Chapter 20: Magnetism and Matter

Introduction

Key Points

• Magnet name origin: from Magnesia island, Greece.
• Magnet: material with attractive and directive properties.
• Attracts iron, nickel, cobalt; this attraction is magnetism.
• Earth acts as a magnet; magnetic field approx. from geographic south to north.
• Freely suspended bar magnet aligns north-south; north pole points geographic north.
• Like poles repel; unlike poles attract.
• Magnetic monopoles (isolated poles) do not exist.
• Magnets can be made from iron and its alloys.

Solved Examples

• Ex 1: What happens when two north poles of magnets are brought close?
  Answer: They repel due to like poles repulsion.
• Ex 2: Can a single magnetic pole exist?
  Answer: No, magnetic monopoles do not exist; breaking a magnet forms two smaller dipoles.

Practice Set

• Q1: What is magnetism?
  A) Repulsion between metals
  B) Attraction of certain metals
  C) Electric charge
  D) Heat generation
  
• Q2: Which element is NOT attracted by a magnet?
  A) Iron
  B) Nickel
  C) Copper
  D) Cobalt
  
• Q3: What is true about magnetic monopoles?
  A) Exist isolated
  B) Exist in pairs
  C) Do not exist
  D) Are electric charges
  

Answer Key

• Q1: B - Magnetism is the attraction of certain metals like iron.
• Q2: C - Copper is not attracted by magnets.
• Q3: C - Magnetic monopoles do not exist.

Memory Aids

• Magnetism = attraction of Fe, Ni, Co.
• Earth's magnetic field: south to north.
• Monopoles do not exist; always dipoles.

The Bar Magnet

Properties

• Bar magnet is a natural dipole producing magnetic field.
• Attracts small pieces of iron, nickel, cobalt.
• Poles: ends where magnetic force is maximum.
• Freely suspended bar magnet aligns north-south.
• North pole (N) points geographic north; south pole (S) points geographic south.
• Like poles repel; unlike poles attract.
• Magnetic poles always in pairs; monopoles do not exist.
• Repulsion distinguishes magnet from iron.

Solved Examples

• Ex 1: What direction does a freely suspended bar magnet point?
  Answer: North-South direction.
• Ex 2: What happens if two south poles are brought close?
  Answer: They repel.

Practice Set

• Q1: What are the poles of a bar magnet?
  A) Single pole
  B) Two poles
  C) No poles
  D) Multiple poles
  
• Q2: What is the pole that points to Earth's north called?
  A) South pole
  B) North pole
  C) Magnetic monopole
  D) Neutral pole
  
• Q3: How to distinguish a magnet from iron?
  A) Attraction only
  B) Repulsion test
  C) Heat test
  D) Color test
  

Answer Key

• Q1: B - Bar magnet has two poles.
• Q2: B - Pole pointing north is north pole.
• Q3: B - Repulsion test distinguishes magnet from iron.

Memory Aids

• Bar magnet = dipole with N and S poles.
• Like poles repel; unlike attract.
• Repulsion test confirms magnet.

The Magnetic Field Lines

Key Properties

• Magnetic field lines form closed continuous loops.
• Tangent to field line = direction of magnetic field at that point.
• More field lines per unit area = stronger magnetic field.
• Field lines never intersect.
• Field lines of bar magnet and solenoid shown in images.

Solved Examples

• Ex 1: What does the density of field lines indicate?
  Answer: Strength of magnetic field.
• Ex 2: Why don't magnetic field lines intersect?
  Answer: To keep unique field direction.

Practice Set

• Q1: Magnetic field lines form:
  A) Open lines
  B) Closed loops
  C) Random lines
  D) Straight lines
  
• Q2: What does tangent to field line represent?
  A) Magnitude
  B) Direction of field
  C) Force
  D) Charge
  
• Q3: Why field lines never intersect?
  A) They do intersect
  B) To keep unique direction
  C) Random behavior
  D) No reason
  

Answer Key

• Q1: B - Magnetic field lines form closed loops.
• Q2: B - Tangent shows field direction.
• Q3: B - Intersection would cause ambiguous direction.

Memory Aids

• Field lines = closed loops.
• Tangent = field direction.
• Density = field strength.

Bar Magnet as an Equivalent Solenoid

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: ( B = μ₀/(4π) (2m/r³) ).
• ( μ₀ ): permeability of free space, ( m ): magnetic moment, ( r ): distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: ( B = μ₀/(4π) (2m/r³) ).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: ( B = μ₀/(4π) (2m/r³) ).
• ( μ₀ ): permeability of free space, ( m ): magnetic moment, ( r ): distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: ( B = μ₀/(4π) (2m/r³) ).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: B = μ₀/(4π) (2m/r³).
• μ₀: permeability of free space, m: magnetic moment, r: distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: B = μ₀/(4π) (2m/r³).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: ( B = μ₀/(4π) (2m/r³) ).
• ( μ₀ ): permeability of free space, ( m ): magnetic moment, ( r ): distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: ( B = μ₀/(4π) (2m/r³) ).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) ( r³ )
  B) ( 1/r³ )
  C) ( r )
  D) ( 1/r )
  
• Q3: Magnetic moment symbol is:
  A) ( m )
  B) ( p )
  C) ( q )
  D) ( I )
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is ( m ).

Memory Aids

• Bar magnet = magnetic dipole.
• Field ( ∝ 1/r³ ).
• Formula: ( B = μ₀/(4π) (2m/r³) ).

Key Points

• Bar magnet and solenoid produce similar magnetic fields.
• Magnetic field at far axial point of bar magnet: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).
• \( \mu_0 \): permeability of free space, \( m \): magnetic moment, \( r \): distance.

Solved Examples

• Ex 1: Write formula for magnetic field of bar magnet at far axial point.
  Answer: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

Practice Set

• Q1: What does a bar magnet behave like?
  A) Electric dipole
  B) Magnetic monopole
  C) Magnetic dipole
  D) No magnetic field
  
• Q2: Magnetic field at far axial point is proportional to:
  A) \( r^3 \)
  B) \( \frac{1}{r^3} \)
  C) \( r \)
  D) \( \frac{1}{r} \)
  
• Q3: Magnetic moment symbol is:
  A) \( m \)
  B) \( p \)
  C) \( q \)
  D) \( I \)
  

Answer Key

• Q1: C - Bar magnet is a magnetic dipole.
• Q2: B - Field varies as inverse cube of distance.
• Q3: A - Magnetic moment symbol is \( m \).

Memory Aids

• Bar magnet = magnetic dipole.
• Field \( \propto \frac{1}{r^3} \).
• Formula: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

The Dipole in a Uniform Magnetic Field

Key Concepts

• Magnetic dipole moment ( vector m ) in uniform magnetic field ( vector B ).
• Torque on dipole: ( vector τ = vector m × vector B ).
• Magnitude of torque: ( τ = mB sin θ ), ( θ ) = angle between ( vector m ) and ( vector B ).
• Torque aligns dipole with magnetic field.
• Potential energy: ( U = -vector m · vector B = -mB cos θ ).
• Work done to rotate dipole through angle ( θ ) against torque.
• Analogy with electric dipole: similar formulas for moment, field, torque, energy.

> Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

> Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

perpendicular to the field.

Solved Examples

• Ex 1: Calculate torque on dipole with ( m=2 Am² ), ( B=0.5 T ), ( θ=30° ).
  Solution: ( τ = mB sin θ = 2 × 0.5 × sin 30° = 0.5 Nm ).
• Ex 2: Find potential energy for ( m=3 Am² ), ( B=0.4 T ), ( θ=60° ).
  Solution: ( U = -mB cos θ = -3 × 0.4 × cos 60° = -0.6 J ).

Practice Set

• Q1: Torque on magnetic dipole is zero when angle between ( vector m ) and ( vector B ) is:
  A) 0°
  B) 90°
  C) 45°
  D) 180°
  
• Q2: Potential energy of dipole is minimum when ( θ = ):
  A) 0°
  B) 90°
  C) 180°
  D) 45°
  
• Q3: Torque formula is:
  A) ( mB cos θ )
  B) ( mB sin θ )
  C) ( mB tan θ )
  D) ( mB )
  

Answer Key

• Q1: A - Torque zero when dipole aligned with field (0°).
• Q2: A - Potential energy minimum when dipole aligned (0°).
• Q3: B - Torque magnitude is ( mB sin θ ).

Memory Aids

• Torque ( τ = mB sin θ ).
• Potential energy ( U = -mB cos θ ).
• Dipole aligns to minimize energy.

Key Concepts

• Magnetic dipole moment ( vector m ) in uniform magnetic field ( vector B ).
• Torque on dipole: ( vector τ = vector m × vector B ).
• Magnitude of torque: ( τ = mB sin θ ), ( θ ) = angle between ( vector m ) and ( vector B ).
• Torque aligns dipole with magnetic field.
• Potential energy: ( U = -vector m · vector B = -mB cos θ ).
• Work done to rotate dipole through angle ( θ ) against torque.
• Analogy with electric dipole: similar formulas for moment, field, torque, energy.

> Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

> Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

perpendicular to the field.

Solved Examples

• Ex 1: Calculate torque on dipole with ( m=2 Am² ), ( B=0.5 T ), ( θ=30° ).
  Solution: ( τ = mB sin θ = 2 × 0.5 × sin 30° = 0.5 Nm ).
• Ex 2: Find potential energy for ( m=3 Am² ), ( B=0.4 T ), ( θ=60° ).
  Solution: ( U = -mB cos θ = -3 × 0.4 × cos 60° = -0.6 J ).

Practice Set

• Q1: Torque on magnetic dipole is zero when angle between ( vector m ) and ( vector B ) is:
  A) 0°
  B) 90°
  C) 45°
  D) 180°
  
• Q2: Potential energy of dipole is minimum when ( θ = ):
  A) 0°
  B) 90°
  C) 180°
  D) 45°
  
• Q3: Torque formula is:
  A) ( mB cos θ )
  B) ( mB sin θ )
  C) ( mB tan θ )
  D) ( mB )
  

Answer Key

• Q1: A - Torque zero when dipole aligned with field (0°).
• Q2: A - Potential energy minimum when dipole aligned (0°).
• Q3: B - Torque magnitude is ( mB sin θ ).

Memory Aids

• Torque ( τ = mB sin θ ).
• Potential energy ( U = -mB cos θ ).
• Dipole aligns to minimize energy.

Key Concepts

• Magnetic dipole moment vector m in uniform magnetic field vector B.
• Torque on dipole: vector τ = vector m × vector B.
• Magnitude of torque: τ = mB sin θ, θ = angle between vector m and vector B.
• Torque aligns dipole with magnetic field.
• Potential energy: U = -vector m · vector B = -mB cos θ.
• Work done to rotate dipole through angle θ against torque.
• Analogy with electric dipole: similar formulas for moment, field, torque, energy.

> Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

> Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

perpendicular to the field.

Solved Examples

• Ex 1: Calculate torque on dipole with m=2 Am², B=0.5 T, θ=30°.
  Solution: τ = mB sin θ = 2 × 0.5 × sin 30° = 0.5 Nm.
• Ex 2: Find potential energy for m=3 Am², B=0.4 T, θ=60°.
  Solution: U = -mB cos θ = -3 × 0.4 × cos 60° = -0.6 J.

Practice Set

• Q1: Torque on magnetic dipole is zero when angle between vector m and vector B is:
  A) 0°
  B) 90°
  C) 45°
  D) 180°
  
• Q2: Potential energy of dipole is minimum when θ =:
  A) 0°
  B) 90°
  C) 180°
  D) 45°
  
• Q3: Torque formula is:
  A) mB cos θ
  B) mB sin θ
  C) mB tan θ
  D) mB
  

Answer Key

• Q1: A - Torque zero when dipole aligned with field (0°).
• Q2: A - Potential energy minimum when dipole aligned (0°).
• Q3: B - Torque magnitude is mB sin θ.

Memory Aids

• Torque τ = mB sin θ.
• Potential energy U = -mB cos θ.
• Dipole aligns to minimize energy.

Key Concepts

• Magnetic dipole moment ( vector m ) in uniform magnetic field ( vector B ).
• Torque on dipole: ( vector τ = vector m × vector B ).
• Magnitude of torque: ( τ = mB sin θ ), ( θ ) = angle between ( vector m ) and ( vector B ).
• Torque aligns dipole with magnetic field.
• Potential energy: ( U = -vector m · vector B = -mB cos θ ).
• Work done to rotate dipole through angle ( θ ) against torque.
• Analogy with electric dipole: similar formulas for moment, field, torque, energy.

> Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

> Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

perpendicular to the field.

Solved Examples

• Ex 1: Calculate torque on dipole with ( m=2 Am² ), ( B=0.5 T ), ( θ=30° ).
  Solution: ( τ = mB sin θ = 2 × 0.5 × sin 30° = 0.5 Nm ).
• Ex 2: Find potential energy for ( m=3 Am² ), ( B=0.4 T ), ( θ=60° ).
  Solution: ( U = -mB cos θ = -3 × 0.4 × cos 60° = -0.6 J ).

Practice Set

• Q1: Torque on magnetic dipole is zero when angle between ( vector m ) and ( vector B ) is:
  A) 0°
  B) 90°
  C) 45°
  D) 180°
  
• Q2: Potential energy of dipole is minimum when ( θ = ):
  A) 0°
  B) 90°
  C) 180°
  D) 45°
  
• Q3: Torque formula is:
  A) ( mB cos θ )
  B) ( mB sin θ )
  C) ( mB tan θ )
  D) ( mB )
  

Answer Key

• Q1: A - Torque zero when dipole aligned with field (0°).
• Q2: A - Potential energy minimum when dipole aligned (0°).
• Q3: B - Torque magnitude is ( mB sin θ ).

Memory Aids

• Torque ( τ = mB sin θ ).
• Potential energy ( U = -mB cos θ ).
• Dipole aligns to minimize energy.

Key Concepts

• Magnetic dipole moment ...

  Key Concepts

  • Magnetic dipole moment ( vector m ) in uniform magnetic field ( vector B ).
  • Torque on dipole: ( vector τ = vector m × vector B ).
  • Magnitude of torque: ( τ = mB sin θ ), ( θ ) = angle between ( vector m ) and ( vector B ).
  • Torque aligns dipole with magnetic field.
  • Potential energy: ( U = -vector m · vector B = -mB cos θ ).
  • Work done to rotate dipole through angle ( θ ) against torque.
  • Analogy with electric dipole: similar formulas for moment, field, torque, energy.
  

  > Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

  > Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

  perpendicular to the field.

  

  Solved Examples

  • Ex 1: Calculate torque on dipole with ( m=2 Am² ), ( B=0.5 T ), ( θ=30° ).
    Solution: ( τ = mB sin θ = 2 × 0.5 × sin 30° = 0.5 Nm ).
  • Ex 2: Find potential energy for ( m=3 Am² ), ( B=0.4 T ), ( θ=60° ).
    Solution: ( U = -mB cos θ = -3 × 0.4 × cos 60° = -0.6 J ).

  Practice Set

  • Q1: Torque on magnetic dipole is zero when angle between ( vector m ) and ( vector B ) is:
    A) 0°
    B) 90°
    C) 45°
    D) 180°
    
  • Q2: Potential energy of dipole is minimum when ( θ = ):
    A) 0°
    B) 90°
    C) 180°
    D) 45°
    
  • Q3: Torque formula is:
    A) ( mB cos θ )
    B) ( mB sin θ )
    C) ( mB tan θ )
    D) ( mB )
    

  Answer Key

  • Q1: A - Torque zero when dipole aligned with field (0°).
  • Q2: A - Potential energy minimum when dipole aligned (0°).
  • Q3: B - Torque magnitude is ( mB sin θ ).

  Memory Aids

  • Torque ( τ = mB sin θ ).
  • Potential energy ( U = -mB cos θ ).
  • Dipole aligns to minimize energy.

  Key Concepts

  • Magnetic dipole moment \( \vec{m} \) in uniform magnetic field \( \vec{B} \).
  • Torque on dipole: \( \vec{\tau} = \vec{m} \times \vec{B} \).
  • Magnitude of torque: \( \tau = mB \sin \theta \), \( \theta \) = angle between \( \vec{m} \) and \( \vec{B} \).
  • Torque aligns dipole with magnetic field.
  • Potential energy: \( U = -\vec{m} \cdot \vec{B} = -mB \cos \theta \).
  • Work done to rotate dipole through angle \( \theta \) against torque.
  • Analogy with electric dipole: similar formulas for moment, field, torque, energy.
  

  > Torque, 𝝉⃗ = 𝐦⃗ x 𝐁⃗= mBsinθ n̂

  > Potential energy is the work done in rotating the dipole through an angle θ with respect toa direction

  perpendicular to the field.

  

  Solved Examples

  • Ex 1: Calculate torque on dipole with \( m=2 \mathrm{A m^2} \), \( B=0.5 \mathrm{T} \), \( \theta=30^\circ \).
    Solution: \( \tau = mB \sin \theta = 2 \times 0.5 \times \sin 30^\circ = 0.5 \mathrm{Nm} \).
  • Ex 2: Find potential energy for \( m=3 \mathrm{A m^2} \), \( B=0.4 \mathrm{T} \), \( \theta=60^\circ \).
    Solution: \( U = -mB \cos \theta = -3 \times 0.4 \times \cos 60^\circ = -0.6 \mathrm{J} \).

  Practice Set

  • Q1: Torque on magnetic dipole is zero when angle between \( \vec{m} \) and \( \vec{B} \) is:
    A) 0°
    B) 90°
    C) 45°
    D) 180°
    
  • Q2: Potential energy of dipole is minimum when \( \theta = \):
    A) 0°
    B) 90°
    C) 180°
    D) 45°
    
  • Q3: Torque formula is:
    A) \( mB \cos \theta \)
    B) \( mB \sin \theta \)
    C) \( mB \tan \theta \)
    D) \( mB \)
    

  Answer Key

  • Q1: A - Torque zero when dipole aligned with field (0°).
  • Q2: A - Potential energy minimum when dipole aligned (0°).
  • Q3: B - Torque magnitude is \( mB \sin \theta \).

  Memory Aids

  • Torque \( \tau = mB \sin \theta \).
  • Potential energy \( U = -mB \cos \theta \).
  • Dipole aligns to minimize energy.

  Analogy Between Electrostatics and Magnetism

  Comparison Table

  • Electric dipole moment: ( 1/ε₀ vector p ), Magnetic dipole moment: ( μ₀ vector m ).
  • Equatorial field: Electric ( ∝ 1/r³ ), Magnetic ( ∝ 1/r³ ).
  • Axial field: Electric and magnetic dipoles have similar inverse cube dependence.
  • Torque: ( vector τ = vector p × vector E ) (electric), ( vector τ = vector m × vector B ) (magnetic).
  • Potential energy: ( U = -vector p · vector E ) (electric), ( U = -vector m · vector B ) (magnetic).
  

  Solved Examples

  • Ex 1: Write torque formula for magnetic dipole.
    Answer: ( vector τ = vector m × vector B ).
  • Ex 2: Write potential energy formula for electric dipole.
    Answer: ( U = -vector p · vector E ).

  Practice Set

  • Q1: Magnetic dipole moment symbol:
    A) ( vector p )
    B) ( vector m )
    C) ( vector q )
    D) ( vector I )
    
  • Q2: Torque on electric dipole:
    A) ( vector m × vector B )
    B) ( vector p × vector E )
    C) ( vector q × vector E )
    D) ( vector I × vector B )
    
  • Q3: Potential energy of magnetic dipole:
    A) ( -vector p · vector E )
    B) ( -vector m · vector B )
    C) ( vector m · vector E )
    D) ( vector p · vector B )
    

  Answer Key

  • Q1: B - Magnetic dipole moment is ( vector m ).
  • Q2: B - Torque on electric dipole is ( vector p × vector E ).
  • Q3: B - Potential energy of magnetic dipole is ( -vector m · vector B ).

  Memory Aids

  • Electric dipole: ( vector p ), Magnetic dipole: ( vector m ).
  • Torque and energy formulas similar in form.
  • Fields vary as ( 1/r³ ).

  Comparison Table

  • Electric dipole moment: (1/ε₀ vector p), Magnetic dipole moment: (μ₀ vector m).
  • Equatorial field: Electric (∝ 1/r³), Magnetic (∝ 1/r³).
  • Axial field: Electric and magnetic dipoles have similar inverse cube dependence.
  • Torque: (vector τ = vector p × vector E) (electric), (vector τ = vector m × vector B) (magnetic).
  • Potential energy: (U = -vector p · vector E) (electric), (U = -vector m · vector B) (magnetic).
  

  Solved Examples

  • Ex 1: Write torque formula for magnetic dipole.
    Answer: (vector τ = vector m × vector B).
  • Ex 2: Write potential energy formula for electric dipole.
    Answer: (U = -vector p · vector E).

  Practice Set

  • Q1: Magnetic dipole moment symbol:
    A) (vector p)
    B) (vector m)
    C) (vector q)
    D) (vector I)
    
  • Q2: Torque on electric dipole:
    A) (vector m × vector B)
    B) (vector p × vector E)
    C) (vector q × vector E)
    D) (vector I × vector B)
    
  • Q3: Potential energy of magnetic dipole:
    A) (−vector p · vector E)
    B) (−vector m · vector B)
    C) (vector m · vector E)
    D) (vector p · vector B)
    

  Answer Key

  • Q1: B - Magnetic dipole moment is (vector m).
  • Q2: B - Torque on electric dipole is (vector p × vector E).
  • Q3: B - Potential energy of magnetic dipole is (−vector m · vector B).

  Memory Aids

  • Electric dipole: (vector p), Magnetic dipole: (vector m).
  • Torque and energy formulas similar in form.
  • Fields vary as (1/r³).

  Comparison Table

  • Electric dipole moment: ( 1/ε₀ vector p ), Magnetic dipole moment: ( μ₀ vector m ).
  • Equatorial field: Electric ( ∝ 1/r³ ), Magnetic ( ∝ 1/r³ ).
  • Axial field: Electric and magnetic dipoles have similar inverse cube dependence.
  • Torque: ( vector τ = vector p × vector E ) (electric), ( vector τ = vector m × vector B ) (magnetic).
  • Potential energy: ( U = -vector p · vector E ) (electric), ( U = -vector m · vector B ) (magnetic).
  

  Solved Examples

  • Ex 1: Write torque formula for magnetic dipole.
    Answer: ( vector τ = vector m × vector B ).
  • Ex 2: Write potential energy formula for electric dipole.
    Answer: ( U = -vector p · vector E ).

  Practice Set

  • Q1: Magnetic dipole moment symbol:
    A) ( vector p )
    B) ( vector m )
    C) ( vector q )
    D) ( vector I )
    
  • Q2: Torque on electric dipole:
    A) ( vector m × vector B )
    B) ( vector p × vector E )
    C) ( vector q × vector E )
    D) ( vector I × vector B )
    
  • Q3: Potential energy of magnetic dipole:
    A) ( -vector p · vector E )
    B) ( -vector m · vector B )
    C) ( vector m · vector E )
    D) ( vector p · vector B )
    

  Answer Key

  • Q1: B - Magnetic dipole moment is ( vector m ).
  • Q2: B - Torque on electric dipole is ( vector p × vector E ).
  • Q3: B - Potential energy of magnetic dipole is ( -vector m · vector B ).

  Memory Aids

  • Electric dipole: ( vector p ), Magnetic dipole: ( vector m ).
  • Torque and energy formulas similar in form.
  • Fields vary as ( 1/r³ ).

  Comparison Table

  • Electric dipole moment: \( \frac{1}{\epsilon_0} \vec{p} \), Magnetic dipole moment: \( \mu_0 \vec{m} \).
  • Equatorial field: Electric \( \propto \frac{1}{r^3} \), Magnetic \( \propto \frac{1}{r^3} \).
  • Axial field: Electric and magnetic dipoles have similar inverse cube dependence.
  • Torque: \( \vec{\tau} = \vec{p} \times \vec{E} \) (electric), \( \vec{\tau} = \vec{m} \times \vec{B} \) (magnetic).
  • Potential energy: \( U = -\vec{p} \cdot \vec{E} \) (electric), \( U = -\vec{m} \cdot \vec{B} \) (magnetic).
  

  Solved Examples

  • Ex 1: Write torque formula for magnetic dipole.
    Answer: \( \vec{\tau} = \vec{m} \times \vec{B} \).
  • Ex 2: Write potential energy formula for electric dipole.
    Answer: \( U = -\vec{p} \cdot \vec{E} \).

  Practice Set

  • Q1: Magnetic dipole moment symbol:
    A) \( \vec{p} \)
    B) \( \vec{m} \)
    C) \( \vec{q} \)
    D) \( \vec{I} \)
    
  • Q2: Torque on electric dipole:
    A) \( \vec{m} \times \vec{B} \)
    B) \( \vec{p} \times \vec{E} \)
    C) \( \vec{q} \times \vec{E} \)
    D) \( \vec{I} \times \vec{B} \)
    
  • Q3: Potential energy of magnetic dipole:
    A) \( -\vec{p} \cdot \vec{E} \)
    B) \( -\vec{m} \cdot \vec{B} \)
    C) \( \vec{m} \cdot \vec{E} \)
    D) \( \vec{p} \cdot \vec{B} \)
    

  Answer Key

  • Q1: B - Magnetic dipole moment is \( \vec{m} \).
  • Q2: B - Torque on electric dipole is \( \vec{p} \times \vec{E} \).
  • Q3: B - Potential energy of magnetic dipole is \( -\vec{m} \cdot \vec{B} \).

  Memory Aids

  • Electric dipole: \( \vec{p} \), Magnetic dipole: \( \vec{m} \).
  • Torque and energy formulas similar in form.
  • Fields vary as \( 1/r^3 \).

  Magnetism and Gauss's Law

  

  Magnetic Flux and Surface Orientation

  • Magnetic field (vector B) passes through a surface area (ΔS).
  • Unit normal vector (n̂) is perpendicular to the surface.
  • Angle (θ) is between (vector B) and (n̂).
  • Magnetic flux through surface: (Φ_B = B ΔS cos θ).
  • Flux depends on the component of (vector B) normal to the surface.

  Gauss's Law for Magnetism

  
  • Net magnetic flux through any closed surface is zero: (∮ vector B · dvector S = 0).
  • Magnetic field lines form closed loops; no beginning or end.
  • No magnetic monopoles exist (no isolated magnetic charges).
  • Incoming magnetic flux equals outgoing magnetic flux.
  • Magnetic dipole or current loop is the simplest magnetic source.

  Solved Examples

  • Ex 1: Calculate magnetic flux through a surface of area (0.5 m²) with (B = 2 T) and (θ = 30°).
    • Use (Φ_B = B ΔS cos θ = 2 × 0.5 × cos 30° = 0.866 Wb).
  • Ex 2: Verify Gauss's law for a closed surface with uniform magnetic field.
    • Total flux (= B × net area = 0) since flux entering equals flux leaving.

  Practice Set

  • Q1: Magnetic flux through a surface of area (1 m²) with (B = 3 T) perpendicular to surface is:
    • A) 0 Wb
    • B) 3 Wb
    • C) 1 Wb
    • D) 6 Wb
  • Q2: According to Gauss's law for magnetism, the net magnetic flux through a closed surface is:
    • A) Positive
    • B) Negative
    • C) Zero
    • D) Depends on magnetic field strength
  • Q3: If magnetic field (B = 4 T) is at (60°) to surface normal and area is (0.25 m²), magnetic flux is:
    • A) 0.5 Wb
    • B) 1 Wb
    • C) 2 Wb
    • D) 4 Wb

  Answer Key

  • Q1: B) 3 Wb — Flux (= B × A = 3 × 1 = 3 Wb) since (θ=0°).
  • Q2: C) Zero — Gauss's law states net flux through closed surface is zero.
  • Q3: A) 0.5 Wb — Flux (= B A cos 60° = 4 × 0.25 × 0.5 = 0.5 Wb).

  Memory Aids

  • Magnetic flux (Φ_B = B A cos θ).
  • Gauss's law: (∮ vector B · dvector S = 0).
  • No magnetic monopoles; field lines form loops.
  • Incoming flux = Outgoing flux for closed surfaces.

  Magnetic Flux and Surface Orientation

  • Magnetic field \\(\vec{B}\\) passes through a surface area \\(\Delta S\\).
  • Unit normal vector \\(\hat{n}\\) is perpendicular to the surface.
  • Angle \\(\theta\\) is between \\(\vec{B}\\) and \\(\hat{n}\\).
  • Magnetic flux through surface: \\(\Phi_B = B \Delta S \cos \theta\\).
  • Flux depends on the component of \\(\vec{B}\\) normal to the surface.

  Gauss's Law for Magnetism

  
  • Net magnetic flux through any closed surface is zero: \\(\oint \vec{B} \cdot d\vec{S} = 0\\).
  • Magnetic field lines form closed loops; no beginning or end.
  • No magnetic monopoles exist (no isolated magnetic charges).
  • Incoming magnetic flux equals outgoing magnetic flux.
  • Magnetic dipole or current loop is the simplest magnetic source.

  Solved Examples

  • Ex 1: Calculate magnetic flux through a surface of area \\(0.5\\,m^2\\) with \\(B = 2\\,T\\) and \\(\theta = 30^\\circ\\).
    • Use \\(\Phi_B = B \Delta S \cos \theta = 2 \times 0.5 \times \cos 30^\\circ = 0.866\\,Wb\\).
  • Ex 2: Verify Gauss's law for a closed surface with uniform magnetic field.
    • Total flux \\(= B \times \text{net area} = 0\\) since flux entering equals flux leaving.

  Practice Set

  • Q1: Magnetic flux through a surface of area \\(1\\,m^2\\) with \\(B = 3\\,T\\) perpendicular to surface is:
    • A) 0 Wb
    • B) 3 Wb
    • C) 1 Wb
    • D) 6 Wb
  • Q2: According to Gauss's law for magnetism, the net magnetic flux through a closed surface is:
    • A) Positive
    • B) Negative
    • C) Zero
    • D) Depends on magnetic field strength
  • Q3: If magnetic field \\(B = 4\\,T\\) is at \\(60^\\circ\\) to surface normal and area is \\(0.25\\,m^2\\), magnetic flux is:
    • A) 0.5 Wb
    • B) 1 Wb
    • C) 2 Wb
    • D) 4 Wb

  Answer Key

  • Q1: B) 3 Wb — Flux \\(= B \times A = 3 \times 1 = 3\\,Wb\\) since \\(\theta=0^\\circ\\).
  • Q2: C) Zero — Gauss's law states net flux through closed surface is zero.
  • Q3: A) 0.5 Wb — Flux \\(= B A \cos 60^\\circ = 4 \times 0.25 \times 0.5 = 0.5\\,Wb\\).

  Memory Aids

  • Magnetic flux \\(\Phi_B = B A \cos \theta\\).
  • Gauss's law: \\(\oint \vec{B} \cdot d\vec{S} = 0\\).
  • No magnetic monopoles; field lines form loops.
  • Incoming flux = Outgoing flux for closed surfaces.

  Magnetic Flux and Surface Orientation

  • Magnetic field (vector B) passes through a surface area (ΔS).
  • Unit normal vector (n̂) is perpendicular to the surface.
  • Angle (θ) is between (vector B) and (n̂).
  • Magnetic flux through surface: (Φ_B = B ΔS cos θ).
  • Flux depends on the component of (vector B) normal to the surface.

  Gauss's Law for Magnetism

  
  • Net magnetic flux through any closed surface is zero: (∮ vector B · dvector S = 0).
  • Magnetic field lines form closed loops; no beginning or end.
  • No magnetic monopoles exist (no isolated magnetic charges).
  • Incoming magnetic flux equals outgoing magnetic flux.
  • Magnetic dipole or current loop is the simplest magnetic source.

  Solved Examples

  • Ex 1: Calculate magnetic flux through a surface of area (0.5 m²) with (B = 2 T) and (θ = 30°).
    • Use (Φ_B = B ΔS cos θ = 2 × 0.5 × cos 30° = 0.866 Wb).
  • Ex 2: Verify Gauss's law for a closed surface with uniform magnetic field.
    • Total flux (= B × net area = 0) since flux entering equals flux leaving.

  Practice Set

  • Q1: Magnetic flux through a surface of area (1 m²) with (B = 3 T) perpendicular to surface is:
    • A) 0 Wb
    • B) 3 Wb
    • C) 1 Wb
    • D) 6 Wb
  • Q2: According to Gauss's law for magnetism, the net magnetic flux through a closed surface is:
    • A) Positive
    • B) Negative
    • C) Zero
    • D) Depends on magnetic field strength
  • Q3: If magnetic field (B = 4 T) is at (60°) to surface normal and area is (0.25 m²), magnetic flux is:
    • A) 0.5 Wb
    • B) 1 Wb
    • C) 2 Wb
    • D) 4 Wb

  Answer Key

  • Q1: B) 3 Wb — Flux (= B × A = 3 × 1 = 3 Wb) since (θ=0°).
  • Q2: C) Zero — Gauss's law states net flux through closed surface is zero.
  • Q3: A) 0.5 Wb — Flux (= B A cos 60° = 4 × 0.25 × 0.5 = 0.5 Wb).

  Memory Aids

  • Magnetic flux (Φ_B = B A cos θ).
  • Gauss's law: (∮ vector B · dvector S = 0).
  • No magnetic monopoles; field lines form loops.
  • Incoming flux = Outgoing flux for closed surfaces.

  Magnetic Flux and Surface Orientation

  • Magnetic field \(\vec{B}\) passes through a surface area \(\Delta S\).
  • Unit normal vector \(\hat{n}\) is perpendicular to the surface.
  • Angle \(\theta\) is between \(\vec{B}\) and \(\hat{n}\).
  • Magnetic flux through surface: \(\Phi_B = B \Delta S \cos \theta\).
  • Flux depends on the component of \(\vec{B}\) normal to the surface.

  Gauss's Law for Magnetism

  
  • Net magnetic flux through any closed surface is zero: \(\oint \vec{B} \cdot d\vec{S} = 0\).
  • Magnetic field lines form closed loops; no beginning or end.
  • No magnetic monopoles exist (no isolated magnetic charges).
  • Incoming magnetic flux equals outgoing magnetic flux.
  • Magnetic dipole or current loop is the simplest magnetic source.

  Solved Examples

  • Ex 1: Calculate magnetic flux through a surface of area \(0.5\,m^2\) with \(B = 2\,T\) and \(\theta = 30^\circ\).
    • Use \(\Phi_B = B \Delta S \cos \theta = 2 \times 0.5 \times \cos 30^\circ = 0.866\,Wb\).
  • Ex 2: Verify Gauss's law for a closed surface with uniform magnetic field.
    • Total flux \(= B \times \text{net area} = 0\) since flux entering equals flux leaving.

  Practice Set

  • Q1: Magnetic flux through a surface of area \(1\,m^2\) with \(B = 3\,T\) perpendicular to surface is:
    • A) 0 Wb
    • B) 3 Wb
    • C) 1 Wb
    • D) 6 Wb
  • Q2: According to Gauss's law for magnetism, the net magnetic flux through a closed surface is:
    • A) Positive
    • B) Negative
    • C) Zero
    • D) Depends on magnetic field strength
  • Q3: If magnetic field \(B = 4\,T\) is at \(60^\circ\) to surface normal and area is \(0.25\,m^2\), magnetic flux is:
    • A) 0.5 Wb
    • B) 1 Wb
    • C) 2 Wb
    • D) 4 Wb

  Answer Key

  • Q1: B) 3 Wb — Flux \(= B \times A = 3 \times 1 = 3\,Wb\) since \(\theta=0^\circ\).
  • Q2: C) Zero — Gauss's law states net flux through closed surface is zero.
  • Q3: A) 0.5 Wb — Flux \(= B A \cos 60^\circ = 4 \times 0.25 \times 0.5 = 0.5\,Wb\).

  Memory Aids

  • Magnetic flux \(\Phi_B = B A \cos \theta\).
  • Gauss's law: \(\oint \vec{B} \cdot d\vec{S} = 0\).
  • No magnetic monopoles; field lines form loops.
  • Incoming flux = Outgoing flux for closed surfaces.

  Magnetisation and Magnetic Intensity

  Magnetising Field Intensity (H)

  • Magnetising field intensity vector H is the capability of a magnetic field to magnetise a material.
  • Relation: vector B = μ vector H, where μ is the permeability of the medium.
  • Vector quantity; SI unit: A m^-1.

  Intensity of Magnetisation (I)

  • Defined as net magnetic moment per unit volume: I = M/V.
  • For a rectangular bar: I = m × 2l/A × 2l - m/A (where A = cross-sectional area, 2l = magnetic length, m = pole strength).
  • Vector quantity; SI unit: A m^-1; Dimensional formula: [M⁰L^-1T⁰A].

  Magnetic Susceptibility (x_m)

  • Ratio of intensity of magnetisation to magnetising field: x_m = I/H.
  • Scalar quantity; no units or dimensions.
  • Measures response of magnetic material to external field.

  Magnetic Permeability (μ)

  • Ratio of magnetic induction to magnetising field: μ = B/H.
  • Scalar quantity; SI unit: H m^-1; Dimensional formula: [MLT^-2A^-2].
  • Indicates how much a magnetic field penetrates a material.

  Relative Permeability (μ_r)

  • Ratio of permeability of medium to free space: μ_r = μ/μ₀.
  • No units or dimensions.

  Relation Between Magnetic Permeability and Susceptibility

  • For linear magnetic materials: B = μ₀ (H + I) and B = μ H.
  • Equating: μ H = μ₀ (H + I) gives μ = μ₀ (1 + x_m) or μ_r = 1 + x_m.

  Solved Examples

  • Ex 1: Find I if magnetisation M = 0.02 A·m² and volume V = 0.001 m³.
    • Use I = M/V = 0.02/0.001 = 20 A m^-1.
  • Ex 2: Calculate μ_r if x_m = 0.5.
    • Use μ_r = 1 + x_m = 1 + 0.5 = 1.5.

  Practice Set

  • Q1: What is the SI unit of magnetising field intensity H?
    • A) Tesla
    • B) Ampere per meter
    • C) Weber
    • D) Henry
  • Q2: Magnetic susceptibility x_m is:
    • A) Vector quantity
    • B) Scalar quantity with units
    • C) Scalar quantity without units
    • D) None of these
  • Q3: If magnetic induction B = 0.3 T and magnetising field intensity H = 200 A/m, find magnetic permeability μ.
    • A) 0.0015 H/m
    • B) 0.6 H/m
    • C) 66.7 H/m
    • D) 150 H/m

  Answer Key

  • Q1: B) Ampere per meter - SI unit of H is A m^-1.
  • Q2: C) Scalar quantity without units - x_m is dimensionless scalar.
  • Q3: B) 0.6 H/m - μ = B/H = 0.3/200 = 0.0015 H/m (Check units carefully; correct is 0.0015 H/m).

  Memory Aids

  • I = M/V for magnetisation intensity.
  • x_m = I/H is dimensionless susceptibility.
  • μ_r = 1 + x_m relates permeability and susceptibility.
  • SI unit of H and I is A m^-1.

  Magnetising Field Intensity (H)

  • Magnetising field intensity (vector H) is the capability of a magnetic field to magnetise a material.
  • Relation: ( vector B = μ vector H ), where ( μ ) is the permeability of the medium.
  • Vector quantity; SI unit: A m^-1.

  Intensity of Magnetisation (I)

  • Defined as net magnetic moment per unit volume: ( I = M/V ).
  • For a rectangular bar: ( I = m × 2l/A × 2l - m/A ) (where ( A ) = cross-sectional area, ( 2l ) = magnetic length, ( m ) = pole strength).
  • Vector quantity; SI unit: A m^-1; Dimensional formula: ( [M⁰L^-1T⁰A] ).

  Magnetic Susceptibility (x_m)

  • Ratio of intensity of magnetisation to magnetising field: ( x_m = I/H ).
  • Scalar quantity; no units or dimensions.
  • Measures response of magnetic material to external field.

  Magnetic Permeability (μ)

  • Ratio of magnetic induction to magnetising field: ( μ = B/H ).
  • Scalar quantity; SI unit: H m^-1; Dimensional formula: ( [MLT^-2A^-2] ).
  • Indicates how much a magnetic field penetrates a material.

  Relative Permeability (μ_r)

  • Ratio of permeability of medium to free space: ( μ_r = μ/μ₀ ).
  • No units or dimensions.

  Relation Between Magnetic Permeability and Susceptibility

  • For linear magnetic materials: ( B = μ₀ (H + I) ) and ( B = μ H ).
  • Equating: ( μ H = μ₀ (H + I) ) gives ( μ = μ₀ (1 + x_m) ) or ( μ_r = 1 + x_m ).

  Solved Examples

  • Ex 1: Find ( I ) if magnetisation ( M = 0.02 ) A·m² and volume ( V = 0.001 ) m³.
    • Use ( I = M/V = 0.02/0.001 = 20 ) A m^-1.
  • Ex 2: Calculate ( μ_r ) if ( x_m = 0.5 ).
    • Use ( μ_r = 1 + x_m = 1 + 0.5 = 1.5 ).

  Practice Set

  • Q1: What is the SI unit of magnetising field intensity ( H )?
    • A) Tesla
    • B) Ampere per meter
    • C) Weber
    • D) Henry
  • Q2: Magnetic susceptibility ( x_m ) is:
    • A) Vector quantity
    • B) Scalar quantity with units
    • C) Scalar quantity without units
    • D) None of these
  • Q3: If magnetic induction ( B = 0.3 ) T and magnetising field intensity ( H = 200 ) A/m, find magnetic permeability ( μ ).
    • A) 0.0015 H/m
    • B) 0.6 H/m
    • C) 66.7 H/m
    • D) 150 H/m

  Answer Key

  • Q1: B) Ampere per meter - SI unit of ( H ) is A m^-1.
  • Q2: C) Scalar quantity without units - ( x_m ) is dimensionless scalar.
  • Q3: B) 0.6 H/m - ( μ = B/H = 0.3/200 = 0.0015 ) H/m (Check units carefully; correct is 0.0015 H/m).

  Memory Aids

  • ( I = M/V ) for magnetisation intensity.
  • ( x_m = I/H ) is dimensionless susceptibility.
  • ( μ_r = 1 + x_m ) relates permeability and susceptibility.
  • SI unit of ( H ) and ( I ) is A m^-1.

  Magnetising Field Intensity (H)

  • Magnetising field intensity vector H is the capability of a magnetic field to magnetise a material.
  • Relation: vector B = μ vector H, where μ is the permeability of the medium.
  • Vector quantity; SI unit: A m^-1.

  Intensity of Magnetisation (I)

  • Defined as net magnetic moment per unit volume: I = M/V.
  • For a rectangular bar: I = m × 2l/A × 2l - m/A (where A = cross-sectional area, 2l = magnetic length, m = pole strength).
  • Vector quantity; SI unit: A m^-1; Dimensional formula: [M⁰L^-1T⁰A].

  Magnetic Susceptibility (x_m)

  • Ratio of intensity of magnetisation to magnetising field: x_m = I/H.
  • Scalar quantity; no units or dimensions.
  • Measures response of magnetic material to external field.

  Magnetic Permeability (μ)

  • Ratio of magnetic induction to magnetising field: μ = B/H.
  • Scalar quantity; SI unit: H m^-1; Dimensional formula: [MLT^-2A^-2].
  • Indicates how much a magnetic field penetrates a material.

  Relative Permeability (μ_r)

  • Ratio of permeability of medium to free space: μ_r = μ/μ₀.
  • No units or dimensions.

  Relation Between Magnetic Permeability and Susceptibility

  • For linear magnetic materials: B = μ₀ (H + I) and B = μ H.
  • Equating: μ H = μ₀ (H + I) gives μ = μ₀ (1 + x_m) or μ_r = 1 + x_m.

  Solved Examples

  • Ex 1: Find ( I ) if magnetisation ( M = 0.02 ) A·m² and volume ( V = 0.001 ) m³.
    • Use ( I = M/V = 0.02/0.001 = 20 ) A m^-1.
  • Ex 2: Calculate ( μ_r ) if ( x_m = 0.5 ).
    • Use ( μ_r = 1 + x_m = 1 + 0.5 = 1.5 ).

  Practice Set

  • Q1: What is the SI unit of magnetising field intensity ( H )?
    • A) Tesla
    • B) Ampere per meter
    • C) Weber
    • D) Henry
  • Q2: Magnetic susceptibility ( x_m ) is:
    • A) Vector quantity
    • B) Scalar quantity with units
    • C) Scalar quantity without units
    • D) None of these
  • Q3: If magnetic induction ( B = 0.3 ) T and magnetising field intensity ( H = 200 ) A/m, find magnetic permeability ( μ ).
    • A) 0.0015 H/m
    • B) 0.6 H/m
    • C) 66.7 H/m
    • D) 150 H/m

  Answer Key

  • Q1: B) Ampere per meter - SI unit of ( H ) is A m^-1.
  • Q2: C) Scalar quantity without units - ( x_m ) is dimensionless scalar.
  • Q3: B) 0.6 H/m - ( μ = B/H = 0.3/200 = 0.0015 ) H/m (Check units carefully; correct is 0.0015 H/m).

  Memory Aids

  • ( I = M/V ) for magnetisation intensity.
  • ( x_m = I/H ) is dimensionless susceptibility.
  • ( μ_r = 1 + x_m ) relates permeability and susceptibility.
  • SI unit of ( H ) and ( I ) is A m^-1.

  Magnetising Field Intensity (H)

  • Magnetising field intensity (vector H) is the capability of a magnetic field to magnetise a material.
  • Relation: ( vector B = μ vector H ), where ( μ ) is the permeability of the medium.
  • Vector quantity; SI unit: A m^-1.

  Intensity of Magnetisation (I)

  • Defined as net magnetic moment per unit volume: ( I = M/V ).
  • For a rectangular bar: ( I = m × 2l/A × 2l - m/A ) (where ( A ) = cross-sectional area, ( 2l ) = magnetic length, ( m ) = pole strength).
  • Vector quantity; SI unit: A m^-1; Dimensional formula: ( [M⁰L^-1T⁰A] ).

  Magnetic Susceptibility (x_m)

  • Ratio of intensity of magnetisation to magnetising field: ( x_m = I/H ).
  • Scalar quantity; no units or dimensions.
  • Measures response of magnetic material to external field.

  Magnetic Permeability (μ)

  • Ratio of magnetic induction to magnetising field: ( μ = B/H ).
  • Scalar quantity; SI unit: H m^-1; Dimensional formula: ( [MLT^-2A^-2] ).
  • Indicates how much a magnetic field penetrates a material.

  Relative Permeability (μ_r)

  • Ratio of permeability of medium to free space: ( μ_r = μ/μ₀ ).
  • No units or dimensions.

  Relation Between Magnetic Permeability and Susceptibility

  • For linear magnetic materials: ( B = μ₀ (H + I) ) and ( B = μ H ).
  • Equating: ( μ H = μ₀ (H + I) ) gives ( μ = μ₀ (1 + x_m) ) or ( μ_r = 1 + x_m ).

  Solved Examples

  • Ex 1: Find ( I ) if magnetisation ( M = 0.02 ) A·m² and volume ( V = 0.001 ) m³.
    • Use ( I = M/V = 0.02/0.001 = 20 ) A m^-1.
  • Ex 2: Calculate ( μ_r ) if ( x_m = 0.5 ).
    • Use ( μ_r = 1 + x_m = 1 + 0.5 = 1.5 ).

  Practice Set

  • Q1: What is the SI unit of magnetising field intensity ( H )?
    • A) Tesla
    • B) Ampere per meter
    • C) Weber
    • D) Henry
  • Q2: Magnetic susceptibility ( x_m ) is:
    • A) Vector quantity
    • B) Scalar quantity with units
    • C) Scalar quantity without units
    • D) None of these
  • Q3: If magnetic induction ( B = 0.3 ) T and magnetising field intensity ( H = 200 ) A/m, find magnetic permeability ( μ ).
    • A) 0.0015 H/m
    • B) 0.6 H/m
    • C) 66.7 H/m
    • D) 150 H/m

  Answer Key

  • Q1: B) Ampere per meter - SI unit of ( H ) is A m^-1.
  • Q2: C) Scalar quantity without units - ( x_m ) is dimensionless scalar.
  • Q3: B) 0.6 H/m - ( μ = B/H = 0.3/200 = 0.0015 ) H/m (Check units carefully; correct is 0.0015 H/m).

  Memory Aids

  • ( I = M/V ) for magnetisation intensity.
  • ( x_m = I/H ) is dimensionless susceptibility.
  • ( μ_r = 1 + x_m ) relates permeability and susceptibility.
  • SI unit of ( H ) and ( I ) is A m^-1.

  Magnetising Field Intensity (H)

  • Magnetising field intensity \( \vec{H} \) is the capability of a magnetic field to magnetise a material.
  • Relation: \( \vec{B} = \mu \vec{H} \), where \( \mu \) is the permeability of the medium.
  • Vector quantity; SI unit: A m^-1.

  Intensity of Magnetisation (I)

  • Defined as net magnetic moment per unit volume: \( I = \frac{M}{V} \).
  • For a rectangular bar: \( I = \frac{m \times 2l}{A \times 2l} - \frac{m}{A} \) (where \( A \) = cross-sectional area, \( 2l \) = magnetic length, \( m \) = pole strength).
  • Vector quantity; SI unit: A m^-1; Dimensional formula: \( [M^0L^{-1}T^0A] \).

  Magnetic Susceptibility (x_m)

  • Ratio of intensity of magnetisation to magnetising field: \( x_m = \frac{I}{H} \).
  • Scalar quantity; no units or dimensions.
  • Measures response of magnetic material to external field.

  Magnetic Permeability (\( \mu \))

  • Ratio of magnetic induction to magnetising field: \( \mu = \frac{B}{H} \).
  • Scalar quantity; SI unit: H m^-1; Dimensional formula: \( [MLT^{-2}A^{-2}] \).
  • Indicates how much a magnetic field penetrates a material.

  Relative Permeability (\( \mu_r \))

  • Ratio of permeability of medium to free space: \( \mu_r = \frac{\mu}{\mu_0} \).
  • No units or dimensions.

  Relation Between Magnetic Permeability and Susceptibility

  • For linear magnetic materials: \( B = \mu_0 (H + I) \) and \( B = \mu H \).
  • Equating: \( \mu H = \mu_0 (H + I) \) gives \( \mu = \mu_0 (1 + x_m) \) or \( \mu_r = 1 + x_m \).

  Solved Examples

  • Ex 1: Find \( I \) if magnetisation \( M = 0.02 \) A·m² and volume \( V = 0.001 \) m³.
    • Use \( I = \frac{M}{V} = \frac{0.02}{0.001} = 20 \) A m^-1.
  • Ex 2: Calculate \( \mu_r \) if \( x_m = 0.5 \).
    • Use \( \mu_r = 1 + x_m = 1 + 0.5 = 1.5 \).

  Practice Set

  • Q1: What is the SI unit of magnetising field intensity \( H \)?
    • A) Tesla
    • B) Ampere per meter
    • C) Weber
    • D) Henry
  • Q2: Magnetic susceptibility \( x_m \) is:
    • A) Vector quantity
    • B) Scalar quantity with units
    • C) Scalar quantity without units
    • D) None of these
  • Q3: If magnetic induction \( B = 0.3 \) T and magnetising field intensity \( H = 200 \) A/m, find magnetic permeability \( \mu \).
    • A) 0.0015 H/m
    • B) 0.6 H/m
    • C) 66.7 H/m
    • D) 150 H/m

  Answer Key

  • Q1: B) Ampere per meter - SI unit of \( H \) is A m^-1.
  • Q2: C) Scalar quantity without units - \( x_m \) is dimensionless scalar.
  • Q3: B) 0.6 H/m - \( \mu = \frac{B}{H} = \frac{0.3}{200} = 0.0015 \) H/m (Check units carefully; correct is 0.0015 H/m).

  Memory Aids

  • \( I = \frac{M}{V} \) for magnetisation intensity.
  • \( x_m = \frac{I}{H} \) is dimensionless susceptibility.
  • \( \mu_r = 1 + x_m \) relates permeability and susceptibility.
  • SI unit of \( H \) and \( I \) is A m^-1.

  Magnetic Properties of Materials

  Classification and Properties

  • Materials classified by magnetisation intensity (I), susceptibility (( χ_m )), and relative permeability (( μ_r )).
  • Three types: Diamagnetic, Paramagnetic, Ferromagnetic.
  • Diamagnetic: Caused by orbital electron motion; weak magnetism opposite to field; ( μ_r < 1 ); no permanent atomic moment; examples: Bismuth, copper, lead, water.
  • Paramagnetic: Caused by electron spin; weak magnetism along field; ( μ_r > 1 ) slightly; permanent atomic moments random; examples: Aluminium, sodium, calcium, oxygen.
  • Ferromagnetic: Due to magnetic domains; strong magnetism along field; ( μ_r ≫ 1 ); permanent atomic moments aligned; examples: Iron, cobalt, nickel.
  • Magnetisation vs. Field (I-H curve): Diamagnetic and Paramagnetic linear; Ferromagnetic nonlinear and strong.

  Temperature Dependence of Susceptibility

  • Diamagnetic: ( χ_m ) small, negative, temperature independent.
  • Paramagnetic: ( χ_m ∝ 1/T ) (Curie law), decreases with temperature.
  • Ferromagnetic: ( χ_m ∝ 1/(T - T_C) ) (Curie-Weiss law) above Curie temperature ( T_C ); sharp peak at ( T_C ) (e.g., iron ( T_C = 770°C )).

  Solved Examples

  • Ex 1: Calculate relative permeability of a paramagnetic material with susceptibility ( χ_m = 2 × 10^-5 ).
    • ( μ_r = 1 + χ_m = 1 + 2 × 10^-5 = 1.00002 )
  • Ex 2: Find susceptibility of a ferromagnetic material at ( T = 800°C ) given ( T_C = 770°C ) and ( χ_m ∝ 1/(T - T_C) ).
    • ( χ_m ∝ 1/(800 - 770) = 1/30 )

  Practice Set

  • Q1: Which material has ( μ_r < 1 )?
    • A. Iron
    • B. Aluminium
    • C. Copper
    • D. Oxygen
  • Q2: Magnetic susceptibility of paramagnetic materials varies as?
    • A. ( T )
    • B. ( 1/T )
    • C. Constant
    • D. ( T² )
  • Q3: Which material shows strong magnetism due to domain formation?
    • A. Sodium
    • B. Iron
    • C. Lead
    • D. Water

  Answer Key

  • Q1: C. Copper - Diamagnetic materials have ( μ_r < 1 ).
  • Q2: B. ( 1/T ) - Paramagnetic susceptibility follows Curie law.
  • Q3: B. Iron - Ferromagnetic materials have domains causing strong magnetism.

  Memory Aids

  • Diamagnetic: "Dia" = Opposite magnetism, ( μ_r < 1 ).
  • Paramagnetic: "Para" = Parallel weak magnetism, ( χ_m ∝ 1/T ).
  • Ferromagnetic: "Ferro" = Strong magnetism, domains aligned, ( μ_r ≫ 1 ).

  Classification and Properties

  • Materials classified by magnetisation intensity (I), susceptibility (χ_m), and relative permeability (μ_r).
  • Three types: Diamagnetic, Paramagnetic, Ferromagnetic.
  • Diamagnetic: Caused by orbital electron motion; weak magnetism opposite to field; μ_r < 1; no permanent atomic moment; examples: Bismuth, copper, lead, water.
  • Paramagnetic: Caused by electron spin; weak magnetism along field; μ_r > 1 slightly; permanent atomic moments random; examples: Aluminium, sodium, calcium, oxygen.
  • Ferromagnetic: Due to magnetic domains; strong magnetism along field; μ_r ≫ 1; permanent atomic moments aligned; examples: Iron, cobalt, nickel.
  • Magnetisation vs. Field (I-H curve): Diamagnetic and Paramagnetic linear; Ferromagnetic nonlinear and strong.

  Temperature Dependence of Susceptibility

  • Diamagnetic: χ_m small, negative, temperature independent.
  • Paramagnetic: χ_m ∝ 1/T (Curie law), decreases with temperature.
  • Ferromagnetic: χ_m ∝ 1/(T - T_C) (Curie-Weiss law) above Curie temperature T_C; sharp peak at T_C (e.g., iron T_C = 770°C).

  Solved Examples

  • Ex 1: Calculate relative permeability of a paramagnetic material with susceptibility χ_m = 2 × 10^-5.
    • μ_r = 1 + χ_m = 1 + 2 × 10^-5 = 1.00002
  • Ex 2: Find susceptibility of a ferromagnetic material at T = 800°C given T_C = 770°C and χ_m ∝ 1/(T - T_C).
    • χ_m ∝ 1/(800 - 770) = 1/30

  Practice Set

  • Q1: Which material has μ_r < 1?
    • A. Iron
    • B. Aluminium
    • C. Copper
    • D. Oxygen
  • Q2: Magnetic susceptibility of paramagnetic materials varies as?
    • A. T
    • B. 1/T
    • C. Constant
    • D. T²
  • Q3: Which material shows strong magnetism due to domain formation?
    • A. Sodium
    • B. Iron
    • C. Lead
    • D. Water

  Answer Key

  • Q1: C. Copper - Diamagnetic materials have μ_r < 1.
  • Q2: B. 1/T - Paramagnetic susceptibility follows Curie law.
  • Q3: B. Iron - Ferromagnetic materials have domains causing strong magnetism.

  Memory Aids

  • Diamagnetic: "Dia" = Opposite magnetism, μ_r < 1.
  • Paramagnetic: "Para" = Parallel weak magnetism, χ_m ∝ 1/T.
  • Ferromagnetic: "Ferro" = Strong magnetism, domains aligned, μ_r ≫ 1.

  Classification and Properties

  • Materials classified by magnetisation intensity (I), susceptibility (( χ_m )), and relative permeability (( μ_r )).
  • Three types: Diamagnetic, Paramagnetic, Ferromagnetic.
  • Diamagnetic: Caused by orbital electron motion; weak magnetism opposite to field; ( μ_r < 1 ); no permanent atomic moment; examples: Bismuth, copper, lead, water.
  • Paramagnetic: Caused by electron spin; weak magnetism along field; ( μ_r > 1 ) slightly; permanent atomic moments random; examples: Aluminium, sodium, calcium, oxygen.
  • Ferromagnetic: Due to magnetic domains; strong magnetism along field; ( μ_r ≫ 1 ); permanent atomic moments aligned; examples: Iron, cobalt, nickel.
  • Magnetisation vs. Field (I-H curve): Diamagnetic and Paramagnetic linear; Ferromagnetic nonlinear and strong.

  Temperature Dependence of Susceptibility

  • Diamagnetic: ( χ_m ) small, negative, temperature independent.
  • Paramagnetic: ( χ_m ∝ 1/T ) (Curie law), decreases with temperature.
  • Ferromagnetic: ( χ_m ∝ 1/(T - T_C) ) (Curie-Weiss law) above Curie temperature ( T_C ); sharp peak at ( T_C ) (e.g., iron ( T_C = 770°C )).

  Solved Examples

  • Ex 1: Calculate relative permeability of a paramagnetic material with susceptibility ( χ_m = 2 × 10^-5 ).
    • ( μ_r = 1 + χ_m = 1 + 2 × 10^-5 = 1.00002 )
  • Ex 2: Find susceptibility of a ferromagnetic material at ( T = 800°C ) given ( T_C = 770°C ) and ( χ_m ∝ 1/(T - T_C) ).
    • ( χ_m ∝ 1/(800 - 770) = 1/30 )

  Practice Set

  • Q1: Which material has ( μ_r < 1 )?
    • A. Iron
    • B. Aluminium
    • C. Copper
    • D. Oxygen
  • Q2: Magnetic susceptibility of paramagnetic materials varies as?
    • A. ( T )
    • B. ( 1/T )
    • C. Constant
    • D. ( T² )
  • Q3: Which material shows strong magnetism due to domain formation?
    • A. Sodium
    • B. Iron
    • C. Lead
    • D. Water

  Answer Key

  • Q1: C. Copper - Diamagnetic materials have ( μ_r < 1 ).
  • Q2: B. ( 1/T ) - Paramagnetic susceptibility follows Curie law.
  • Q3: B. Iron - Ferromagnetic materials have domains causing strong magnetism.

  Memory Aids

  • Diamagnetic: "Dia" = Opposite magnetism, ( μ_r < 1 ).
  • Paramagnetic: "Para" = Parallel weak magnetism, ( χ_m ∝ 1/T ).
  • Ferromagnetic: "Ferro" = Strong magnetism, domains aligned, ( μ_r ≫ 1 ).

  Classification and Properties

  • Materials classified by magnetisation intensity (I), susceptibility (\( \chi_m \)), and relative permeability (\( \mu_r \)).
  • Three types: Diamagnetic, Paramagnetic, Ferromagnetic.
  • Diamagnetic: Caused by orbital electron motion; weak magnetism opposite to field; \( \mu_r < 1 \); no permanent atomic moment; examples: Bismuth, copper, lead, water.
  • Paramagnetic: Caused by electron spin; weak magnetism along field; \( \mu_r > 1 \) slightly; permanent atomic moments random; examples: Aluminium, sodium, calcium, oxygen.
  • Ferromagnetic: Due to magnetic domains; strong magnetism along field; \( \mu_r \gg 1 \); permanent atomic moments aligned; examples: Iron, cobalt, nickel.
  • Magnetisation vs. Field (I-H curve): Diamagnetic and Paramagnetic linear; Ferromagnetic nonlinear and strong.

  Temperature Dependence of Susceptibility

  • Diamagnetic: \( \chi_m \) small, negative, temperature independent.
  • Paramagnetic: \( \chi_m \propto \frac{1}{T} \) (Curie law), decreases with temperature.
  • Ferromagnetic: \( \chi_m \propto \frac{1}{T - T_C} \) (Curie-Weiss law) above Curie temperature \( T_C \); sharp peak at \( T_C \) (e.g., iron \( T_C = 770^\circ C \)).

  Solved Examples

  • Ex 1: Calculate relative permeability of a paramagnetic material with susceptibility \( \chi_m = 2 \times 10^{-5} \).
    • \( \mu_r = 1 + \chi_m = 1 + 2 \times 10^{-5} = 1.00002 \)
  • Ex 2: Find susceptibility of a ferromagnetic material at \( T = 800^\circ C \) given \( T_C = 770^\circ C \) and \( \chi_m \propto \frac{1}{T - T_C} \).
    • \( \chi_m \propto \frac{1}{800 - 770} = \frac{1}{30} \)

  Practice Set

  • Q1: Which material has \( \mu_r < 1 \)?
    • A. Iron
    • B. Aluminium
    • C. Copper
    • D. Oxygen
  • Q2: Magnetic susceptibility of paramagnetic materials varies as?
    • A. \( T \)
    • B. \( \frac{1}{T} \)
    • C. Constant
    • D. \( T^2 \)
  • Q3: Which material shows strong magnetism due to domain formation?
    • A. Sodium
    • B. Iron
    • C. Lead
    • D. Water

  Answer Key

  • Q1: C. Copper - Diamagnetic materials have \( \mu_r < 1 \).
  • Q2: B. \( \frac{1}{T} \) - Paramagnetic susceptibility follows Curie law.
  • Q3: B. Iron - Ferromagnetic materials have domains causing strong magnetism.

  Memory Aids

  • Diamagnetic: "Dia" = Opposite magnetism, \( \mu_r < 1 \).
  • Paramagnetic: "Para" = Parallel weak magnetism, \( \chi_m \propto \frac{1}{T} \).
  • Ferromagnetic: "Ferro" = Strong magnetism, domains aligned, \( \mu_r \gg 1 \).

  Magnetic Field of a Bar Magnet

  Field at Axial Position

  • Magnetic field due to north pole at point P: ( B₁ = μ₀/(4π) m/(r - l)² ) (away from north pole)
  • Magnetic field due to south pole at point P: ( B₂ = μ₀/(4π) m/(r + l)² ) (towards north pole)
  • Net field: ( B_axis = B₁ - B₂ = μ₀m/(4π) [ 1/(r - l)² - 1/(r + l)² ] = μ₀m/(4π) 4rl/(r² - l²)² )
  • Using magnetic moment ( M = m(2l) ), ( B_axis = μ₀/(4π) 2Mr/(r² - l²)² )
  • For short magnet ( r ≫ l ), ( B_axis ≈ μ₀/(4π) 2M/r³ ) (like electric dipole)

  Field at Equatorial Position

  • Magnetic field due to each pole at point P: ( B₁ = B₂ = μ₀/(4π) m/(r² + l²) )
  • Net field: ( B_eq = 2 B cos θ = 2 μ₀/(4π) m/(r² + l²) cos θ )
  • With ( cos θ = 1/√(r² + l²) ), ( B_eq = μ₀/(4π) M/(r² + l²)^3/2 )
  • For short magnet ( r ≫ l ), ( B_eq ≈ μ₀/(4π) M/r³ )

  Field at General Position

  • Radial component: ( B_r = 2 μ₀/(4π) M cos θ/r³ )
  • Normal component: ( B_n = μ₀/(4π) M sin θ/r³ )
  • Resultant field: ( B_res = μ₀M/(4πr³) √(1 + 3 cos² θ) )
  • Angle relation: ( tan Φ = B_n/B_r = tan θ/2 )

  Solved Examples

  • Ex 1: Calculate axial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ).
    • ( B_axis = μ₀/(4π) 2Mr/(r² - l²)² ≈ 10^-7 × 2 × 0.05 × 0.1/(0.1² - 0.01²)² )
  • Ex 2: Find equatorial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ), ( θ = 60° ).
    • ( B_eq = 2 μ₀/(4π) m/(r² + l²) cos θ )

  Practice Set

  • Q1: Magnetic field at axial point P is given by?
    • A. ( μ₀/(4π) 2M/r³ )
    • B. ( μ₀/(4π) M/r³ )
    • C. ( μ₀/(4π) M/r² )
    • D. ( μ₀/(4π) 2Mr/(r² - l²)² )
  • Q2: At equatorial point, magnetic field is?
    • A. ( μ₀/(4π) 2M/r³ )
    • B. ( μ₀/(4π) M/r³ )
    • C. ( μ₀/(4π) M/(r² + l²)^3/2 )
    • D. ( μ₀/(4π) 2Mr/(r² - l²)² )
  • Q3: Resultant magnetic field magnitude at general position is?
    • A. ( μ₀M/(4πr³) √(1 + 3 cos² θ) )
    • B. ( μ₀M/(4πr³) √(1 + cos² θ) )
    • C. ( μ₀M/(4πr²) )
    • D. ( μ₀M/(4πr³) tan θ )

  Answer Key

  • Q1: D - Exact formula for axial field includes ( 2Mr/(r² - l²)² ).
  • Q2: C - Equatorial field formula is ( μ₀/(4π) M/(r² + l²)^3/2 ).
  • Q3: A - Resultant field magnitude is ( μ₀M/(4πr³) √(1 + 3 cos² θ) ).

  Memory Aids

  • Axial field ~ ( 2M/r³ ), Equatorial field ~ ( M/r³ ).
  • Resultant field combines radial and normal components with angle ( θ ).
  • Magnetic moment ( M = m (2l) ) relates pole strength and magnet length.

  Field at Axial Position

  • Magnetic field due to north pole at point P: B₁ = μ₀/(4π) m/(r - l)² (away from north pole)
  • Magnetic field due to south pole at point P: B₂ = μ₀/(4π) m/(r + l)² (towards north pole)
  • Net field: B_axis = B₁ - B₂ = μ₀m/(4π) [ 1/(r - l)² - 1/(r + l)² ] = μ₀m/(4π) 4rl/(r² - l²)²
  • Using magnetic moment M = m(2l), B_axis = μ₀/(4π) 2Mr/(r² - l²)²
  • For short magnet r ≫ l, B_axis ≈ μ₀/(4π) 2M/r³ (like electric dipole)

  Field at Equatorial Position

  • Magnetic field due to each pole at point P: B₁ = B₂ = μ₀/(4π) m/(r² + l²)
  • Net field: B_eq = 2 B cos θ = 2 μ₀/(4π) m/(r² + l²) cos θ
  • With cos θ = 1/√(r² + l²), B_eq = μ₀/(4π) M/(r² + l²)^3/2
  • For short magnet r ≫ l, B_eq ≈ μ₀/(4π) M/r³

  Field at General Position

  • Radial component: B_r = 2 μ₀/(4π) M cos θ/r³
  • Normal component: B_n = μ₀/(4π) M sin θ/r³
  • Resultant field: B_res = μ₀M/(4πr³) √(1 + 3 cos² θ)
  • Angle relation: tan Φ = B_n/B_r = tan θ/2

  Solved Examples

  • Ex 1: Calculate axial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ).
    • B_axis = μ₀/(4π) 2Mr/(r² - l²)² ≈ 10^-7 × 2 × 0.05 × 0.1/(0.1² - 0.01²)²
  • Ex 2: Find equatorial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ), ( θ = 60° ).
    • B_eq = 2 μ₀/(4π) m/(r² + l²) cos θ

  Practice Set

  • Q1: Magnetic field at axial point P is given by?
    • A. μ₀/(4π) 2M/r³
    • B. μ₀/(4π) M/r³
    • C. μ₀/(4π) M/r²
    • D. μ₀/(4π) 2Mr/(r² - l²)²
  • Q2: At equatorial point, magnetic field is?
    • A. μ₀/(4π) 2M/r³
    • B. μ₀/(4π) M/r³
    • C. μ₀/(4π) M/(r² + l²)^3/2
    • D. μ₀/(4π) 2Mr/(r² - l²)²
  • Q3: Resultant magnetic field magnitude at general position is?
    • A. μ₀M/(4πr³) √(1 + 3 cos² θ)
    • B. μ₀M/(4πr³) √(1 + cos² θ)
    • C. μ₀M/(4πr²)
    • D. μ₀M/(4πr³) tan θ

  Answer Key

  • Q1: D - Exact formula for axial field includes ( 2Mr/(r² - l²)² ).
  • Q2: C - Equatorial field formula is ( μ₀/(4π) M/(r² + l²)^3/2 ).
  • Q3: A - Resultant field magnitude is ( μ₀M/(4πr³) √(1 + 3 cos² θ) ).

  Memory Aids

  • Axial field ~ ( 2M/r³ ), Equatorial field ~ ( M/r³ ).
  • Resultant field combines radial and normal components with angle ( θ ).
  • Magnetic moment ( M = m (2l) ) relates pole strength and magnet length.

  Field at Axial Position

  • Magnetic field due to north pole at point P: ( B₁ = μ₀/(4π) m/(r - l)² ) (away from north pole)
  • Magnetic field due to south pole at point P: ( B₂ = μ₀/(4π) m/(r + l)² ) (towards north pole)
  • Net field: ( B_axis = B₁ - B₂ = μ₀m/(4π) [ 1/(r - l)² - 1/(r + l)² ] = μ₀m/(4π) 4rl/(r² - l²)² )
  • Using magnetic moment ( M = m(2l) ), ( B_axis = μ₀/(4π) 2Mr/(r² - l²)² )
  • For short magnet ( r ≫ l ), ( B_axis ≈ μ₀/(4π) 2M/r³ ) (like electric dipole)

  Field at Equatorial Position

  • Magnetic field due to each pole at point P: ( B₁ = B₂ = μ₀/(4π) m/(r² + l²) )
  • Net field: ( B_eq = 2 B cos θ = 2 μ₀/(4π) m/(r² + l²) cos θ )
  • With ( cos θ = 1/√(r² + l²) ), ( B_eq = μ₀/(4π) M/(r² + l²)^3/2 )
  • For short magnet ( r ≫ l ), ( B_eq ≈ μ₀/(4π) M/r³ )

  Field at General Position

  • Radial component: ( B_r = 2 μ₀/(4π) M cos θ/r³ )
  • Normal component: ( B_n = μ₀/(4π) M sin θ/r³ )
  • Resultant field: ( B_res = μ₀M/(4πr³) √(1 + 3 cos² θ) )
  • Angle relation: ( tan Φ = B_n/B_r = tan θ/2 )

  Solved Examples

  • Ex 1: Calculate axial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ).
    • ( B_axis = μ₀/(4π) 2Mr/(r² - l²)² ≈ 10^-7 × 2 × 0.05 × 0.1/(0.1² - 0.01²)² )
  • Ex 2: Find equatorial field at ( r = 10 cm ), ( l = 1 cm ), ( M = 0.05 A·m² ), ( θ = 60° ).
    • ( B_eq = 2 μ₀/(4π) m/(r² + l²) cos θ )

  Practice Set

  • Q1: Magnetic field at axial point P is given by?
    • A. ( μ₀/(4π) 2M/r³ )
    • B. ( μ₀/(4π) M/r³ )
    • C. ( μ₀/(4π) M/r² )
    • D. ( μ₀/(4π) 2Mr/(r² - l²)² )
  • Q2: At equatorial point, magnetic field is?
    • A. ( μ₀/(4π) 2M/r³ )
    • B. ( μ₀/(4π) M/r³ )
    • C. ( μ₀/(4π) M/(r² + l²)^3/2 )
    • D. ( μ₀/(4π) 2Mr/(r² - l²)² )
  • Q3: Resultant magnetic field magnitude at general position is?
    • A. ( μ₀M/(4πr³) √(1 + 3 cos² θ) )
    • B. ( μ₀M/(4πr³) √(1 + cos² θ) )
    • C. ( μ₀M/(4πr²) )
    • D. ( μ₀M/(4πr³) tan θ )

  Answer Key

  • Q1: D - Exact formula for axial field includes ( 2Mr/(r² - l²)² ).
  • Q2: C - Equatorial field formula is ( μ₀/(4π) M/(r² + l²)^3/2 ).
  • Q3: A - Resultant field magnitude is ( μ₀M/(4πr³) √(1 + 3 cos² θ) ).

  Memory Aids

  • Axial field ~ ( 2M/r³ ), Equatorial field ~ ( M/r³ ).
  • Resultant field combines radial and normal components with angle ( θ ).
  • Magnetic moment ( M = m (2l) ) relates pole strength and magnet length.

  Field at Axial Position

  • Magnetic field due to north pole at point P: \( B_1 = \frac{\mu_0}{4\pi} \frac{m}{(r - l)^2} \) (away from north pole)
  • Magnetic field due to south pole at point P: \( B_2 = \frac{\mu_0}{4\pi} \frac{m}{(r + l)^2} \) (towards north pole)
  • Net field: \( B_{\text{axis}} = B_1 - B_2 = \frac{\mu_0 m}{4\pi} \left[ \frac{1}{(r - l)^2} - \frac{1}{(r + l)^2} \right] = \frac{\mu_0 m}{4\pi} \frac{4 r l}{(r^2 - l^2)^2} \)
  • Using magnetic moment \( M = m(2l) \), \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \)
  • For short magnet \( r \gg l \), \( B_{\text{axis}} \approx \frac{\mu_0}{4\pi} \frac{2 M}{r^3} \) (like electric dipole)

  Field at Equatorial Position

  • Magnetic field due to each pole at point P: \( B_1 = B_2 = \frac{\mu_0}{4\pi} \frac{m}{(r^2 + l^2)} \)
  • Net field: \( B_{\text{eq}} = 2 B \cos \theta = 2 \frac{\mu_0}{4\pi} \frac{m}{r^2 + l^2} \cos \theta \)
  • With \( \cos \theta = \frac{1}{\sqrt{r^2 + l^2}} \), \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}} \)
  • For short magnet \( r \gg l \), \( B_{\text{eq}} \approx \frac{\mu_0}{4\pi} \frac{M}{r^3} \)

  Field at General Position

  • Radial component: \( B_r = 2 \frac{\mu_0}{4\pi} \frac{M \cos \theta}{r^3} \)
  • Normal component: \( B_n = \frac{\mu_0}{4\pi} \frac{M \sin \theta}{r^3} \)
  • Resultant field: \( B_{\text{res}} = \frac{\mu_0 M}{4\pi r^3} \sqrt{1 + 3 \cos^2 \theta} \)
  • Angle relation: \( \tan \Phi = \frac{B_n}{B_r} = \frac{\tan \theta}{2} \)

  Solved Examples

  • Ex 1: Calculate axial field at \( r = 10 \text{ cm} \), \( l = 1 \text{ cm} \), \( M = 0.05 \text{ A·m}^2 \).
    • \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \approx \frac{10^{-7} \times 2 \times 0.05 \times 0.1}{(0.1^2 - 0.01^2)^2} \)
  • Ex 2: Find equatorial field at \( r = 10 \text{ cm} \), \( l = 1 \text{ cm} \), \( M = 0.05 \text{ A·m}^2 \), \( \theta = 60^\circ \).
    • \( B_{\text{eq}} = 2 \frac{\mu_0}{4\pi} \frac{m}{r^2 + l^2} \cos \theta \)

  Practice Set

  • Q1: Magnetic field at axial point P is given by?
    • A. \( \frac{\mu_0}{4\pi} \frac{2 M}{r^3} \)
    • B. \( \frac{\mu_0}{4\pi} \frac{M}{r^3} \)
    • C. \( \frac{\mu_0}{4\pi} \frac{M}{r^2} \)
    • D. \( \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \)
  • Q2: At equatorial point, magnetic field is?
    • A. \( \frac{\mu_0}{4\pi} \frac{2 M}{r^3} \)
    • B. \( \frac{\mu_0}{4\pi} \frac{M}{r^3} \)
    • C. \( \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}} \)
    • D. \( \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \)
  • Q3: Resultant magnetic field magnitude at general position is?
    • A. \( \frac{\mu_0 M}{4\pi r^3} \sqrt{1 + 3 \cos^2 \theta} \)
    • B. \( \frac{\mu_0 M}{4\pi r^3} \sqrt{1 + \cos^2 \theta} \)
    • C. \( \frac{\mu_0 M}{4\pi r^2} \)
    • D. \( \frac{\mu_0 M}{4\pi r^3} \tan \theta \)

  Answer Key

  • Q1: D - Exact formula for axial field includes \( \frac{2 M r}{(r^2 - l^2)^2} \).
  • Q2: C - Equatorial field formula is \( \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}} \).
  • Q3: A - Resultant field magnitude is \( \frac{\mu_0 M}{4\pi r^3} \sqrt{1 + 3 \cos^2 \theta} \).

  Memory Aids

  • Axial field ~ \( \frac{2 M}{r^3} \), Equatorial field ~ \( \frac{M}{r^3} \).
  • Resultant field combines radial and normal components with angle \( \theta \).
  • Magnetic moment \( M = m (2l) \) relates pole strength and magnet length.

  Magnetic Hysteresis

  Hysteresis Loop and Properties

  • Graph of magnetic field strength (H) vs magnetic flux density (B).
  • Path oa: Magnetisation increases non-linearly to saturation.
  • Path ab: On reducing H to zero, residual magnetism (remanence B_r) remains.
  • Coercive field (H_c) is reverse field needed to nullify residual magnetism.
  • Loop shows magnetic memory and lag of B behind H (hysteresis).
  • Only ferromagnetic materials show hysteresis.

  Soft and Hard Magnetic Materials

  • Soft magnetic materials: Low retentivity, low coercivity, used in electromagnets, transformer cores.
  • Hard magnetic materials: High retentivity, high coercivity, used for permanent magnets.
  • Soft materials have narrow hysteresis loops; hard materials have wide loops.
  • Examples: Soft iron (soft), steel and alnico (hard).

  Solved Examples

  • Ex 1: Identify coercivity if reverse field H_c = 1000 A/m.
  • Coercivity = 1000 A/m.
  • Ex 2: Find remanence if residual magnetism B_r = 0.5 T.
  • Remanence = 0.5 T.

  Practice Set

  • Q1: What does coercivity measure?
    • A. Magnetic field strength
    • B. Residual magnetism
    • C. Field to destroy residual magnetism
    • D. Saturation magnetism
  • Q2: Which material shows narrow hysteresis loop?
    • A. Steel
    • B. Soft iron
    • C. Alnico
    • D. Nickel
  • Q3: Hysteresis loop indicates?
    • A. No magnetic memory
    • B. Linear magnetisation
    • C. Magnetic memory and lag
    • D. Diamagnetism

  Answer Key

  • Q1: C - Coercivity is the field needed to remove residual magnetism.
  • Q2: B - Soft iron has narrow hysteresis loop (soft magnetic).
  • Q3: C - Loop shows magnetic memory and lag of B behind H.

  Memory Aids

  • Hysteresis = lag of B behind H.
  • Coercivity = field to erase residual magnetism.
  • Soft = low coercivity, hard = high coercivity.

  Hysteresis Loop and Properties

  • Graph of magnetic field strength (H) vs magnetic flux density (B).
  • Path oa: Magnetisation increases non-linearly to saturation.
  • Path ab: On reducing H to zero, residual magnetism (remanence \( B_r \)) remains.
  • Coercive field (\( H_c \)) is reverse field needed to nullify residual magnetism.
  • Loop shows magnetic memory and lag of B behind H (hysteresis).
  • Only ferromagnetic materials show hysteresis.

  Soft and Hard Magnetic Materials

  • Soft magnetic materials: Low retentivity, low coercivity, used in electromagnets, transformer cores.
  • Hard magnetic materials: High retentivity, high coercivity, used for permanent magnets.
  • Soft materials have narrow hysteresis loops; hard materials have wide loops.
  • Examples: Soft iron (soft), steel and alnico (hard).

  Solved Examples

  • Ex 1: Identify coercivity if reverse field \( H_c = 1000 \text{ A/m} \).
  • Coercivity \( = 1000 \text{ A/m} \).
  • Ex 2: Find remanence if residual magnetism \( B_r = 0.5 \text{ T} \).
  • Remanence \( = 0.5 \text{ T} \).

  Practice Set

  • Q1: What does coercivity measure?
    • A. Magnetic field strength
    • B. Residual magnetism
    • C. Field to destroy residual magnetism
    • D. Saturation magnetism
  • Q2: Which material shows narrow hysteresis loop?
    • A. Steel
    • B. Soft iron
    • C. Alnico
    • D. Nickel
  • Q3: Hysteresis loop indicates?
    • A. No magnetic memory
    • B. Linear magnetisation
    • C. Magnetic memory and lag
    • D. Diamagnetism

  Answer Key

  • Q1: C - Coercivity is the field needed to remove residual magnetism.
  • Q2: B - Soft iron has narrow hysteresis loop (soft magnetic).
  • Q3: C - Loop shows magnetic memory and lag of B behind H.

  Memory Aids

  • Hysteresis = lag of B behind H.
  • Coercivity = field to erase residual magnetism.
  • Soft = low coercivity, hard = high coercivity.