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.
5.3 MAGNETISM AND GAUSS'S LAW
Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. This is a fundamental difference from Gauss's law in electrostatics, where the net electric flux through a closed surface equals the enclosed electric charge divided by the permittivity of free space. Magnetic field lines are continuous and form closed loops, meaning they do not begin or end on any magnetic monopole, as such monopoles do not exist. This is visually demonstrated by examining Gaussian surfaces around bar magnets and current-carrying solenoids, where the number of magnetic field lines entering the surface equals the number leaving it, resulting in zero net flux. Mathematically, the magnetic flux through a small vector area element ΔS of a closed surface S is Δφ_B = B · ΔS, and the total flux φ_B through the surface is the sum over all elements, which equals zero. This contrasts with the electric flux, which depends on enclosed charge. The absence of magnetic monopoles implies that magnetic fields are generated by dipoles or current loops, and no isolated magnetic charges exist. This principle is essential in understanding magnetic field behavior and is foundational in electromagnetism. Example 5.3 presents several diagrams showing magnetic field lines, some of which violate Gauss's law for magnetism by showing field lines emanating from or terminating on points, crossing lines, or not forming closed loops. The correct magnetic field lines always form closed loops and never cross, unlike electric field lines which begin and end on charges.
📊 Diagram: FIGURE 5.5 shows a small vector area element ΔS of a closed surface S with magnetic field B passing through it, illustrating the concept of magnetic flux; FIGURE 5.6 shows various diagrams of field lines, some incorrect for magnetic fields but correct for electrostatics.
🧪 Activity: Analyzing diagrams of field lines to identify which represent magnetic fields correctly and which do not, reinforcing understanding of Gauss's law for magnetism.
🔗 Connection: Leads to the study of magnetisation and magnetic intensity in materials.
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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