Class 12 Physics Notes – Chapter 5: Magnetism and Matter

1. Introduction to Magnetism

The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north. When a bar magnet is freely suspended, it points in the north-south direction. Similar to electric charges, like magnetic poles repel and unlike poles attract each other. However, unlike electric charges, isolated magnetic north and south poles (magnetic monopoles) do not exist.

2. The Bar Magnet and Magnetic Field Lines

The pattern of iron filings around a magnet suggests that the magnet has two poles: North and South. The magnetic field lines represent the magnetic field visually.

Properties of Magnetic Field Lines:

• The magnetic field lines of a magnet form continuous closed loops. They emerge from the North pole and enter the South pole externally, and go from South to North inside the magnet.
• The tangent to the field line at a given point represents the direction of the net magnetic field B at that point.
• The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
• Two magnetic field lines never intersect each other. If they did, the direction of the magnetic field would not be unique at the point of intersection.

3. Magnetic Dipole in a Uniform Magnetic Field

When a magnetic dipole (like a small compass needle or a bar magnet) of magnetic moment m is placed in a uniform magnetic field B, it experiences a restoring torque.

Torque Formula: τ = m × B ⇒ τ = mB sinθ

Where θ is the angle between the magnetic moment (m) and the magnetic field (B).

Magnetic Potential Energy (U_m): The potential energy of a magnetic dipole in a uniform magnetic field is given by:

Formula: U_m = – m · B = – mB cosθ

4. The Electrostatic Analog

The magnetic field of a bar magnet at a large distance is analogous to the electric field of an electric dipole. We can replace electric dipole moment p with magnetic dipole moment m, and the constant 1 / (4πε₀) with μ₀ / 4π.

• Axial Field (B_A): B_A = (μ₀ / 4π) · (2m / r³)
• Equatorial Field (B_E): B_E = – (μ₀ / 4π) · (m / r³)

5. Magnetisation and Magnetic Intensity

Magnetisation (M): It is defined as the net magnetic moment per unit volume of a material.

M = m_net / V. Its SI unit is A m^-1.

Magnetic Intensity (H): It is the measure of the external magnetic field that induces magnetisation in a material.

The total magnetic field B inside the material is: B = μ₀ (H + M)

Magnetic Susceptibility and Permeability

For linear materials, magnetisation is proportional to magnetic intensity: M = χ H

Where χ (chi) is the magnetic susceptibility.

The relative magnetic permeability is given by: μ_r = 1 + χ

The magnetic permeability of the substance is: μ = μ₀ μ_r

6. Magnetic Properties of Materials

Based on magnetic susceptibility (χ), materials are classified into three categories:

(i) Diamagnetic Materials

• These materials have a tendency to move from stronger to the weaker part of the external magnetic field.
• They are feebly repelled by magnets.
• Examples: Bismuth, copper, lead, silicon, water.
• Susceptibility (χ) is small and negative (-1 ≤ χ < 0).
• Superconductors exhibit perfect diamagnetism (Meissner effect) where χ = -1.

(ii) Paramagnetic Materials

• These materials get weakly magnetised in the direction of the external field.
• They have a tendency to move from a region of weak magnetic field to strong magnetic field (feebly attracted).
• Examples: Aluminium, sodium, calcium, oxygen.
• Susceptibility (χ) is small and positive.

(iii) Ferromagnetic Materials

• These materials get strongly magnetised in the direction of the external field.
• They are strongly attracted to a magnet.
• Examples: Iron, cobalt, nickel, gadolinium.
• Susceptibility (χ) is very large and positive (χ >> 1).
• The ferromagnetic property depends on temperature. At high enough temperatures, a ferromagnet becomes a paramagnet.

Multiple Choice Questions (MCQs)

1. Isolated magnetic monopoles:
a) Exist only in ferromagnets
b) Exist only in space
c) Do not exist
d) Exist at high temperatures
Answer: (c) Do not exist

2. Magnetic field lines inside a bar magnet travel from:
a) North pole to South pole
b) South pole to North pole
c) Centre to the poles
d) They do not exist inside
Answer: (b) South pole to North pole

3. The SI unit of magnetic dipole moment is:
a) A m
b) A m²
c) T m / A
d) Weber / m²
Answer: (b) A m²

4. A material has magnetic susceptibility χ = -0.00015. It is a:
a) Paramagnetic material
b) Ferromagnetic material
c) Diamagnetic material
d) Superconductor
Answer: (c) Diamagnetic material

5. The perfect diamagnetism shown by superconductors is called:
a) Faraday effect
b) Meissner effect
c) Curie effect
d) Lorentz effect
Answer: (b) Meissner effect

6. The relation between relative permeability (μ_r) and magnetic susceptibility (χ) is:
a) μ_r = χ – 1
b) μ_r = 1 + χ
c) μ_r = 1 / χ
d) μ_r = 1 – χ
Answer: (b) μ_r = 1 + χ

7. The torque experienced by a magnetic dipole in a uniform magnetic field is maximum when the angle between m and B is:
a) 0°
b) 45°
c) 90°
d) 180°
Answer: (c) 90°

8. The potential energy of a magnetic dipole in a uniform magnetic field is given by:
a) mB sinθ
b) -mB cosθ
c) mB cosθ
d) -mB sinθ
Answer: (b) -mB cosθ

9. Which of the following is a ferromagnetic material?
a) Copper
b) Aluminium
c) Water
d) Iron
Answer: (d) Iron

10. As temperature increases, a ferromagnetic material transforms into:
a) Diamagnetic material
b) Paramagnetic material
c) Superconductor
d) Insulator
Answer: (b) Paramagnetic material

Textbook Exercise Solutions

Q 5.1: A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^-2 J. What is the magnitude of magnetic moment of the magnet?
Solution:
Given: θ = 30°, B = 0.25 T, τ = 4.5 × 10^-2 J.
We know that Torque τ = mB sinθ
4.5 × 10^-2 = m × 0.25 × sin(30°)
4.5 × 10^-2 = m × 0.25 × 0.5
m = (4.5 × 10^-2) / (0.125) = 0.36 A m² (or J T^-1).

Q 5.2: A short bar magnet of magnetic moment m = 0.32 J T^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution:
Given: m = 0.32 J T^-1, B = 0.15 T.
(a) Stable equilibrium: The magnet is in stable equilibrium when its magnetic moment is parallel to the magnetic field (θ = 0°).
Potential Energy U = -mB cos(0°) = – 0.32 × 0.15 × 1 = – 4.8 × 10^-2 J.
(b) Unstable equilibrium: The magnet is in unstable equilibrium when its magnetic moment is anti-parallel to the magnetic field (θ = 180°).
Potential Energy U = -mB cos(180°) = – 0.32 × 0.15 × (-1) = + 4.8 × 10^-2 J.

Q 5.3: A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^-4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Solution:
Given: N = 800, A = 2.5 × 10^-4 m², I = 3.0 A.
A current-carrying solenoid acts as a bar magnet because its magnetic field lines resemble those of a bar magnet (forming continuous closed loops). One end acts as a North pole and the other as a South pole, depending on the direction of current.
Magnetic moment m = N · I · A
m = 800 × 3.0 × 2.5 × 10^-4 = 6000 × 10^-4 = 0.60 A m² (or J T^-1).

Q 5.4: If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Solution:
Given: m = 0.60 J T^-1 (from Q 5.3), B = 0.25 T, θ = 30°.
Torque τ = mB sinθ
τ = 0.60 × 0.25 × sin(30°)
τ = 0.15 × 0.5 = 7.5 × 10^-2 J.

Q 5.5: A bar magnet of magnetic moment 1.5 J T^-1 lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Solution:
Given: m = 1.5 J T^-1, B = 0.22 T. Initial angle θ₁ = 0°.
(a) Work done W = -mB(cosθ₂ – cosθ₁)
(i) Normal to field (θ₂ = 90°):
W = – 1.5 × 0.22 × (cos 90° – cos 0°) = – 0.33 × (0 – 1) = 0.33 J.
(ii) Opposite to field (θ₂ = 180°):
W = – 1.5 × 0.22 × (cos 180° – cos 0°) = – 0.33 × (-1 – 1) = – 0.33 × (-2) = 0.66 J.

(b) Torque τ = mB sinθ
(i) Case (i) θ = 90°: τ = 1.5 × 0.22 × sin 90° = 0.33 N m.
(ii) Case (ii) θ = 180°: τ = 1.5 × 0.22 × sin 180° = 0 N m.

Q 5.6: A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 T is set up at an angle of 30° with the axis of the solenoid?
Solution:
Given: N = 2000, A = 1.6 × 10^-4 m², I = 4.0 A, B = 7.5 × 10^-2 T, θ = 30°.
(a) Magnetic moment m = N · I · A = 2000 × 4.0 × 1.6 × 10^-4 = 1.28 A m² (or J T^-1).
(b) Since the magnetic field is uniform, the net force on the solenoid is Zero.
Torque τ = mB sinθ = 1.28 × 7.5 × 10^-2 × sin 30°
τ = 0.096 × 0.5 = 4.8 × 10^-2 N m.

Q 5.7: A short bar magnet has a magnetic moment of 0.48 J T^-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Solution:
Given: m = 0.48 J T^-1, d = 10 cm = 0.1 m.
(a) On the axis: B_A = (μ₀ / 4π) × (2m / d³)
B_A = 10^-7 × (2 × 0.48) / (0.1)³ = 10^-7 × (0.96 / 0.001) = 10^-7 × 960 = 9.6 × 10^-5 T.
The direction of the field on the axial line is along the magnetic moment (i.e., from South to North pole).
(b) On the equatorial line: B_E = (μ₀ / 4π) × (m / d³)
B_E = 10^-7 × (0.48) / (0.1)³ = 10^-7 × (0.48 / 0.001) = 10^-7 × 480 = 4.8 × 10^-5 T.
The direction of the field on the equatorial line is opposite to the magnetic moment (i.e., from North to South pole).