NCERT Exemplar Solutions Class 12 Physics Magnetism and Matter — Multiple Choice Questions Q1–10

NCERT Exemplar Problems, Class 12 Physics — Magnetism and Matter, MCQ I and MCQ II. All 10 questions solved.

The book numbers these questions 5.1 to 5.10; here they are Questions 1 to 10. Questions 1 to 5 are MCQ I (exactly one option is correct) and Questions 6 to 10 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Gauss’s law for magnetism. $\nabla\cdot\mathbf{B}=0$, so lines of $\mathbf{B}$ never begin or end. This decides Questions 4, 6 and 9.
• Dipole fields. Any non-zero magnetic moment gives a field that falls as $1/r^3$ far away, which settles the toroid in Question 1.
• Origin and types of magnetism. Orbital currents and electron spin are the sources; domains, Curie’s law and the Curie temperature explain permanent magnets and temperature effects (Questions 3, 5, 7, 8).
• Earth as a dipole. The tilt of the dipole axis fixes the range of declination and dip (Questions 2 and 10).

Key insight. Most of these are not calculations but a decision about what is continuous, what can end, and what is merely an idealisation. Questions 4 and 9 turn on it.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

A toroid of $n$ turns, mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $x$-$y$ plane. Its magnetic moment $\mathbf{m}$

Solution. The field of an ideal toroid is confined entirely inside the winding; outside it is zero. Now suppose the toroid had a non-zero moment $\mathbf{m}$, in whatever direction. Any current distribution with a net moment produces, far away, a dipole field that falls as $1/r^3$. That contradicts the zero field outside, so the net moment must be zero. This is also reasonable physically: the current runs round the tube in loops whose individual moments, pointing round the ring, cancel in pairs on opposite sides of the toroid. Options (a), (b) and (d) each assign a definite non-zero direction and so each would give an external field.

Question 2

The magnetic field of Earth can be modelled by that of a point dipole placed at the centre of the Earth. The dipole axis makes an angle of $11.3^\circ$ with the axis of Earth. At Mumbai, declination is nearly zero. Then,

Earth’s magnetism is no longer in the current NCERT textbook or the JEE Main syllabus; this question is solved for completeness.

Solution. Declination is the angle between geographic north and the horizontal direction of the magnetic field at a place. If the dipole axis is tilted $11.3^\circ$ from the rotation axis, the largest angle by which the magnetic meridian can swing away from the geographic meridian is $11.3^\circ$, to the west on one side of the Earth and to the east on the other. So the declination ranges from $11.3^\circ$ W to $11.3^\circ$ E: (a) is true.

Declination at Mumbai is nearly zero, which means that Mumbai lies in the plane containing the two axes, the dipole axis and the rotation axis; there the two meridians coincide. That says nothing about Greenwich, so (c) is not implied. Option (b) is false because the declination takes negative (west) as well as positive (east) values, so its least value is $-11.3^\circ$, not $0^\circ$. Option (d) is false because, by the symmetry of the model, the declination is as often west as east and its average over the Earth does not have to be negative.

Question 3

In a permanent magnet at room temperature

Solution. A permanent magnet is a ferromagnet, so its atoms do carry moments, which rules out (a). Inside it the moments are lined up within small regions called domains, each of which is saturated. Whether the whole piece is magnetised depends on how the domains are arranged. In a permanent magnet, they are only partially aligned: more domains point one way than the other, which leaves a net moment, but the alignment is far from perfect. Thermal agitation at room temperature prevents perfect alignment, and the domains and moments are never all lined up (that would require absolute zero and no defects), so (b) and (d) are false.

Question 4

Consider the two idealised systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length $L \gg R$, radius of cross-section. In (i) $\mathbf{E}$ is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below:

Solution. Take each in turn.

• (a) In the capacitor, the field lines begin and end on the charges on the plates, so a closed surface that cuts through the field region encloses charge. Gauss’s law $\oint\mathbf{E}\cdot d\mathbf{S}=q/\varepsilon_0$ is satisfied. False.
• (b) In the solenoid, $\mathbf{B}$ is uniform inside and zero outside. Take a closed surface round one end of the solenoid: flux enters the surface through the inside face and nothing leaves through the outside, so $\oint\mathbf{B}\cdot d\mathbf{S}\neq 0$. That is impossible for magnetic fields, since lines of $\mathbf{B}$ never end. In a real solenoid the lines spread out and close round outside. True.
• (c) A loop lying partly between the plates and partly outside would give $\oint\mathbf{E}\cdot d\mathbf{l}=Ed\neq0$, because $\mathbf{E}$ is zero on the outside part. So case (i) contradicts the law $\oint\mathbf{E}\cdot d\mathbf{l}=0$ rather than agreeing with it. False.
• (d) A loop passing inside the solenoid and returning outside threads the windings, and $\oint\mathbf{H}\cdot d\mathbf{l}=nIl=I_{en}$ holds. No contradiction. False.

Question 5

A paramagnetic sample shows a net magnetisation of $8\ \text{A m}^{-1}$ when placed in an external magnetic field of $0.6$ T at a temperature of $4$ K. When the same sample is placed in an external magnetic field of $0.2$ T at a temperature of $16$ K, the magnetisation will be

Solution. Curie’s law for a paramagnet gives $M = C\,B/T$, so for the same sample $M\propto B/T$. Therefore

$$M_2 = M_1\cdot\frac{B_2}{B_1}\cdot\frac{T_1}{T_2} = 8\times\frac{0.2}{0.6}\times\frac{4}{16} = 8\times\frac13\times\frac14=\frac23\ \text{A m}^{-1}.$$

The field is three times smaller and the temperature four times higher, so the magnetisation falls by a factor of $12$, from $8$ to $\tfrac23$.

Question 6

$S$ is the surface of a lump of magnetic material.

Solution. Gauss’s law for magnetism, $\nabla\cdot\mathbf{B}=0$, says that lines of $\mathbf{B}$ are closed and never begin or end, so they cannot break at a surface: (a) is true and (b) is false. The auxiliary field is defined by $\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}$. Across the surface of the lump, $\mathbf{B}$ is continuous but $\mathbf{M}$ drops abruptly from its value inside to zero outside, so $\mathbf{H}$ jumps. Lines of $\mathbf{H}$ therefore cannot all be continuous across $S$: (d) is true and (c) is false. Equivalently, the lines of $\mathbf{H}$ start on the north-pole surface and end on the south-pole surface, like the lines of $\mathbf{E}$ from surface charges.

Question 7

The primary origin(s) of magnetism lies in

Solution. Magnetism arises from moving charge. The electrons’ orbital motion around the nucleus acts as a tiny current loop, the atomic current, with an orbital magnetic moment: (a). In addition, an electron has an intrinsic spin that carries its own magnetic moment: (d). The Pauli exclusion principle governs how electrons fill levels and so influences which atoms end up with unpaired spins, but it is not itself a source of magnetism. The polar nature of molecules concerns electric dipoles and has nothing to do with magnetic moments.

Question 8

A long solenoid has $1000$ turns per metre and carries a current of $1$ A. It has a soft iron core of $\mu_r = 1000$. The core is heated beyond the Curie temperature, $T_c$.

Solution. The field $\mathbf{H}=nI=1000\ \text{A m}^{-1}$ is fixed by the free current alone, so it does not change when the core is heated: nearly unchanged. Below $T_c$ the iron is ferromagnetic with $\mu_r=1000$, so $\chi\approx10^3$ and $B=\mu_0\mu_rH$ is large. Above $T_c$ the domains break up and the iron turns paramagnetic, with $\chi$ of only about $10^{-5}$ and $\mu_r\approx1$. Then $B$ falls to about $\mu_0H$, a drop by a factor of about $1000$, so (a) is true and (b) is false.

The magnetisation is $M=\chi H$. It falls from $\chi H\approx10^3H$ to about $10^{-5}H$, a factor of $10^3/10^{-5}=10^8$: (d) is true. It diminishes but does not change sign, since a paramagnet magnetises along the field, so (c) is false.

Question 9

Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to

Solution. In a conductor the free charges rearrange themselves until the field inside is zero. This is possible because electric field lines end on charges, so the induced surface charges terminate the lines at the shell: (a) is true. Magnetic field lines are different: they never end, since there are no free magnetic charges, so nothing on the shell can terminate them, and a perfect shield of the interior is not possible: (c) is true and (b) is false. The best that can be done is to give the lines an easier path, by making the shell of a material of high permeability, such as soft iron. The lines crowd into the shell wall and are diverted round the interior region, which greatly reduces the field inside: (d) is true.

Question 10

Let the magnetic field on earth be modelled by that of a point magnetic dipole at the centre of earth. The angle of dip at a point on the geographical equator

Earth’s magnetism is no longer in the current NCERT textbook or the JEE Main syllabus; this question is solved for completeness.

Solution. For a dipole, the dip $\delta$ at a place whose magnetic latitude is $\lambda$ satisfies $\tan\delta=2\tan\lambda$. The dip is zero only on the magnetic equator. The magnetic axis is tilted $11.3^\circ$ from the rotation axis, so the magnetic equator is tilted by the same angle to the geographical equator, and the two great circles cross at two points. At those two points the dip is zero: (b) is true, and (a) is false since the dip is not zero elsewhere on the geographical equator.

Between the crossing points, half of the geographical equator lies on the north side of the magnetic equator and the other half on the south side. The dip is positive (field pointing into the ground) on one side and negative on the other: (c) is true. Along the geographical equator the magnetic latitude never exceeds $11.3^\circ$ in magnitude, so $|\tan\delta|\le2\tan11.3^\circ$, which keeps the dip within a small range: (d) is true.

Common mistakes

• Giving a toroid a moment in Question 1. Its field is confined inside, so its net moment is zero; a non-zero moment would send a $1/r^3$ field outside.
• Saying the least declination is $0^\circ$ in Question 2. It ranges to the west as well, down to $-11.3^\circ$ (that is, $11.3^\circ$ W).
• Thinking the idealised capacitor breaks Gauss’s law in Question 4. Its field lines end on plate charges; it is the solenoid, whose lines would have to end at its mouth, that breaks Gauss’s law for $\mathbf{B}$.
• Forgetting that $M\propto B/T$ in Question 5. Changing both field and temperature changes the magnetisation by the product of the two ratios.
• Treating $\mathbf{H}$ as continuous like $\mathbf{B}$ in Question 6. $\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}$ jumps where $\mathbf{M}$ does.
• Saying $\mathbf{B}$ is unchanged above the Curie temperature in Question 8. $\mathbf{H}=nI$ is fixed by the current, but $\mathbf{B}$ falls with the permeability.
• Believing a thick conducting shell shields against static magnetic fields in Question 9. Lines of $\mathbf{B}$ cannot be terminated; they can only be diverted through a high-permeability shell.

Practise next

• Exemplar Magnetism and Matter: Very Short, Short and Long Answer Questions — where the same ideas are applied to magnetisation currents, susceptibility, dipole fields and the oscillating magnet.
• NCERT textbook Chapter 5 — Magnetism and Matter for the theory of the bar magnet as a solenoid, dipole field and torque, and para, dia and ferromagnetism.
• Exemplar Moving Charges and Magnetism — the previous unit, whose currents and fields underlie the magnetic moment.
• Exemplar Electromagnetic Induction — the next unit, where changing magnetic flux drives an emf.