B l2 ν = - 3.14 × 0.5 × 0.3^2 × 60 = -8.5 V

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf ε = 15 sin 200 t. If N=100, A=0.01 m^2, B=0.3 T, find angular velocity ω.
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — ε0 = N B A ω = 50 × 0.3 × 0.02 × 40 = 12 V
• Q2: B) 10 V — ε = π B l2 ν = 3.14 × 0.4 × 0.4^2 × 100 = 10 V
• Q3: C) 200 rad/s — ε0 = N B A ω = 15 ⇒ ω = 15/(100 × 0.01 × 0.3) = 200

Memory Aids

• Peak emf ε0 = N B A ω
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf ∝ B l2 ν

By Changing Magnetic Field

• Changing magnetic field intensity B in a closed circuit changes flux ϕ, inducing emf.
• Formula: ε = - dϕ/dt = - d/dt (B A cos θ)
• Or ε = - A cos θ dB/dt, where dB/dt is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area A inside magnetic field changes flux, inducing emf.
• Formula: ε = - dϕ/dt = - B cos θ dA/dt
• For rectangular loop moving out of field with velocity v: ε = B l v cos θ

By Changing Orientation of Coil or Angle θ

• Coil of area A, turns N, rotating with angular velocity ω in magnetic field B.
• Flux linked: ϕ = N B A cos (ω t)
• Induced emf: ε = - dϕ/dt = N B A ω sin (ω t) = ε0 sin (ω t)
• ε0 = N B A ω is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length l rotates with angular velocity ω in perpendicular magnetic field B.
• Time period T = 2π/ω.
• Induced emf: ε = - B dA/dt = - B π l2/T = - B l2 ω/2 = - π B l2 ν, where ν = ω/2π is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: ε0 = N B A ω = 100 × 0.2 × 0.01 × 50 = 10 V
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: ε = - π B l2 ν = - 3.14 × 0.5 × 0.3^2 × 60 = -8.5 V

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf ε = 15 sin 200 t. If N=100, A=0.01 m^2, B=0.3 T, find angular velocity ω.
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — ε0 = N B A ω = 50 × 0.3 × 0.02 × 40 = 12 V
• Q2: B) 10 V — ε = π B l2 ν = 3.14 × 0.4 × 0.4^2 × 100 = 10 V
• Q3: C) 200 rad/s — ε0 = N B A ω = 15 ⇒ ω = 15/(100 × 0.01 × 0.3) = 200

Memory Aids

• Peak emf ε0 = N B A ω
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf ∝ B l2 ν

By Changing Magnetic Field

• Changing magnetic field intensity B in a closed circuit changes flux ϕ, inducing emf.
• Formula: ε = - dϕ/dt = - d/dt (B A cos θ)
• Or ε = - A cos θ dB/dt, where dB/dt is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area A inside magnetic field changes flux, inducing emf.
• Formula: ε = - dϕ/dt = - B cos θ dA/dt
• For rectangular loop moving out of field with velocity v: ε = B l v cos θ

By Changing Orientation of Coil or Angle θ

• Coil of area A, turns N, rotating with angular velocity ω in magnetic field B.
• Flux linked: ϕ = N B A cos (ω t)
• Induced emf: ε = - dϕ/dt = N B A ω sin (ω t) = ε0 sin (ω t)
• ε0 = N B A ω is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length l rotates with angular velocity ω in perpendicular magnetic field B.
• Time period T = 2π/ω.
• Induced emf: ε = - B dA/dt = - B π l2/T = - B l2 ω/2 = - π B l2 ν, where ν = ω/2π is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: ε0 = N B A ω = 100 × 0.2 × 0.01 × 50 = 10 V
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: ε = - π B l2 ν = - 3.14 × 0.5 × 0.3^2 × 60 = -8.5 V

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf ε = 15 sin 200 t. If N=100, A=0.01 m^2, B=0.3 T, find angular velocity ω.
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — ε0 = N B A ω = 50 × 0.3 × 0.02 × 40 = 12 V
• Q2: B) 10 V — ε = π B l2 ν = 3.14 × 0.4 × 0.4^2 × 100 = 10 V
• Q3: C) 200 rad/s — ε0 = N B A ω = 15 ⇒ ω = 15/(100 × 0.01 × 0.3) = 200

Memory Aids

• Peak emf ε0 = N B A ω
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf ∝ B l2 ν

By Changing Magnetic Field

• Changing magnetic field intensity B in a closed circuit changes flux ϕ, inducing emf.
• Formula: ε = - dϕ/dt = - d/dt (B A cos θ)
• Or ε = - A cos θ dB/dt, where dB/dt is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area A inside magnetic field changes flux, inducing emf.
• Formula: ε = - dϕ/dt = - B cos θ dA/dt
• For rectangular loop moving out of field with velocity v: ε = B l v cos θ

By Changing Orientation of Coil or Angle θ

• Coil of area A, turns N, rotating with angular velocity ω in magnetic field B.
• Flux linked: ϕ = N B A cos (ω t)
• Induced emf: ε = - dϕ/dt = N B A ω sin (ω t) = ε0 sin (ω t)
• ε0 = N B A ω is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length l rotates with angular velocity ω in perpendicular magnetic field B.
• Time period T = 2π/ω.
• Induced emf: ε = - B dA/dt = - B π l2/T = - B l2 ω/2 = - π B l2 ν, where ν = ω/2π is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: ε0 = N B A ω = 100 × 0.2 × 0.01 × 50 = 10 V
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: ε = - π B l2 ν = - 3.14 × 0.5 × 0.3^2 × 60 = -8.5 V

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf ε = 15 sin 200 t. If N=100, A=0.01 m^2, B=0.3 T, find angular velocity ω.
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — ε0 = N B A ω = 50 × 0.3 × 0.02 × 40 = 12 V
• Q2: B) 10 V — ε = π B l2 ν = 3.14 × 0.4 × 0.4^2 × 100 = 10 V
• Q3: C) 200 rad/s — ε0 = N B A ω = 15 ⇒ ω = 15/(100 × 0.01 × 0.3) = 200

Memory Aids

• Peak emf ε0 = N B A ω
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf ∝ B l2 ν

By Changing Magnetic Field

• Changing magnetic field intensity \( B \) in a closed circuit changes flux \( \phi \), inducing emf.
• Formula: \( \varepsilon = - \frac{d\phi}{dt} = - \frac{d}{dt} (B A \cos \theta) \)
• Or \( \varepsilon = - A \cos \theta \frac{dB}{dt} \), where \( \frac{dB}{dt} \) is rate of change of magnetic field.

By Changing Area of Circuit Inside Magnetic Field

• Changing area \( A \) inside magnetic field changes flux, inducing emf.
• Formula: \( \varepsilon = - \frac{d\phi}{dt} = - B \cos \theta \frac{dA}{dt} \)
• For rectangular loop moving out of field with velocity \( v \): \( \varepsilon = B l v \cos \theta \)

By Changing Orientation of Coil or Angle \( \theta \)

• Coil of area \( A \), turns \( N \), rotating with angular velocity \( \omega \) in magnetic field \( B \).
• Flux linked: \( \phi = N B A \cos (\omega t) \)
• Induced emf: \( \varepsilon = - \frac{d\phi}{dt} = N B A \omega \sin (\omega t) = \varepsilon_0 \sin (\omega t) \)
• \( \varepsilon_0 = N B A \omega \) is peak emf; emf varies sinusoidally (AC emf).

Revolving Rod in Magnetic Field

• Rod PQ of length \( l \) rotates with angular velocity \( \omega \) in perpendicular magnetic field \( B \).
• Time period \( T = \frac{2\pi}{\omega} \).
• Induced emf: \( \varepsilon = - B \frac{dA}{dt} = - \frac{B \pi l^2}{T} = - \frac{B l^2 \omega}{2} = - \pi B l^2 \nu \), where \( \nu = \frac{\omega}{2\pi} \) is frequency.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  Solution: \( \varepsilon_0 = N B A \omega = 100 \times 0.2 \times 0.01 \times 50 = 10 \text{ V} \)
• Ex 2: Rod of length 0.3 m rotates at 60 Hz in 0.5 T field. Find emf.
  Solution: \( \varepsilon = - \pi B l^2 \nu = - 3.14 \times 0.5 \times 0.3^2 \times 60 = -8.5 \text{ V} \)

Practice Set

• Q1: A coil of 50 turns, area 0.02 m² rotates at 40 rad/s in 0.3 T field. Peak emf is:
  A) 12 V
  B) 24 V
  C) 8 V
  D) 6 V
• Q2: A rod of length 0.4 m rotates at 100 Hz in 0.4 T field. Induced emf is:
  A) 20 V
  B) 10 V
  C) 5 V
  D) 15 V
• Q3: A coil rotates in magnetic field producing emf \( \varepsilon = 15 \sin 200 t \). If \( N=100, A=0.01 \text{ m}^2, B=0.3 \text{ T} \), find angular velocity \( \omega \).
  A) 50 rad/s
  B) 150 rad/s
  C) 200 rad/s
  D) 100 rad/s

Answer Key

• Q1: A) 12 V — \( \varepsilon_0 = N B A \omega = 50 \times 0.3 \times 0.02 \times 40 = 12 \text{ V} \)
• Q2: B) 10 V — \( \varepsilon = \pi B l^2 \nu = 3.14 \times 0.4 \times 0.4^2 \times 100 = 10 \text{ V} \)
• Q3: C) 200 rad/s — \( \varepsilon_0 = N B A \omega = 15 \Rightarrow \omega = \frac{15}{100 \times 0.01 \times 0.3} = 200 \)

Memory Aids

• Peak emf \( \varepsilon_0 = N B A \omega \)
• Emf varies sinusoidally with time in rotating coil
• Revolving rod emf \( \propto B l^2 \nu \)

Eddy Currents

Concept and Applications

• Eddy currents are currents induced in the bulk of a conductor when magnetic flux linked with it changes.
• Direction found by Lenz’s law or Fleming’s right hand rule.
• Applications:
• Induction furnace (heating effect)
• Speedometer (measures vehicle speed)
• Energy meter (records electricity consumption)

Inductance

Concepts

• Inductance opposes change in current, analogous to inertia in mechanics.
• Scalar quantity; SI unit: henry (H), dimensions: [ML²T^-2A^-2].
• Self Inductance: Change in current in a coil induces emf opposing the change: ε = -L dI/dt.
• When current is constant, no emf induced.
• Self inductance formula: L = N Φ/I = NBA/I = φtotal/I.
• Real inductor modeled as resistor R in series with ideal inductor L.
• Self inductance of long solenoid: L = μ0 μr N² A/l = μ0 μr n² A l = μ0 μr n² V, where V = Al.
• Mutual Inductance: emf induced in one coil due to change in current in another coupled coil.
• Mutual inductance formula: M = N₂ Φ₂/I₁ = N₂ B₁ A₂/I₁.
• Mutual inductance of two coaxial solenoids: M = μ0 N₁ N₂ A/l.
• Mutual inductance of two concentric coplanar coils: M_{c1 c2} = N₂ B₁ A₂/I₁ = μ0 N₁ N₂ π r²/2 π l.
• Factors affecting mutual inductance: number of turns, common cross-sectional area, separation distance, and relative orientation.
• Energy stored in inductor: U = 1/2 L I².
• Combination of inductors:
  • Series: L_s = L₁ + L₂ + ··· + L_n.
  • Parallel: 1/L_p = 1/L₁ + 1/L₂ + ··· + 1/L_n.
• Effective inductance of two inductors in series with mutual inductance M: V = (L₁ + L₂ + 2M) di/dt.

Solved Examples

• Ex 1: Calculate self inductance L of a solenoid with N=500, area A=1 × 10^-4 m², length l=0.5 m, and permeability μr=1.
  • Use L = μ0 μr N² A/l.
  • Calculate L = 4π × 10^-7 × 1 × 500² × 1 × 10^-4/0.5 = 6.28 × 10^-3 H.
• Ex 2: Find mutual inductance M of two coaxial solenoids with N₁=200, N₂=300, area A=2 × 10^-4 m², length l=0.4 m.
  • Use M = μ0 N₁ N₂ A/l.
  • Calculate M = 4π × 10^-7 × 200 × 300 × 2 × 10^-4/0.4 = 3.77 × 10^-3 H.

Practice Set

• Q1: What is the SI unit of inductance?
  • A) Ohm
  • B) Henry
  • C) Tesla
  • D) Weber
• Q2: The self-induced emf in a coil is given by?
  • A) ε = L dI/dt
  • B) ε = -L dI/dt
  • C) ε = L I
  • D) ε = -I dL/dt
• Q3: Mutual inductance depends on?
  • A) Number of turns
  • B) Cross-sectional area
  • C) Distance between coils
  • D) All of the above

Answer Key

• Q1: B - Henry is the SI unit of inductance.
• Q2: B - Self-induced emf opposes change in current, hence negative sign.
• Q3: D - Mutual inductance depends on turns, area, and separation.

Memory Aids

• Inductance opposes current change like inertia opposes motion change.
• Self inductance formula: ε = -L dI/dt.
• Mutual inductance increases with turns and area, decreases with distance.
• Energy stored: U = 1/2 L I².

Concepts

• Inductance opposes change in current, analogous to inertia in mechanics.
• Scalar quantity; SI unit: henry (H), dimensions: [ML²T^-2A^-2].
• Self Inductance: Change in current in a coil induces emf opposing the change: ε = -L dI/dt.
• When current is constant, no emf induced.
• Self inductance formula: L = N Φ/I = NBA/I = φtotal/I.
• Real inductor modeled as resistor R in series with ideal inductor L.
• Self inductance of long solenoid: L = μ0 μr N2 A/l = μ0 μr n2 A l = μ0 μr n2 V, where V = Al.
• Mutual Inductance: emf induced in one coil due to change in current in another coupled coil.
• Mutual inductance formula: M = N2 Φ2/I1 = N2 B1 A2/I1.
• Mutual inductance of two coaxial solenoids: M = μ0 N1 N2 A/l.
• Mutual inductance of two concentric coplanar coils: M c1 c2 = N2 B1 A2/I1 = μ0 N1 N2 π r2/2 π l.
• Factors affecting mutual inductance: number of turns, common cross-sectional area, separation distance, and relative orientation.
• Energy stored in inductor: U = 1/2 L I2.
• Combination of inductors:
  • Series: Ls = L1 + L2 + ··· + Ln.
  • Parallel: 1/Lp = 1/L1 + 1/L2 + ··· + 1/Ln.
• Effective inductance of two inductors in series with mutual inductance M: V = (L1 + L2 + 2M) di/dt.

Solved Examples

• Ex 1: Calculate self inductance L of a solenoid with N=500, area A=1 × 10^-4 m², length l=0.5 m, and permeability μr=1.
  • Use L = μ0 μr N2 A/l.
  • Calculate L = 4π × 10^-7 × 1 × 500² × 1 × 10^-4/0.5 = 6.28 × 10^-3 H.
• Ex 2: Find mutual inductance M of two coaxial solenoids with N1=200, N2=300, area A=2 × 10^-4 m², length l=0.4 m.
  • Use M = μ0 N1 N2 A/l.
  • Calculate M = 4π × 10^-7 × 200 × 300 × 2 × 10^-4/0.4 = 3.77 × 10^-3 H.

Practice Set

• Q1: What is the SI unit of inductance?
  • A) Ohm
  • B) Henry
  • C) Tesla
  • D) Weber
• Q2: The self-induced emf in a coil is given by?
  • A) ε = L dI/dt
  • B) ε = -L dI/dt
  • C) ε = L I
  • D) ε = -I dL/dt
• Q3: Mutual inductance depends on?
  • A) Number of turns
  • B) Cross-sectional area
  • C) Distance between coils
  • D) All of the above

Answer Key

• Q1: B - Henry is the SI unit of inductance.
• Q2: B - Self-induced emf opposes change in current, hence negative sign.
• Q3: D - Mutual inductance depends on turns, area, and separation.

Memory Aids

• Inductance opposes current change like inertia opposes motion change.
• Self inductance formula: ε = -L dI/dt.
• Mutual inductance increases with turns and area, decreases with distance.
• Energy stored: U = 1/2 L I2.

Concepts

• Inductance opposes change in current, analogous to inertia in mechanics.
• Scalar quantity; SI unit: henry (H), dimensions: [ML²T^-2A^-2].
• Self Inductance: Change in current in a coil induces emf opposing the change: ε = -L dI/dt.
• When current is constant, no emf induced.
• Self inductance formula: L = N Φ/I = NBA/I = φtotal/I.
• Real inductor modeled as resistor R in series with ideal inductor L.
• Self inductance of long solenoid: L = μ0 μr N2 A/l = μ0 μr n2 A l = μ0 μr n2 V, where V = Al.
• Mutual Inductance: emf induced in one coil due to change in current in another coupled coil.
• Mutual inductance formula: M = N2 Φ2/I1 = N2 B1 A2/I1.
• Mutual inductance of two coaxial solenoids: M = μ0 N1 N2 A/l.
• Mutual inductance of two concentric coplanar coils: Mc1 c2 = N2 B1 A2/I1 = μ0 N1 N2 π r2/2 π l.
• Factors affecting mutual inductance: number of turns, common cross-sectional area, separation distance, and relative orientation.
• Energy stored in inductor: U = 1/2 L I2.
• Combination of inductors:
  • Series: Ls = L1 + L2 + ··· + Ln.
  • Parallel: 1/Lp = 1/L1 + 1/L2 + ··· + 1/Ln.
• Effective inductance of two inductors in series with mutual inductance M: V = (L1 + L2 + 2M) di/dt.

Solved Examples

• Ex 1: Calculate self inductance L of a solenoid with N=500, area A=1 × 10^-4 m², length l=0.5 m, and permeability μr=1.
  • Use L = μ0 μr N2 A/l.
  • Calculate L = 4π × 10^-7 × 1 × 500² × 1 × 10^-4/0.5 = 6.28 × 10^-3 H.
• Ex 2: Find mutual inductance M of two coaxial solenoids with N1=200, N2=300, area A=2 × 10^-4 m², length l=0.4 m.
  • Use M = μ0 N1 N2 A/l.
  • Calculate M = 4π × 10^-7 × 200 × 300 × 2 × 10^-4/0.4 = 3.77 × 10^-3 H.

Practice Set

• Q1: What is the SI unit of inductance?
  • A) Ohm
  • B) Henry
  • C) Tesla
  • D) Weber
• Q2: The self-induced emf in a coil is given by?
  • A) ε = L dI/dt
  • B) ε = -L dI/dt
  • C) ε = L I
  • D) ε = -I dL/dt
• Q3: Mutual inductance depends on?
  • A) Number of turns
  • B) Cross-sectional area
  • C) Distance between coils
  • D) All of the above

Answer Key

• Q1: B - Henry is the SI unit of inductance.
• Q2: B - Self-induced emf opposes change in current, hence negative sign.
• Q3: D - Mutual inductance depends on turns, area, and separation.

Memory Aids

• Inductance opposes current change like inertia opposes motion change.
• Self inductance formula: ε = -L dI/dt.
• Mutual inductance increases with turns and area, decreases with distance.
• Energy stored: U = 1/2 L I2.

Concepts

• Inductance opposes change in current, analogous to inertia in mechanics.
• Scalar quantity; SI unit: henry (H), dimensions: [ML²T^-2A^-2].
• Self Inductance: Change in current in a coil induces emf opposing the change: \( \varepsilon = -L \frac{dI}{dt} \).
• When current is constant, no emf induced.
• Self inductance formula: \( L = \frac{N \Phi}{I} = \frac{NBA}{I} = \frac{\phi_{total}}{I} \).
• Real inductor modeled as resistor \(R\) in series with ideal inductor \(L\).
• Self inductance of long solenoid: \( L = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_0 \mu_r n^2 A l = \mu_0 \mu_r n^2 V \), where \(V = Al\).
• Mutual Inductance: emf induced in one coil due to change in current in another coupled coil.
• Mutual inductance formula: \( M = \frac{N_2 \Phi_2}{I_1} = \frac{N_2 B_1 A_2}{I_1} \).
• Mutual inductance of two coaxial solenoids: \( M = \frac{\mu_0 N_1 N_2 A}{l} \).
• Mutual inductance of two concentric coplanar coils: \( M_{c_1 c_2} = \frac{N_2 B_1 A_2}{I_1} = \frac{\mu_0 N_1 N_2 \pi r^2}{2 \pi l} \).
• Factors affecting mutual inductance: number of turns, common cross-sectional area, separation distance, and relative orientation.
• Energy stored in inductor: \( U = \frac{1}{2} L I^2 \).
• Combination of inductors:
  • Series: \( L_s = L_1 + L_2 + \cdots + L_n \).
  • Parallel: \( \frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2} + \cdots + \frac{1}{L_n} \).
• Effective inductance of two inductors in series with mutual inductance \(M\): \( V = (L_1 + L_2 + 2M) \frac{di}{dt} \).

Solved Examples

• Ex 1: Calculate self inductance \(L\) of a solenoid with \(N=500\), area \(A=1 \times 10^{-4} m^2\), length \(l=0.5 m\), and permeability \(\mu_r=1\).
  • Use \( L = \frac{\mu_0 \mu_r N^2 A}{l} \).
  • Calculate \( L = \frac{4\pi \times 10^{-7} \times 1 \times 500^2 \times 1 \times 10^{-4}}{0.5} = 6.28 \times 10^{-3} H \).
• Ex 2: Find mutual inductance \(M\) of two coaxial solenoids with \(N_1=200\), \(N_2=300\), area \(A=2 \times 10^{-4} m^2\), length \(l=0.4 m\).
  • Use \( M = \frac{\mu_0 N_1 N_2 A}{l} \).
  • Calculate \( M = \frac{4\pi \times 10^{-7} \times 200 \times 300 \times 2 \times 10^{-4}}{0.4} = 3.77 \times 10^{-3} H \).

Practice Set

• Q1: What is the SI unit of inductance?
  • A) Ohm
  • B) Henry
  • C) Tesla
  • D) Weber
• Q2: The self-induced emf in a coil is given by?
  • A) \( \varepsilon = L \frac{dI}{dt} \)
  • B) \( \varepsilon = -L \frac{dI}{dt} \)
  • C) \( \varepsilon = L I \)
  • D) \( \varepsilon = -I \frac{dL}{dt} \)
• Q3: Mutual inductance depends on?
  • A) Number of turns
  • B) Cross-sectional area
  • C) Distance between coils
  • D) All of the above

Answer Key

• Q1: B - Henry is the SI unit of inductance.
• Q2: B - Self-induced emf opposes change in current, hence negative sign.
• Q3: D - Mutual inductance depends on turns, area, and separation.

Memory Aids

• Inductance opposes current change like inertia opposes motion change.
• Self inductance formula: \( \varepsilon = -L \frac{dI}{dt} \).
• Mutual inductance increases with turns and area, decreases with distance.
• Energy stored: \( U = \frac{1}{2} L I^2 \).

LR Circuits

Current Growth

• Emf equation: V = IR + L dI/dt.
• Current at time t: I = I0 (1 - e^-t/τ), where I0 = V/R, τ = L/R (time constant).
• At t=0, inductor acts as open circuit (infinite resistance), current I=0.
• At t → ∞, inductor acts as short circuit (zero resistance), current I=I0.
• Peak current independent of inductance.

Current Decay

• Emf equation: IR + L dI/dt = 0.
• Current at time t: I = I0 e^-t/τ, where I0 = V/R, τ = L/R.
• At t=0, current I=I0.
• At t → ∞, current I=0.

Solved Examples

• Ex 1: Calculate time constant τ for L=2H, R=4Ω.
  • τ = L/R = 2/4 = 0.5 s.
• Ex 2: Find current I at t=1s for I0=5A, τ=0.5s.
  • I = I0 (1 - e^-t/τ) = 5 (1 - e^-2) = 4.32 A.

Practice Set

• Q1: What is the time constant τ in an LR circuit?
  • A) L × R
  • B) L/R
  • C) R/L
  • D) L + R
• Q2: At t=0, current in LR circuit is?
  • A) Maximum
  • B) Zero
  • C) Half of maximum
  • D) Infinite
• Q3: Current I at time t during growth is given by?
  • A) I = I0 e^-t/τ
  • B) I = I0 (1 - e^-t/τ)
  • C) I = I0
  • D) I = 0

Answer Key

• Q1: B - Time constant τ = L/R.
• Q2: B - Inductor initially opposes current, so current is zero at t=0.
• Q3: B - Current growth follows I = I0 (1 - e^-t/τ).

Memory Aids

• Time constant τ = L/R controls current change speed.
• Current starts at zero and grows exponentially in LR circuit.
• Inductor acts as open circuit at start, short circuit at steady state.

Current Growth

• Emf equation: V = IR + L dI/dt.
• Current at time t: I = I0 (1 - e-t/τ), where I0 = V/R, τ = L/R (time constant).
• At t=0, inductor acts as open circuit (infinite resistance), current I=0.
• At t → ∞, inductor acts as short circuit (zero resistance), current I=I0.
• Peak current independent of inductance.

Current Decay

• Emf equation: IR + L dI/dt = 0.
• Current at time t: I = I0 e-t/τ, where I0 = V/R, τ = L/R.
• At t=0, current I=I0.
• At t → ∞, current I=0.

Solved Examples

• Ex 1: Calculate time constant τ for L=2H, R=4Ω.
  • τ = L/R = 2/4 = 0.5 s.
• Ex 2: Find current I at t=1s for I0=5A, τ=0.5s.
  • I = I0 (1 - e-t/τ) = 5 (1 - e-2) = 4.32 A.

Practice Set

• Q1: What is the time constant τ in an LR circuit?
  • A) L × R
  • B) L/R
  • C) R/L
  • D) L + R
• Q2: At t=0, current in LR circuit is?
  • A) Maximum
  • B) Zero
  • C) Half of maximum
  • D) Infinite
• Q3: Current I at time t during growth is given by?
  • A) I = I0 e-t/τ
  • B) I = I0 (1 - e-t/τ)
  • C) I = I0
  • D) I = 0

Answer Key

• Q1: B - Time constant τ = L/R.
• Q2: B - Inductor initially opposes current, so current is zero at t=0.
• Q3: B - Current growth follows I = I0 (1 - e-t/τ).

Memory Aids

• Time constant τ = L/R controls current change speed.
• Current starts at zero and grows exponentially in LR circuit.
• Inductor acts as open circuit at start, short circuit at steady state.

Current Growth

• Emf equation: V = IR + L dI/dt.
• Current at time t: I = I0 (1 - e-t/τ), where I0 = V/R, τ = L/R (time constant).
• At t=0, inductor acts as open circuit (infinite resistance), current I=0.
• At t → ∞, inductor acts as short circuit (zero resistance), current I=I0.
• Peak current independent of inductance.

Current Decay

• Emf equation: IR + L dI/dt = 0.
• Current at time t: I = I0 e-t/τ, where I0 = V/R, τ = L/R.
• At t=0, current I=I0.
• At t → ∞, current I=0.

Solved Examples

• Ex 1: Calculate time constant τ for L=2H, R=4Ω.
  • τ = L/R = 2/4 = 0.5 s.
• Ex 2: Find current I at t=1s for I0=5A, τ=0.5s.
  • I = I0 (1 - e-t/τ) = 5 (1 - e-2) = 4.32 A.

Practice Set

• Q1: What is the time constant τ in an LR circuit?
  • A) L × R
  • B) L/R
  • C) R/L
  • D) L + R
• Q2: At t=0, current in LR circuit is?
  • A) Maximum
  • B) Zero
  • C) Half of maximum
  • D) Infinite
• Q3: Current I at time t during growth is given by?
  • A) I = I0 e-t/τ
  • B) I = I0 (1 - e-t/τ)
  • C) I = I0
  • D) I = 0

Answer Key

• Q1: B - Time constant τ = L/R.
• Q2: B - Inductor initially opposes current, so current is zero at t=0.
• Q3: B - Current growth follows I = I0 (1 - e-t/τ).

Memory Aids

• Time constant τ = L/R controls current change speed.
• Current starts at zero and grows exponentially in LR circuit.
• Inductor acts as open circuit at start, short circuit at steady state.

Current Growth

• Emf equation: \( V = IR + L \frac{dI}{dt} \).
• Current at time \(t\): \( I = I_0 (1 - e^{-t/\tau}) \), where \( I_0 = \frac{V}{R} \), \( \tau = \frac{L}{R} \) (time constant).
• At \( t=0 \), inductor acts as open circuit (infinite resistance), current \( I=0 \).
• At \( t \to \infty \), inductor acts as short circuit (zero resistance), current \( I=I_0 \).
• Peak current independent of inductance.

Current Decay

• Emf equation: \( IR + L \frac{dI}{dt} = 0 \).
• Current at time \(t\): \( I = I_0 e^{-t/\tau} \), where \( I_0 = \frac{V}{R} \), \( \tau = \frac{L}{R} \).
• At \( t=0 \), current \( I=I_0 \).
• At \( t \to \infty \), current \( I=0 \).

Solved Examples

• Ex 1: Calculate time constant \( \tau \) for \( L=2H \), \( R=4\Omega \).
  • \( \tau = \frac{L}{R} = \frac{2}{4} = 0.5 s \).
• Ex 2: Find current \( I \) at \( t=1s \) for \( I_0=5A \), \( \tau=0.5s \).
  • \( I = I_0 (1 - e^{-t/\tau}) = 5 (1 - e^{-2}) = 4.32 A \).

Practice Set

• Q1: What is the time constant \( \tau \) in an LR circuit?
  • A) \( L \times R \)
  • B) \( \frac{L}{R} \)
  • C) \( \frac{R}{L} \)
  • D) \( L + R \)
• Q2: At \( t=0 \), current in LR circuit is?
  • A) Maximum
  • B) Zero
  • C) Half of maximum
  • D) Infinite
• Q3: Current \( I \) at time \( t \) during growth is given by?
  • A) \( I = I_0 e^{-t/\tau} \)
  • B) \( I = I_0 (1 - e^{-t/\tau}) \)
  • C) \( I = I_0 \)
  • D) \( I = 0 \)

Answer Key

• Q1: B - Time constant \( \tau = \frac{L}{R} \).
• Q2: B - Inductor initially opposes current, so current is zero at \( t=0 \).
• Q3: B - Current growth follows \( I = I_0 (1 - e^{-t/\tau}) \).

Memory Aids

• Time constant \( \tau = \frac{L}{R} \) controls current change speed.
• Current starts at zero and grows exponentially in LR circuit.
• Inductor acts as open circuit at start, short circuit at steady state.

AC Generator

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: ϕ = N B A cos θ, where N = turns, B = magnetic field, A = coil area.
• With θ = ω t, flux: ϕ = B A N cos ω t.
• Induced emf: ε = -dϕ/dt = B A N ω sin ω t = ε0 sin ω t, where ε0 = B A N ω.
• Current: I = ε/R = I0 sin ω t, where I0 = ε0/R.

Solved Examples

• Ex 1: Calculate maximum emf if N=100, B=0.5 T, A=0.01 m², ω=314 rad/s.
  • ε0 = B A N ω = 0.5 × 0.01 × 100 × 314 = 157 V

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) cos ω t
  • B) sin ω t
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as sin ω t.
• Q3: B - ε0 = B A N ω, doubling ω doubles emf.

Memory Aids

• Emf formula: ε = B A N ω sin ω t
• Current formula: I = ε/R
• Fleming's right-hand rule for current direction.

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Current changes direction periodically.
• Essential parts: armature, field magnets (N and S), slip rings, brushes.

Principle

• Based on electromagnetic induction.
• Changing magnetic flux linked with coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Loop of wire rotates in magnetic field, inducing alternating emf.
• Induced emf varies sinusoidally with rotation angle.
• Magnetic flux linked: \( \phi = N B A \cos \theta \), where \(N\) = turns, \(B\) = magnetic field, \(A\) = coil area.
• With \( \theta = \omega t \), flux: \( \phi = B A N \cos \omega t \).
• Induced emf: \( \varepsilon = -\frac{d\phi}{dt} = B A N \omega \sin \omega t = \varepsilon_0 \sin \omega t \), where \( \varepsilon_0 = B A N \omega \).
• Current: \( I = \frac{\varepsilon}{R} = I_0 \sin \omega t \), where \( I_0 = \frac{\varepsilon_0}{R} \).

Solved Examples

• Ex 1: Calculate maximum emf if \(N=100\), \(B=0.5\,T\), \(A=0.01\,m^2\), \(\omega=314\,rad/s\).
  • \( \varepsilon_0 = B A N \omega = 0.5 \times 0.01 \times 100 \times 314 = 157 V \)

Practice Set

• Q1: What is the principle of AC generator?
  • A) Electrostatic induction
  • B) Electromagnetic induction
  • C) Magnetic attraction
  • D) Thermal conduction
• Q2: The emf induced in an AC generator varies as:
  • A) \( \cos \omega t \)
  • B) \( \sin \omega t \)
  • C) Constant
  • D) Linear with time
• Q3: If the frequency of rotation doubles, the maximum emf will:
  • A) Halve
  • B) Double
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Electromagnetic induction causes emf by changing magnetic flux.
• Q2: B - Emf varies sinusoidally as \( \sin \omega t \).
• Q3: B - \( \varepsilon_0 = B A N \omega \), doubling \( \omega \) doubles emf.

Memory Aids

• Emf formula: \( \varepsilon = B A N \omega \sin \omega t \)
• Current formula: \( I = \frac{\varepsilon}{R} \)
• Fleming's right-hand rule for current direction.

Induced Electric Field

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: ∮ vector E · dvector l = ε = -dϕ/dt ≠ 0.
• Force on charge: vector F = q vector E.
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: E l = |dϕ/dt| = A |dB/dt|, where l = loop length, A = area.

Solved Examples

• Ex 1: Magnetic field changes at rate dB/dt = 0.1 T/s over area A = 0.02 m², loop length l = 0.5 m.
  • E = A |dB/dt|/l = 0.02 × 0.1/0.5 = 0.004 V/m

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf ε = -dϕ/dt.
• Q3: A - E = A |dB/dt|/l = 0.01 × 0.2/1 = 0.002 V/m.

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: E l = A |dB/dt|.
• Force on charge: vector F = q vector E.

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: ∮ vector E · dvector l = ε = -dϕ/dt ≠ 0.
• Force on charge: vector F = q vector E.
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: E l = |dϕ/dt| = A |dB/dt|, where l = loop length, A = area.

Solved Examples

• Ex 1: Magnetic field changes at rate dB/dt = 0.1 T/s over area A = 0.02 m², loop length l = 0.5 m.
  • E = A |dB/dt|/l = 0.02 × 0.1/0.5 = 0.004 V/m

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf ε = -dϕ/dt.
• Q3: A - E = A |dB/dt|/l = 0.01 × 0.2/1 = 0.002 V/m.

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: E l = A |dB/dt|.
• Force on charge: vector F = q vector E.

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: ∮ vector E · dvector l = ε = -dϕ/dt ≠ 0.
• Force on charge: vector F = q vector E.
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: E l = |dϕ/dt| = A |dB/dt|, where l = loop length, A = area.

Solved Examples

• Ex 1: Magnetic field changes at rate dB/dt = 0.1 T/s over area A = 0.02 m², loop length l = 0.5 m.
  • E = A |dB/dt|/l = 0.02 × 0.1/0.5 = 0.004 V/m

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf ε = -dϕ/dt.
• Q3: A - E = A |dB/dt|/l = 0.01 × 0.2/1 = 0.002 V/m.

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: E l = A |dB/dt|.
• Force on charge: vector F = q vector E.

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: \( \oint \vec{E} \cdot d\vec{l} = \varepsilon = -\frac{d\phi}{dt} \neq 0 \).
• Force on charge: \( \vec{F} = q \vec{E} \).
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: \( E l = |\frac{d\phi}{dt}| = A |\frac{dB}{dt}| \), where \(l\) = loop length, \(A\) = area.

Solved Examples

• Ex 1: Magnetic field changes at rate \( \frac{dB}{dt} = 0.1\,T/s \) over area \( A = 0.02\,m^2 \), loop length \( l = 0.5\,m \).
  • \( E = \frac{A |dB/dt|}{l} = \frac{0.02 \times 0.1}{0.5} = 0.004\,V/m \)

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf \( \varepsilon = -d\phi/dt \).
• Q3: A - \( E = \frac{A |dB/dt|}{l} = \frac{0.01 \times 0.2}{1} = 0.002 V/m \).

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: \( E l = A |dB/dt| \).
• Force on charge: \( \vec{F} = q \vec{E} \).

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: ∮ vector E · dvector l = ε = -dϕ/dt ≠ 0.
• Force on charge: vector F = q vector E.
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: E l = |dϕ/dt| = A |dB/dt|, where l = loop length, A = area.

Solved Examples

• Ex 1: Magnetic field changes at rate dB/dt = 0.1 T/s over area A = 0.02 m², loop length l = 0.5 m.
  • E = A |dB/dt|/l = 0.02 × 0.1/0.5 = 0.004 V/m

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf ε = -dϕ/dt.
• Q3: A - E = A |dB/dt|/l = 0.01 × 0.2/1 = 0.002 V/m.

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: E l = A |dB/dt|.
• Force on charge: vector F = q vector E.

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: ∮ vector E · dvector l = ε = -dϕ/dt ≠ 0.
• Force on charge: vector F = q vector E.
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: E l = |dϕ/dt| = A |dB/dt|, where l = loop length, A = area.

Solved Examples

• Ex 1: Magnetic field changes at rate dB/dt = 0.1 T/s over area A = 0.02 m², loop length l = 0.5 m.
  • E = A |dB/dt|/l = 0.02 × 0.1/0.5 = 0.004 V/m

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf ε = -dϕ/dt.
• Q3: A - E = A |dB/dt|/l = 0.01 × 0.2/1 = 0.002 V/m.

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: E l = A |dB/dt|.
• Force on charge: vector F = q vector E.

Concept and Properties

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative: \( \oint \vec{E} \cdot d\vec{l} = \varepsilon = -\frac{d\phi}{dt} \neq 0 \).
• Force on charge: \( \vec{F} = q \vec{E} \).
• Electric field magnitude same at all points on circular loop; tangential direction.
• For symmetric cases: \( E l = |\frac{d\phi}{dt}| = A |\frac{dB}{dt}| \), where \(l\) = loop length, \(A\) = area.

Solved Examples

• Ex 1: Magnetic field changes at rate \( \frac{dB}{dt} = 0.1\,T/s \) over area \( A = 0.02\,m^2 \), loop length \( l = 0.5\,m \).
  • \( E = \frac{A |dB/dt|}{l} = \frac{0.02 \times 0.1}{0.5} = 0.004\,V/m \)

Practice Set

• Q1: Induced electric field is:
  • A) Conservative
  • B) Non-conservative
  • C) Electrostatic
  • D) Zero
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Negative
  • D) Constant
• Q3: For a circular loop of length 1 m and area 0.01 m², if magnetic field changes at 0.2 T/s, induced electric field magnitude is:
  • A) 0.002 V/m
  • B) 0.01 V/m
  • C) 0.2 V/m
  • D) 0.02 V/m

Answer Key

• Q1: B - Induced electric field is non-conservative due to changing magnetic flux.
• Q2: B - Line integral equals emf \( \varepsilon = -d\phi/dt \).
• Q3: A - \( E = \frac{A |dB/dt|}{l} = \frac{0.01 \times 0.2}{1} = 0.002 V/m \).

Memory Aids

• Induced electric field: non-conservative, tangential to loop.
• Formula: \( E l = A |dB/dt| \).
• Force on charge: \( \vec{F} = q \vec{E} \).