Magnetism and Matter | Class 12 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Magnetism and Matter – this guide gives you a concise, exam-ready overview of Magnetism and Matter from Class 12 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
5.5 MAGNETIC PROPERTIES OF MATERIALS
Materials can be classified into diamagnetic, paramagnetic, and ferromagnetic based on their magnetic susceptibility (χ) and relative permeability (μ_r). Diamagnetic materials have negative and small χ, paramagnetic materials have small positive χ, and ferromagnetic materials have large positive χ. Diamagnetic substances tend to move from stronger to weaker parts of an external magnetic field, effectively repelled by magnets. This is due to induced currents opposing the applied field as per Lenz's law, resulting in a net magnetic moment opposite to the applied field. Examples include bismuth, copper, lead, silicon, nitrogen, water, and sodium chloride. Diamagnetism is present in all materials but often overshadowed by stronger magnetic effects. Superconductors exhibit perfect diamagnetism (χ = -1, μ_r = 0), expelling magnetic fields completely, known as the Meissner effect, useful in magnetic levitation applications. Paramagnetic substances have permanent atomic magnetic dipole moments that align weakly with an external field, enhancing the field inside the material. Examples include aluminium, sodium, calcium, oxygen, and copper chloride. Their susceptibility and permeability depend on temperature and field strength, with magnetisation saturating when all dipoles align. Ferromagnetic materials exhibit strong magnetisation due to spontaneous alignment of atomic dipoles into domains, macroscopic regions with uniform magnetisation. Domains align and grow in an external field, concentrating magnetic field lines inside the material. Some ferromagnets retain magnetisation after field removal (hard ferromagnets like Alnico and lodestone), forming permanent magnets, while others (soft ferromagnets like soft iron) do not. Ferromagnetism depends on temperature; above a critical temperature, ferromagnets become paramagnetic as domain structure disintegrates. Examples of ferromagnetic elements include iron, cobalt, nickel, and gadolinium. Table 5.2 summarizes susceptibility and permeability ranges for these materials.
📊 Diagram: FIGURE 5.7 shows magnetic field lines near (a) a diamagnetic material where field lines are repelled, and (b) a paramagnetic material where field lines are concentrated; FIGURE 5.8 shows (a) randomly oriented domains and (b) aligned domains in a ferromagnetic material.
🔗 Connection: Leads to summary and key points consolidating the chapter's concepts.
Table on page 12 (4×3)
| Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|
| -1 ≤ χ < 0 | 0 ≤ χ < ε | χ >> 1 |
| 0 ≤ μr < 1 | 1 < μr ≤ 1 + ε | μr >> 1 |
| μ < μ0 | μ > μ0 | μ >> μ0 |
Frequently asked questions
5.1 A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet?
Given: Torque \( \tau = 4.5 \times 10^{-2} \, \mathrm{J} \), Magnetic field \( B = 0.25 \, \mathrm{T} \), Angle \( \theta = 30^{\circ} \).
Torque on a magnetic dipole in a magnetic field is given by: \[ \tau = m B \sin \theta \] where \( m \) is the magnetic moment.
Rearranging for \( m \): \[ m = \frac{\tau}{B \sin \theta} \]
Calculate \( \sin 30^{\circ} = 0.5 \).
Therefore, \[ m = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36 \, \mathrm{A \cdot m^2} \
5.2 A short bar magnet of magnetic moment $m = 0.32 \, \mathrm{JT}^{-1}$ 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?
(a) Stable equilibrium occurs when the magnetic moment \( \vec{m} \) is aligned with the magnetic field \( \vec{B} \), i.e., the angle \( \theta = 0^{\circ} \).
(b) Unstable equilibrium occurs when \( \vec{m} \) is opposite to \( \vec{B} \), i.e., \( \theta = 180^{\circ} \).
Potential energy of a magnetic dipole in a magnetic field is given by: \[ U = -m B \cos \theta \]
Calculate potential energy for each case:
For stable equilibrium (\( \theta = 0^{\circ} \)): \[ U = -m B \cos 0^{\circ} =
5.3 A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4} \, \mathrm{m}^2$ 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?
A solenoid carrying current produces a magnetic field similar to that of a bar magnet, with a north and south pole at its ends. The magnetic field lines inside the solenoid are nearly uniform and parallel, resembling the field of a bar magnet.
Magnetic moment \( m \) of a current-carrying coil (or solenoid) is given by: \[ m = N I A \] where \( N = 800 \) (number of turns), \( I = 3.0 \, \mathrm{A} \) (current), \( A = 2.5 \times 10^{-4} \, \mathrm{m}^2 \) (area of cross-section).
Calculate: \
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)?
Given: Magnetic moment, m = 1.5 J T⁻¹ Magnetic field, B = 0.22 T
(a) Work done by external torque to rotate the magnet from initial alignment to a new angle θ with the field is given by: W = mB (cos θ_initial - cos θ_final)
(i) To align magnetic moment normal (90°) to the field: Initial angle θ_initial = 0° (aligned) Final angle θ_final = 90°
W = mB (cos 0° - cos 90°) = 1.5 × 0.22 × (1 - 0) = 0.33 J
(ii) To align magnetic moment opposite (180°) to the field: θ_final = 180°
W = mB (cos 0° -
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