Class 12 Physics NCERT REPRINT 2026-27

Chapter 5: Magnetism & Matter

Complete concise study notes, bar magnet as equivalent solenoid, Gauss's law for magnetism, magnetic properties & classification of materials, solved examples 5.1–5.5, and exercises 5.1–5.7.

01 / Exam-Focused Notes

Chapter 5: Magnetism & Matter

Complete official NCERT textbook coverage: The Bar Magnet, Magnetic Field Lines, Equivalent Solenoid, Dipole in Uniform B-Field, Electrostatic Analogy, Gauss's Law for Magnetism, Magnetisation & Magnetic Intensity, Diamagnetism, Paramagnetism, Ferromagnetism & Domains.

100% SYLLABUS
Chapter Navigation • NCERT Sections 5.1 – 5.5
5.1-5.2The Bar Magnet 5.3Gauss's Law of Magnetism 5.4Magnetisation & Intensity 5.5Magnetic Properties
5.1 & 5.2
NCERT Sections

Introduction & The Bar Magnet

Fundamental Properties & Field Lines

5.1.1 Core Principles of Magnetism & Field Lines Properties

  • Universal Nature: From distant galaxies to atoms, magnetic fields permeate nature. Earth's magnetic field points approximately from geographic South to geographic North.
  • Basic Properties of Bar Magnets:
    • Freely suspended magnet aligns in North-South direction (tip pointing to geographic North is the North pole; tip pointing to geographic South is the South pole).
    • Like magnetic poles repel; unlike poles attract.
    • Non-existence of Monopoles: Isolated magnetic poles (monopoles) do not exist. Cutting a bar magnet produces two smaller, complete dipoles.
  • 5.2.1 Magnetic Field Lines — 4 Key Properties:
    1. Continuous Closed Loops: Magnetic field lines of a magnet (or solenoid) form continuous closed loops (running $N \to S$ outside and $S \to N$ inside the magnet). Contrast with electrostatics: Electric field lines begin on positive charges and terminate on negative charges (never forming closed loops).
    2. Direction: The tangent to a field line at any point gives the direction of the net magnetic field $\mathbf{B}$ at that point.
    3. Magnitude/Density: The number of field lines crossing per unit normal area represents the strength of $\mathbf{B}$ (lines crowd where $\mathbf{B}$ is strong).
    4. Non-Intersection: Two magnetic field lines never cross each other (otherwise $\mathbf{B}$ would have two ambiguous directions at the intersection).
Fig 5.3: Magnetic Field Lines for Bar Magnet and Equivalent Current-Carrying Solenoid
S N (a) Bar Magnet S N (b) Solenoid (m = N I A)
Equivalence & Derivations

5.2.2 Bar Magnet as an Equivalent Finite Solenoid & Dipole Fields

  • Ampere's Hypothesis: All magnetic phenomena arise from circulating microscopic current loops. A bar magnet is equivalent to a cylindrical sheet of circulating currents (a solenoid).
  • Axial Field of a Finite Solenoid / Bar Magnet ($r \gg l$):
    For a solenoid of radius $a$, length $2l$, $n$ turns per unit length carrying current $I$:
    Total magnetic moment $m = (n \times 2l) I (\pi a^2) = N I A$.
    The axial field at distance $r$ from center is:
$$\mathbf{B}_A = \frac{\mu_0}{4\pi} \frac{2\mathbf{m}}{r^3} \quad (r \gg l)$$
  • Equatorial Field of a Bar Magnet ($r \gg l$):
$$\mathbf{B}_E = -\frac{\mu_0}{4\pi} \frac{\mathbf{m}}{r^3} \quad (r \gg l)$$
Torque & Potential Energy

5.2.3 Dipole in Uniform Magnetic Field & Electrostatic Analogy

  • Torque on Magnetic Needle: When a needle of magnetic moment $\mathbf{m}$ is placed in uniform $\mathbf{B}$: $$\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}, \qquad \tau = m B \sin\theta$$ Note: Net translatory force is strictly zero ($\mathbf{F}_{\text{net}} = \mathbf{0}$) in a uniform field.
  • Magnetic Potential Energy ($U_m$): $$U_m = \int \tau(\theta)\,d\theta = \int m B \sin\theta\,d\theta = -m B \cos\theta = -\mathbf{m}\cdot\mathbf{B}$$
    • Stable Equilibrium ($\theta = 0^\circ$): $U_m = -mB$ (minimum energy, $\mathbf{m} \parallel \mathbf{B}$).
    • Unstable Equilibrium ($\theta = 180^\circ$): $U_m = +mB$ (maximum energy, $\mathbf{m}$ antiparallel to $\mathbf{B}$).
    • Zero Reference ($\theta = 90^\circ$): $U_m = 0$.
  • Electrostatic vs Magnetism Analogy (Table 5.1):
    Quantity / Relation Electrostatics Magnetism
    Constant Factor 1 / ε₀ μ₀
    Dipole Moment p m
    Axial Field ($r \gg l$) 2p / (4πε₀ r³) μ₀ (2m) / (4π r³)
    Equatorial Field ($r \gg l$) −p / (4πε₀ r³) −μ₀ m / (4π r³)
    Torque in Field p × E m × B
    Potential Energy −p · E −m · B
Example 5.1

Conceptual Questions on Bar Magnets & Poles

(a) What happens if a bar magnet is cut into two pieces: (i) transverse to length, (ii) along length? (b) Why does a magnetized needle in uniform field experience only torque, while an iron nail near a bar magnet experiences both force and torque? (c) Must every magnetic configuration have a North and South pole? What about a toroid? (d) How to determine which of two identical-looking iron bars A and B is magnetized?

(a) Cutting Magnet: In either case (transverse or longitudinal), we obtain two complete magnets, each with its own North and South poles (magnetic monopoles do not exist).
(b) Iron Nail in Non-Uniform Field: The bar magnet produces a non-uniform field. The nail acquires induced dipole moment. The induced pole closer to the magnet experiences stronger attractive force than the farther repelled pole $\implies$ net attractive force in addition to torque.
(c) North/South Poles Requirement: Not necessarily. Poles exist only when the source has a net non-zero dipole moment. An ideal toroid or an infinitely long straight wire has continuous closed circular field lines without any North or South poles.
(d) Identifying Magnetized Bar: Touch an end of bar A to the center of bar B. In a bar magnet, magnetic field is concentrated at the poles and near zero at the center. If A experiences no attraction at the middle of B, then bar B is the magnet. If attraction is felt equally at the end and middle of B, then bar A is the magnet.
Example 5.2

Equilibrium & Potential Energy of Interacting Dipoles

A small magnetized needle P is placed at O. Identical needles Q are placed at positions $Q_1$ to $Q_6$ with different orientations. (a) In which configurations is the system not in equilibrium? (b) Which are in (i) stable, (ii) unstable equilibrium? (c) Which configuration corresponds to lowest potential energy?

Equilibrium Principle: System is in equilibrium when torque $\boldsymbol{\tau} = \mathbf{m}_Q \times \mathbf{B}_P = \mathbf{0}$ ($\mathbf{m}_Q$ is parallel or antiparallel to $\mathbf{B}_P$). Stable when $\mathbf{m}_Q \parallel \mathbf{B}_P$ ($U = -m B_P$), unstable when antiparallel ($U = +m B_P$).
Analysis:
  • (a) Not in equilibrium: $PQ_1$ and $PQ_2$ (torque $\ne 0$).
  • (b) (i) Stable: $PQ_3$ (equatorial, parallel to $\mathbf{B}_P$) and $PQ_6$ (axial, parallel to $\mathbf{B}_P$).
  • (b) (ii) Unstable: $PQ_4$ (equatorial, antiparallel) and $PQ_5$ (axial, antiparallel).
  • (c) Lowest potential energy: $PQ_6$ (axial field is twice as strong as equatorial field, so $U = -m(2B_E)$ is the deepest minimum).
Answers: (a) $PQ_1, PQ_2$; (b) Stable: $PQ_3, PQ_6$, Unstable: $PQ_4, PQ_5$; (c) Lowest PE: $PQ_6$.
5.3
NCERT Section

Magnetism and Gauss's Law

Fundamental Law & Monopole Absence

5.3.1 Gauss's Law for Magnetism & Physical Implications

Gauss's Law for Magnetism: The net magnetic flux through any closed surface $S$ is identically zero:

$$\Phi_B = \oint_S \mathbf{B} \cdot d\mathbf{S} = 0 \quad \text{[Eq. 5.6]}$$
  • Physical Significance:
    • There are no isolated magnetic monopoles (no sources or sinks of magnetic field).
    • Every magnetic field line that enters a closed surface must also exit it.
    • The simplest magnetic entity is a magnetic dipole or current loop.
  • Modification if Monopoles Existed: If magnetic charges (monopoles) $q_m$ existed in nature, Gauss's law would become $\oint \mathbf{B}\cdot d\mathbf{S} = \mu_0 q_m$ (analogous to $\oint \mathbf{E}\cdot d\mathbf{S} = q/\varepsilon_0$).
Example 5.3

Correct vs Incorrect Magnetic Field Line Patterns

Evaluate diagrams showing magnetic field lines in Fig. 5.6 and explain what is wrong with incorrect ones and which represent electrostatic lines.

(a) Radiating outwards from a point: Wrong for $\mathbf{B}$ (violates $\oint \mathbf{B}\cdot d\mathbf{S} = 0$); correctly represents electric field of a positive point charge.
(b) Intersecting & loops in empty space: Wrong. Lines cannot intersect, and static magnetic field lines cannot form closed loops in empty space without enclosing a current.
(c) Toroid: Correct. Field lines form closed concentric loops completely confined within toroidal windings.
(d) Straight finite solenoid ends: Wrong. Lines at solenoid ends must flare outward to form closed loops; straight truncated lines violate Ampere's law.
(e) Bar magnet: Correct. Continuous closed loops through interior and exterior.
(f) Field between plates: Wrong for $\mathbf{B}$ (emanates from plate); correctly represents electrostatic field between capacitor plates.
(g) Straight pole piece edges: Wrong. Magnetic lines must exhibit edge fringing; strictly straight lines at boundaries violate Ampere's circuital law.
Example 5.4

Conceptual Questions on Magnetic Forces & Moments

(a) Do magnetic field lines represent lines of force on a moving charge? (b) How would Gauss's law change if monopoles existed? (c) Does a bar magnet exert torque on itself? (d) Can a system have net magnetic moment if net charge is zero?

(a) Lines of Force: No. Magnetic force $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ is always perpendicular to $\mathbf{B}$, so field lines do not point along force.
(b) Monopole Gauss Law: $\oint \mathbf{B}\cdot d\mathbf{S} = \mu_0 q_m$, where $q_m$ is enclosed magnetic pole strength.
(c) Self-Torque: No, an element cannot exert net force or torque on itself. Different elements of a curved current wire do exert forces on one another, but self-torque of a rigid magnet is zero.
(d) Zero Net Charge with Magnetic Moment: Yes. Neutrons have zero net charge but possess non-zero magnetic moment due to internal constituent quarks. Paramagnetic atoms with neutral electron-proton balance have net orbital and spin magnetic moments.
5.4
NCERT Section

Magnetisation and Magnetic Intensity

Magnetic Vectors & Permeability Relations

5.4.1 Definitions: $\mathbf{M}, \mathbf{H}, \mathbf{B}, \chi, \mu_r, \mu$

  • Magnetisation ($\mathbf{M}$): Net magnetic moment per unit volume of the sample: $$\mathbf{M} \equiv \frac{\mathbf{m}_{\text{net}}}{V} \quad (\text{SI Unit: }\text{A m}^{-1}, \text{ Dimensions: }[L^{-1} A])$$
  • Magnetic Intensity ($\mathbf{H}$): Field contributed purely by external currents (e.g. solenoid windings): $$\mathbf{H} \equiv \frac{\mathbf{B}}{\mu_0} - \mathbf{M} \implies \mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M}) \quad (\text{SI Unit: }\text{A m}^{-1})$$
  • Magnetic Susceptibility ($\chi$): Dimensionless measure of material response to external field: $$\mathbf{M} = \chi \mathbf{H}$$
    • Diamagnetic: $\chi$ is small and negative ($-1 \le \chi < 0$).
    • Paramagnetic: $\chi$ is small and positive ($0 < \chi < \varepsilon$).
    • Ferromagnetic: $\chi$ is very large and positive ($\chi \gg 1$).
  • Relative Permeability ($\mu_r$) & Total Permeability ($\mu$): $$\mathbf{B} = \mu_0(1 + \chi)\mathbf{H} = \mu_0 \mu_r \mathbf{H} = \mu \mathbf{H}$$
$$\mu_r = 1 + \chi \quad (\text{Dimensionless}), \qquad \mu = \mu_0 \mu_r = \mu_0 (1 + \chi) \quad (\text{SI Unit: }\text{T m A}^{-1} \equiv \text{N A}^{-2})$$
Example 5.5

Solenoid with Iron Core — Field Calculations

A solenoid with core of $\mu_r = 400$, $n = 1000\text{ turns/m}$, carries $I = 2\text{ A}$. Calculate: (a) Magnetic intensity $H$, (b) Magnetic field $B$, (c) Magnetisation $M$, (d) Magnetising current $I_M$.

(a) Magnetic Intensity ($H$): Depends only on external winding current:
$H = n I = 1000\text{ m}^{-1} \times 2.0\text{ A} = 2.0 \times 10^3\text{ A m}^{-1}$.
(b) Total Magnetic Field ($B$):
$B = \mu_r \mu_0 H = 400 \times (4\pi \times 10^{-7}\text{ T m A}^{-1}) \times (2 \times 10^3\text{ A m}^{-1}) = 1.005\text{ T} \approx 1.0\text{ T}$.
(c) Magnetisation ($M$):
$M = (\mu_r - 1)H = (400 - 1) \times (2 \times 10^3) = 399 \times 2000 = 7.98 \times 10^5\text{ A m}^{-1} \approx 8.0 \times 10^5\text{ A m}^{-1}$.
(d) Magnetising Current ($I_M$): Additional current without core to achieve same $B$ field:
$B = \mu_0 n (I + I_M) \implies 1.0 = (4\pi \times 10^{-7}) \times 1000 \times (2 + I_M) \implies 2 + I_M = 796 \implies I_M \approx 794\text{ A}$.
Answers: (a) $H = 2 \times 10^3\text{ A/m}$, (b) $B = 1.0\text{ T}$, (c) $M \approx 8 \times 10^5\text{ A/m}$, (d) $I_M = 794\text{ A}$.
5.5
NCERT Section

Magnetic Properties of Materials

Classification & Microscopic Origins

5.5.1 Diamagnetism, Paramagnetism & Ferromagnetism

Property Diamagnetic Paramagnetic Ferromagnetic
Susceptibility ($\chi$) −1 ≤ χ < 0 (small, negative) 0 < χ < ε (small, positive) χ ≫ 1 (very large, positive)
Relative Permeability ($\mu_r$) 0 ≤ μᵣ < 1 1 < μᵣ < 1 + ε μᵣ ≫ 1 (> 1000)
Behavior in Field Repelled weakly; moves from stronger to weaker field Attracted weakly; moves from weaker to stronger field Attracted strongly; moves strongly to high field
Field Lines Pattern Field lines expelled/repelled from material Field lines slightly concentrated inside Field lines highly concentrated inside
Atomic Dipole Moment Zero in absence of field; induced opposite to $\mathbf{B}$ Permanent dipole moment; random thermal orientation Permanent dipole moment; aligned in macroscopic domains
Examples $\text{Bi, Cu, Pb, Si, H}_2\text{O, NaCl, Superconductors}$ $\text{Al, Na, Ca, O}_2\text{ (STP), CuCl}_2$ $\text{Fe, Co, Ni, Gd, Alnico, Lodestone}$

Key Microscopic Mechanisms:

  • Diamagnetism: Present in all materials. In orbiting electrons, applied $\mathbf{B}$ speeds up electrons with opposite magnetic moments and slows down those with parallel moments (Lenz's Law) $\implies$ net induced moment opposite to $\mathbf{B}$.
    Meissner Effect (Superconductors): Perfect diamagnetism ($\chi = -1, \mu_r = 0$). External field is totally expelled from the interior.
  • Paramagnetism: Individual atoms have permanent dipoles. Random thermal motion prevents net magnetization. Applied $\mathbf{B}$ aligns dipoles along field until saturation is reached at low temperatures.
  • Ferromagnetism & Domains: Atoms interact cooperatively to form macroscopic domains ($\sim 1\text{ mm}$, containing $\sim 10^{11}$ atoms). In applied field, domains rotate and grow in size.
    • Hard ferromagnets: Retain magnetization after field removal (e.g. Alnico, permanent magnets).
    • Soft ferromagnets: Magnetization disappears when field is removed (e.g. Soft iron, transformer cores).
    • Temperature effect: At high temperature, thermal agitation disintegrates domain structure and ferromagnet transforms gradually into a paramagnet.
Fig 5.12: Magnetic Field Lines Expulsion in Diamagnetic & Concentration in Paramagnetic / Ferromagnetic Materials
Diamagnetic μ_r < 1 (χ < 0) Paramagnetic μ_r > 1 (χ > 0, small) Ferromagnetic μ_r >> 1000 (Domains)
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering bar magnets, magnetic field lines, dipole fields & torque, Gauss's law of magnetism, magnetic intensity, and material classification with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 5.1 – 5.7

Complete stepwise solutions for every textbook exercise question in NCERT Class 12 Physics Chapter 5 Magnetism and Matter Reprint 2026-27.

7 QUESTIONS
Ex 5.1 Magnetic Moment from Torque

5.1 A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25\text{ T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2}\text{ J}$. What is the magnitude of magnetic moment of the magnet?

Magnetic Moment $m = 0.36\text{ J T}^{-1} = 0.36\text{ A m}^2$.
Formula: Torque on magnetic dipole in uniform field: $$\tau = m B \sin\theta \implies m = \frac{\tau}{B \sin\theta}$$
Given: $\tau = 4.5 \times 10^{-2}\text{ J}$, $B = 0.25\text{ T}$, $\theta = 30^\circ$ ($\sin 30^\circ = 0.5$).
Calculation: $$m = \frac{4.5 \times 10^{-2}\text{ J}}{0.25\text{ T} \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36\text{ J T}^{-1}$$
Ex 5.2 Stable and Unstable Equilibrium

5.2 A short bar magnet of magnetic moment $m = 0.32\text{ J T}^{-1}$ is placed in a uniform magnetic field of $0.15\text{ T}$. 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: $\theta = 0^\circ, U = -4.8 \times 10^{-2}\text{ J}$ • (b) Unstable: $\theta = 180^\circ, U = +4.8 \times 10^{-2}\text{ J}$.
Formula: Magnetic potential energy: $$U = -\mathbf{m}\cdot\mathbf{B} = -m B \cos\theta$$
(a) Stable Equilibrium ($\theta = 0^\circ$):
Magnetic moment $\mathbf{m}$ is parallel to field $\mathbf{B}$. $$U = -m B \cos 0^\circ = - (0.32\text{ J T}^{-1}) \times (0.15\text{ T}) = -0.048\text{ J} = -4.8 \times 10^{-2}\text{ J}$$
(b) Unstable Equilibrium ($\theta = 180^\circ$):
Magnetic moment $\mathbf{m}$ is antiparallel to field $\mathbf{B}$. $$U = -m B \cos 180^\circ = - (0.32) \times (0.15) \times (-1) = +0.048\text{ J} = +4.8 \times 10^{-2}\text{ J}$$
Ex 5.3 Solenoid Magnetic Moment

5.3 A closely wound solenoid of 800 turns and area of cross-section $2.5 \times 10^{-4}\text{ m}^2$ carries a current of $3.0\text{ A}$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Magnetic Moment $m = 0.60\text{ J T}^{-1} = 0.60\text{ A m}^2$ (along axis).
Solenoid as Bar Magnet: A current-carrying solenoid sets up a magnetic field pattern identical to that of a bar magnet, with magnetic field lines emerging from one face (acting as North pole) and entering the other face (acting as South pole), forming continuous closed loops.
Magnetic Moment Formula: $$m = N I A$$
Calculation: $$m = 800 \times (3.0\text{ A}) \times (2.5 \times 10^{-4}\text{ m}^2) = 2400 \times 2.5 \times 10^{-4} = 0.60\text{ A m}^2 = 0.60\text{ J T}^{-1}$$ Directed along the axis of the solenoid according to the right-hand grip rule.
Ex 5.4 Torque on Solenoid

5.4 If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25\text{ T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^\circ$ with the direction of applied field?

Torque $\tau = 7.5 \times 10^{-2}\text{ N m} = 0.075\text{ J}$.
Formula: $$\tau = m B \sin\theta$$
Calculation: $$\tau = (0.60\text{ J T}^{-1}) \times (0.25\text{ T}) \times \sin 30^\circ = 0.15 \times 0.5 = 0.075\text{ N m} = 7.5 \times 10^{-2}\text{ N m}$$ Torque tends to align the axis of the solenoid parallel to the magnetic field.
Ex 5.5 Work Done in Rotating Dipole & Torque

5.5 A bar magnet of magnetic moment $1.5\text{ J T}^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22\text{ T}$.
(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)?

(a) (i) $W = 0.33\text{ J}$, (ii) $W = 0.66\text{ J}$ • (b) (i) $\tau = 0.33\text{ N m}$, (ii) $\tau = 0\text{ N m}$.
Work Done Formula: $W = -m B (\cos\theta_2 - \cos\theta_1) = m B (\cos\theta_1 - \cos\theta_2)$. Initial angle $\theta_1 = 0^\circ$.
$m B = 1.5 \times 0.22 = 0.33\text{ J}$.
(a) (i) Normal to field ($\theta_2 = 90^\circ$):
$W = 0.33 \times (\cos 0^\circ - \cos 90^\circ) = 0.33 \times (1 - 0) = 0.33\text{ J}$.
(a) (ii) Opposite to field ($\theta_2 = 180^\circ$):
$W = 0.33 \times (\cos 0^\circ - \cos 180^\circ) = 0.33 \times (1 - (-1)) = 0.33 \times 2 = 0.66\text{ J}$.
(b) Torque $\tau = m B \sin\theta$:
(i) $\theta = 90^\circ \implies \tau = 0.33 \times \sin 90^\circ = 0.33\text{ N m}$ (tending to align back to field).
(ii) $\theta = 180^\circ \implies \tau = 0.33 \times \sin 180^\circ = 0\text{ N m}$.
Ex 5.6 Suspended Solenoid Force and Torque

5.6 A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4}\text{ m}^2$, carrying a current of $4.0\text{ A}$, 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 $7.5 \times 10^{-2}\text{ T}$ is set up at an angle of $30^\circ$ with the axis of the solenoid?

(a) $m = 1.28\text{ A m}^2 = 1.28\text{ J T}^{-1}$ • (b) Force $F = 0\text{ N}$, Torque $\tau = 0.048\text{ N m} = 4.8 \times 10^{-2}\text{ N m}$.
(a) Magnetic Moment ($m$):
$$m = N I A = 2000 \times 4.0\text{ A} \times (1.6 \times 10^{-4}\text{ m}^2) = 8000 \times 1.6 \times 10^{-4} = 1.28\text{ A m}^2$$ Directed along the magnetic axis of the solenoid.
(b) Net Force ($F$):
In a uniform magnetic field, the net translational force on a magnetic dipole is strictly zero ($F = 0\text{ N}$).
Torque ($\tau$):
$$\tau = m B \sin\theta = 1.28\text{ J T}^{-1} \times (7.5 \times 10^{-2}\text{ T}) \times \sin 30^\circ$$ $$\tau = 0.096 \times 0.5 = 0.048\text{ N m} = 4.8 \times 10^{-2}\text{ N m}$$ Direction: In a sense tending to align the solenoid axis parallel to the magnetic field.
Ex 5.7 Short Bar Magnet Axial and Equatorial Fields

5.7 A short bar magnet has a magnetic moment of $0.48\text{ J T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10\text{ cm}$ from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

(a) $B_A = 0.96 \times 10^{-4}\text{ T} = 0.96\text{ G}$ (along $S \to N$) • (b) $B_E = 0.48 \times 10^{-4}\text{ T} = 0.48\text{ G}$ (along $N \to S$).
Given: $m = 0.48\text{ J T}^{-1}$, $r = 10\text{ cm} = 0.1\text{ m} \implies r^3 = 10^{-3}\text{ m}^3$, $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T m A}^{-1}$.
(a) On the Axis ($B_A$): $$B_A = \frac{\mu_0}{4\pi} \frac{2m}{r^3} = 10^{-7} \times \frac{2 \times 0.48}{10^{-3}} = 10^{-7} \times \frac{0.96}{10^{-3}} = 0.96 \times 10^{-4}\text{ T} = 0.96\text{ G}$$ Direction: Directed along the magnetic moment vector $\mathbf{m}$ (from South to North pole of the magnet).
(b) On the Equatorial Line ($B_E$): $$B_E = \frac{\mu_0}{4\pi} \frac{m}{r^3} = 10^{-7} \times \frac{0.48}{10^{-3}} = 0.48 \times 10^{-4}\text{ T} = 0.48\text{ G}$$ Direction: Directed opposite to the magnetic moment vector $\mathbf{m}$ (from North to South pole, parallel to the magnet axis).
04 / Rapid Reference

Chapter Summary & Points to Ponder

8 core summary principles, Physical Quantities dimension and unit table, and all 7 official NCERT Points to Ponder for Chapter 5.

100% SYLLABUS

Bar Magnet Dipole Field

B_A = 2μ₀m / (4πr³) • B_E = −μ₀m / (4πr³)

Axial field is twice the equatorial field at the same large distance ($r \gg l$).

Dipole Torque & Energy

τ = m × B • U = −m · B

Stable at $\theta = 0^\circ$ ($U = -mB$); Unstable at $\theta = 180^\circ$ ($U = +mB$).

Gauss's Law for Magnetism

∮ B · dS = 0

Continuous closed loops; isolated magnetic monopoles do not exist in nature.

Magnetisation Vector

M = m_net / V (A m⁻¹)

Total magnetic dipole moment per unit volume of material.

Magnetic Intensity & Total B

B = μ₀(H + M) = μH

$H = nI$ (external winding field). Total field $B = \mu_0 \mu_r H$.

Susceptibility & Permeability

μᵣ = 1 + χ • μ = μ₀μᵣ

$\chi < 0$ (Diamagnetic), $\chi > 0$ small (Paramagnetic), $\chi \gg 1$ (Ferromagnetic).

Diamagnetism & Meissner Effect

χ = −1 • μᵣ = 0 (Superconductors)

Universal weak effect; field lines expelled. Superconductors exhibit perfect diamagnetism.

Ferromagnetic Domains

Domains (~1 mm, 10¹¹ atoms)

Hard magnets retain magnetization (Alnico); Soft magnets lose it easily (Soft iron).

NCERT Official Table

Physical Quantities, Symbols, Dimensions & Units

Physical Quantity Symbol Dimensions SI Unit Remark
Permeability of free space μ₀ $[M L T^{-2} A^{-2}]$ T m A−1 (or N A−2) $\mu_0 / 4\pi = 10^{-7}\text{ T m A}^{-1}$
Magnetic Field / Induction B $[M T^{-2} A^{-1}]$ T (Tesla) $1\text{ T} = 10^4\text{ Gauss (G)}$
Magnetic Moment m $[L^2 A]$ A m2 (or J T−1) Vector ($\mathbf{m} = N I \mathbf{A}$)
Magnetic Flux φB $[M L^2 T^{-2} A^{-1}]$ Wb (Weber $\equiv$ T m2) Scalar ($\Phi_B = \mathbf{B}\cdot\mathbf{S}$)
Magnetisation M $[L^{-1} A]$ A m−1 $\mathbf{M} = \mathbf{m}_{\text{net}}/V$
Magnetic Intensity H $[L^{-1} A]$ A m−1 $\mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M})$
Magnetic Susceptibility χ $[M^0 L^0 T^0 A^0]$ Dimensionless $\mathbf{M} = \chi \mathbf{H}$
Relative Permeability μᵣ $[M^0 L^0 T^0 A^0]$ Dimensionless $\mu_r = 1 + \chi = \mu / \mu_0$
Magnetic Permeability μ $[M L T^{-2} A^{-2}]$ T m A−1 (or N A−2) $\mathbf{B} = \mu \mathbf{H} = \mu_0 \mu_r \mathbf{H}$
NCERT Official

Points to Ponder

  1. Engineering Preceding Science: Technological exploitation of the directional properties of magnets (compasses) predated scientific theoretical understanding by over 2000 years.
  2. Non-Existence of Monopoles: Magnetic monopoles do not exist. Slicing a magnet produces two complete magnets. On the other hand, electric charges exist as isolated fundamental monopoles with quantised unit $|e| = 1.6 \times 10^{-19}\text{ C}$.
  3. Continuous Closed Loops: As a direct consequence of the non-existence of magnetic monopoles ($\nabla \cdot \mathbf{B} = 0$), magnetic field lines are continuous and form closed loops (running $S \to N$ inside and $N \to S$ outside). Electrostatic lines terminate on charges.
  4. Sensitivity of $\chi$: A minuscule difference in the value of magnetic susceptibility $\chi$ leads to fundamentally divergent behaviors: $\chi = -10^{-5}$ yields diamagnetism, whereas $\chi = +10^{-5}$ yields paramagnetism.
  5. Superconductors & Meissner Effect: Superconductors are perfect diamagnets ($\chi = -1, \mu_r = 0$) and perfect conductors. Magnetic fields are completely expelled from the interior (Meissner effect), explained by BCS quantum theory (1957).
  6. Universality of Diamagnetism: Diamagnetism is universal and present in all substances due to orbiting atomic electrons, but is easily masked in materials possessing stronger paramagnetism or ferromagnetism.
  7. Exotic Magnetic States: Beyond diamagnetic, paramagnetic, and ferromagnetic classifications, exotic magnetic states like ferrimagnetism, anti-ferromagnetism, and spin glasses exist in specialized materials.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

24 QUESTIONS
Level 1 Active • 8 Questions