Magnetism and Matter | Class 12 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 6 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.
5.2 THE BAR MAGNET
The bar magnet is a fundamental magnetic object exhibiting two poles: north and south. When iron filings are sprinkled on a sheet of glass placed over a bar magnet, they arrange themselves in a pattern that reveals the magnetic field lines. This pattern suggests that the bar magnet behaves like a magnetic dipole, similar to an electric dipole with positive and negative charges. The north pole of the magnet points approximately towards the geographic north when suspended freely, and the south pole points towards the geographic south. A similar pattern of iron filings is observed around a current-carrying solenoid, indicating analogous magnetic behavior. This section introduces the concept of magnetic field lines as a visual representation of the magnetic field around magnets and current-carrying conductors. The magnetic field lines form continuous closed loops, unlike electric field lines which begin and end on charges. The tangent to a magnetic field line at any point gives the direction of the magnetic field vector at that point. The density of field lines indicates the strength of the magnetic field; more lines per unit area mean a stronger field. Importantly, magnetic field lines never intersect, ensuring a unique direction of the magnetic field at every point. The bar magnet can be thought of as an equivalent solenoid, consisting of many tiny current loops producing a magnetic dipole field. This analogy is supported by the similarity in magnetic field patterns and can be tested by observing the deflection of a small compass needle near a bar magnet and a current-carrying solenoid, which are similar. The axial magnetic field of a finite solenoid at large distances resembles that of a bar magnet, with the magnetic moment of the bar magnet equal to that of the equivalent solenoid producing the same field. When a small magnetic dipole with magnetic moment m is placed in a uniform magnetic field B, it experiences a torque τ = m × B that tends to align the dipole with the field. The magnetic potential energy of the dipole in the field is U_m = -m · B, with the zero of potential energy chosen when the dipole is perpendicular to the field. The potential energy is minimum when the dipole aligns with the field (stable equilibrium) and maximum when it is opposite (unstable equilibrium).
📊 Diagram: FIGURE 5.1 The arrangement of iron filings surrounding a bar magnet. The pattern mimics magnetic field lines. The pattern suggests that the bar magnet is a magnetic dipole; (a) Bar magnet field lines; (b) Current-carrying finite solenoid field lines; (c) Electric dipole field lines. At large distances, the field lines are very similar. The curves labelled ① and ⑪ are closed Gaussian surfaces; (a) Calculation of the axial field of a finite solenoid to demonstrate similarity to a bar magnet; (b) A magnetic needle in a uniform magnetic field B used to determine B or magnetic moment m of the needle.
🧪 Activity: Placing a small magnetic compass needle at various points around a bar magnet and a solenoid to observe and plot magnetic field lines.
🔗 Connection: Leads to the electrostatic analog of magnetic dipoles and Gauss's law for magnetism.
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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