NCERT Exemplar Solutions Class 12 Physics Magnetism and Matter — Very Short, Short and Long Answer Questions Q11–25

NCERT Exemplar Problems, Class 12 Physics — Magnetism and Matter, Very Short Answer Questions 11 to 15, Short Answer Questions 16 to 20 and Long Answer Questions 21 to 25. All 15 questions solved. The book numbers them 5.11 to 5.25.

Three ideas carry this page. A magnetised body is a set of atomic currents, so magnetisation $M$ acts like a surface current $Ml$ and a bar magnet behaves like a solenoid. Magnetic field lines never end, which gives $\oint\mathbf B\cdot d\mathbf S=0$ and forces $\mathbf B$ lines to close inside a magnet while $\mathbf H$ lines run the other way. The dipole field and torque mirror the electric case, so oscillation periods, torques and energies carry over once $\mathbf p\to\mathbf m$ and $\mathbf E\to\mathbf B$ are matched.

Key insight. Always ask which quantity is conserved or closed. In Questions 16, 20 and 21 it is the flux or circulation of $\mathbf B$ and $\mathbf H$; in Questions 19 and 25 it is the oscillation formula $\omega^2 = mB/I$, where both the moment $m$ and the moment of inertia $I$ must be found from the geometry.

Very Short Answer Questions

Question 11

A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?

Solution. The magnetic moment of a particle of charge $e$ and mass $M$ is of order $e\hbar/2M$. It is inversely proportional to the mass. A proton is about $1836$ times heavier than an electron, so $\mu_p \approx e\hbar/2m_p$ is about $1/1836$ of $\mu_e \approx e\hbar/2m_e$. The nuclear contribution to the magnetisation of a material is therefore thousands of times smaller than the electronic one and is neglected.

Question 12

A permanent magnet in the shape of a thin cylinder of length 10 cm has $M = 10^{6}$ A/m. Calculate the magnetisation current $I_M$.

Solution. Magnetisation is magnetic moment per unit volume. For a thin cylinder of cross-section $A$ and length $l$, the total moment is $MAl$. The same moment is produced by a surface current $I_M$ circulating round the curved surface, which behaves like a one-turn-per-length solenoid of moment $I_M A$. So $I_M A = MAl$ and

$$I_M = Ml = 10^{6}\times 0.1 = 10^{5}\ \text{A}.$$

This is a bound (Amperian) surface current. There is no free current, so $H=0$ for the magnet and all of $B$ comes from $I_M$.

Question 13

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of $\text{N}_2$ ($\sim 5\times10^{-9}$) (at STP) and Cu ($\sim 10^{-5}$).

Solution. Diamagnetic susceptibility is the sum of tiny induced moments of the atoms or molecules, so $\chi$ is proportional to the number of them per unit volume. Compare number densities.

For $\text{N}_2$ at STP, one mole of $28$ g occupies $22400\ \text{cm}^3$, so the density is $28/22400 = 1.25\times10^{-3}\ \text{g cm}^{-3}$. Copper has density about $8\ \text{g cm}^{-3}$. Per unit mass the two are comparable (molar mass $28$ against $64$ is within a factor of about two), so the ratio of number densities is about

$$\frac{\rho_{N_2}}{\rho_{Cu}} = \frac{28/22400}{8} \approx 1.6\times10^{-4}.$$

The ratio of the susceptibilities is

$$\frac{\chi_{N_2}}{\chi_{Cu}} = \frac{5\times10^{-9}}{10^{-5}} = 5\times10^{-4}.$$

The two ratios are of the same order, so the difference in density accounts for the difference in $\chi$.

Question 14

From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism.

Solution.

• Diamagnetism. The moment is induced in every atom by the field, as in Lenz’s law, and opposes it. It does not depend on the thermal motion of the atoms, so $\chi$ is nearly independent of temperature.
• Paramagnetism. Atoms carry permanent moments that the field tries to line up while thermal motion tries to randomise them. A higher temperature gives less alignment, so $\chi$ falls as $T$ rises: Curie’s law, $\chi\propto 1/T$.
• Ferromagnetism. Below the Curie temperature the moments are locked parallel within domains by exchange interaction, and $\chi$ is large and decreases with temperature as the alignment weakens. Above the Curie temperature $T_c$ the domains break up and the substance becomes paramagnetic, with $\chi\propto 1/(T-T_c)$ (Curie–Weiss law).

Question 15

A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. (i) In which direction will it move? (ii) What will be the direction of its magnetic moment?

Solution. A superconductor is a perfect diamagnet ($\chi=-1$, the Meissner effect): it excludes the field from its interior by setting up surface currents.

(i) The induced moment is opposite to the field of the magnet, so the ball is repelled. It moves away from the magnet, towards the region of weaker field.

(ii) The induced moment is opposite to the field of the magnet at the ball. If the ball faces the north pole of the magnet, the field at the ball points away from the magnet’s north pole, so the ball’s moment points back towards the magnet. Its near face becomes a north pole, which is repelled by the magnet’s north pole.

Short Answer Questions

Question 16

Verify the Gauss’s law for magnetic field of a point dipole of dipole moment $\mathbf m$ at the origin for the surface which is a sphere of radius $R$.

Solution. Take $\mathbf m = m\hat{\mathbf k}$. The dipole field has radial and tangential parts,

$$B_r=\frac{\mu_0}{4\pi}\frac{2m\cos\theta}{r^3},\qquad B_\theta=\frac{\mu_0}{4\pi}\frac{m\sin\theta}{r^3}.$$

On the sphere only $B_r$ is normal to the surface. The flux is

$$\oint\mathbf B\cdot d\mathbf S=\int_0^{\pi}\frac{\mu_0}{4\pi}\frac{2m\cos\theta}{R^3}\,(2\pi R^2\sin\theta\,d\theta)=\frac{\mu_0}{4\pi}\frac{2m}{R}\,2\pi\int_0^{\pi}\cos\theta\sin\theta\,d\theta.$$

The integral is $\tfrac12\sin^2\theta\big|_0^\pi = 0$, so the flux is zero. Field lines leave through the upper hemisphere and enter through the lower one in equal measure, as Gauss’s law for magnetism ($\oint\mathbf B\cdot d\mathbf S=0$) requires.

Printed key error. Minor: the key writes the field as $\frac{\mu_0}{4\pi}\frac{3(\mathbf m\cdot\hat r)}{r^3}$ with a factor $3\cos\theta$. The full field is $\frac{\mu_0}{4\pi}\frac{3(\mathbf m\cdot\hat r)\hat r-\mathbf m}{r^3}$, whose radial component is $2m\cos\theta/r^3$. The result is zero either way.

Question 17

Three identical bar magnets are rivetted together at centre in the same plane as shown in Fig. 5.1. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the Fig. 5.1. Determine the poles of the remaining two.

Solution. The figure shows three magnets crossing at their centres at $60^\circ$ to one another. The vertical one has N at the top and S at the bottom. The other two run from upper-left to lower-right and from upper-right to lower-left, with their four ends marked “?”.

If the system shows no motion in a slowly varying field, no torque acts on it in any field, so $\boldsymbol\tau=\mathbf m_{net}\times\mathbf B=0$ for every $\mathbf B$. That needs $\mathbf m_{net}=0$. The three moments have equal magnitude $m$, so they must add to zero, which is only possible if they are at $120^\circ$ to one another as vectors.

The given magnet has its moment pointing up. The other two moments must each make $120^\circ$ with it, so each points downwards, one to the lower left and one to the lower right, at $30^\circ$ below the horizontal. A moment points from S to N inside a magnet, so the lower ends of these two magnets are N and their upper ends are S. Check: the horizontal parts cancel by symmetry and the vertical parts are $2\times(-m\sin30^\circ)=-m$, which cancels $+m$.

Question 18

Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole $\mathbf p$ in an electrostatic field $\mathbf E$ and (ii) magnetic dipole $\mathbf m$ in a magnetic field $\mathbf B$. Write down a set of conditions on $\mathbf E$, $\mathbf B$, $\mathbf p$, $\mathbf m$ so that the two motions are verified to be identical. (Assume identical initial conditions.)

Solution. The torque and energy are $\boldsymbol\tau=\mathbf p\times\mathbf E$, $U=-\mathbf p\cdot\mathbf E$ for the electric dipole, and $\boldsymbol\tau=\mathbf m\times\mathbf B$, $U=-\mathbf m\cdot\mathbf B$ for the magnetic one. In a uniform field the two dipoles then move identically if the torques and energies are equal at every instant, so that the equation of rotation is the same. Take a constant $c$ and require

$$\mathbf E(\mathbf r)=c\,\mathbf B(\mathbf r)\ \text{at every point},\qquad \mathbf p=\frac{\mathbf m}{c}.$$

Then $\mathbf p\times\mathbf E=\mathbf m\times\mathbf B$ and $-\mathbf p\cdot\mathbf E=-\mathbf m\cdot\mathbf B$. Also the two dipoles must have the same mass and the same moment of inertia, and the same initial position, orientation and velocities. (Any constant $k$ can replace $c$.)

Question 19

A bar magnet of magnetic moment $m$ and moment of inertia $I$ (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let $T$ be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field $\mathbf B$. What would be the similar period $T’$ for each piece?

Solution. For the original magnet $T=2\pi\sqrt{\dfrac{I}{mB}}$.

Cutting perpendicular to the length keeps the pole strength but halves the length, so each piece has moment $m’=\dfrac m2$. Its mass is $M/2$ and its length is $L/2$, so about its own centre

$$I’=\frac{1}{12}\cdot\frac M2\cdot\Big(\frac L2\Big)^2=\frac{I}{8}.$$

Therefore

$$T’=2\pi\sqrt{\frac{I’}{m’B}}=2\pi\sqrt{\frac{I/8}{(m/2)B}}=2\pi\sqrt{\frac{I}{4mB}}=\frac T2.$$

Question 20

Use (i) the Ampere’s law for $\mathbf H$ and (ii) continuity of lines of $\mathbf B$, to conclude that inside a bar magnet, (a) lines of $\mathbf H$ run from the N pole to S pole, while (b) lines of $\mathbf B$ must run from the S pole to N pole.

Solution. (a) Take a closed loop C that runs along a line of $\mathbf B$ outside the magnet from N to S and returns through the magnet. There is no free current, so Ampere’s law for $\mathbf H$ gives $\oint\mathbf H\cdot d\mathbf l=0$. Outside the magnet $\mathbf H=\mathbf B/\mu_0$ and points along the line from N to S, so the outside part of the integral is positive. For the total to vanish, the part inside must be negative, so $\mathbf H$ inside points along the path from N to S of the magnet. That is, $\mathbf H$ lines run from the N pole to the S pole inside the magnet.

(b) Lines of $\mathbf B$ are continuous and form closed loops. Outside the magnet they run from N to S, so to close they must run from S to N through the inside.

Long Answer Questions

Question 21

Verify the Ampere’s law for magnetic field of a point dipole of dipole moment $\mathbf m=m\hat{\mathbf k}$. Take C as the closed curve running clockwise along (i) the $z$-axis from $z=a>0$ to $z=R$; (ii) along the quarter circle of radius $R$ and centre at the origin, in the first quadrant of $x$–$z$ plane; (iii) along the $x$-axis from $x=R$ to $x=a$, and (iv) along the quarter circle of radius $a$ and centre at the origin in the first quadrant of $x$–$z$ plane.

Solution. The dipole field is $B_r=\dfrac{\mu_0}{4\pi}\dfrac{2m\cos\theta}{r^3}$, $B_\theta=\dfrac{\mu_0}{4\pi}\dfrac{m\sin\theta}{r^3}$, with $\theta$ measured from the $z$-axis. Evaluate $\int\mathbf B\cdot d\mathbf l$ along each piece.

(i) On the $z$-axis $\theta=0$, the field is radial, $B=\dfrac{\mu_0}{4\pi}\dfrac{2m}{z^3}$, and $dl=dz$: $$\int_a^R\frac{\mu_0}{4\pi}\frac{2m}{z^3}\,dz=\frac{\mu_0m}{4\pi}\Big(\frac1{a^2}-\frac1{R^2}\Big).$$

(ii) On the arc of radius $R$ we move in the direction of increasing $\theta$, from $0$ to $\pi/2$, with $dl=R\,d\theta$ along $\hat\theta$. Only $B_\theta$ contributes: $$\int_0^{\pi/2}\frac{\mu_0}{4\pi}\frac{m\sin\theta}{R^3}R\,d\theta=\frac{\mu_0m}{4\pi R^2}.$$

(iii) On the $x$-axis $\theta=90^\circ$, where the field is $B_\theta$, which is along $-\hat{\mathbf k}$. The path is along $x$, perpendicular to the field, so the contribution is $0$.

(iv) The arc of radius $a$ is traversed from $\theta=\pi/2$ back to $0$, so the sign is reversed: $$-\frac{\mu_0m}{4\pi a^2}.$$

The sum is $$\frac{\mu_0m}{4\pi}\Big(\frac1{a^2}-\frac1{R^2}\Big)+\frac{\mu_0m}{4\pi R^2}+0-\frac{\mu_0m}{4\pi a^2}=0.$$ No current passes through the loop, so $\mu_0I_{enc}=0$ and Ampere’s law holds.

Question 22

What are the dimensions of $\chi$, the magnetic susceptibility? Consider an H-atom. Guess an expression for $\chi$, up to a constant by constructing a quantity of dimensions of $\chi$, out of parameters of the atom: $e$, $m$, $v$, $R$ and $\mu_0$. Here, $m$ is the electronic mass, $v$ is electronic velocity, $R$ is Bohr radius. Estimate the number so obtained and compare with the value of $|\chi|\sim10^{-5}$ for many solid materials.

Solution. Since $\mathbf M=\chi\mathbf H$ and $\mathbf M$, $\mathbf H$ have the same units (A m$^{-1}$), $\chi$ is dimensionless.

Write $\chi=\mu_0^{\,a}\,e^{b}\,m^{\alpha}v^{\beta}R^{\gamma}$ and use dimensions. The dimensions of $\mu_0$ are $\mathrm{kg\,m\,C^{-2}}$, so $\mu_0e^2$ has dimensions $\mathrm{kg\,m}$, the same as $mR$. Hence $\dfrac{\mu_0e^2}{mR}$ is dimensionless, while $v$ cannot appear without bringing in a time dimension, so $\beta=0$. Solving gives $a=1$, $b=2$, $\alpha=-1$, $\beta=0$, $\gamma=-1$:

$$\chi\sim\frac{\mu_0e^2}{mR}.$$

Numerically, with $R=0.53\times10^{-10}$ m, $$\chi\sim\frac{(4\pi\times10^{-7})(1.6\times10^{-19})^2}{(9.1\times10^{-31})(0.53\times10^{-10})}\approx6.7\times10^{-4}\sim10^{-4}.$$

This is within about an order of magnitude of $10^{-5}$. (Including the natural factor $1/4\pi$ gives $r_e/R\approx5\times10^{-5}$, which is closer still.)

Question 23

Assume the dipole model for earth’s magnetic field $B$ which is given by $B_V=$ vertical component of magnetic field $=\dfrac{\mu_0}{4\pi}\dfrac{2m\cos\theta}{r^3}$, $B_H=$ horizontal component of magnetic field $=\dfrac{\mu_0}{4\pi}\dfrac{\sin\theta\,m}{r^3}$, $\theta=90^\circ-$ latitude as measured from magnetic equator. Find loci of points for which (i) $|\mathbf B|$ is minimum; (ii) dip angle is zero; and (iii) dip angle is $\pm45^\circ$.

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

The magnitude is $$|\mathbf B|=\sqrt{B_V^2+B_H^2}=\frac{\mu_0m}{4\pi r^3}\sqrt{4\cos^2\theta+\sin^2\theta}=\frac{\mu_0m}{4\pi r^3}\sqrt{1+3\cos^2\theta}.$$

(i) At a given distance $r$ this is least when $\cos\theta=0$, that is $\theta=90^\circ$: the magnetic equator.

(ii) The dip angle $\delta$ satisfies $\tan\delta=\dfrac{B_V}{B_H}=2\cot\theta$. It is zero when $\cot\theta=0$, again $\theta=90^\circ$: the magnetic equator.

(iii) $\delta=\pm45^\circ$ means $2\cot\theta=\pm1$, so $\tan\theta=\pm2$. This gives $\theta=\tan^{-1}2\approx63.4^\circ$ and $\theta=180^\circ-63.4^\circ=116.6^\circ$. These are the two circles of magnetic latitude $\pm26.6^\circ$ (given by $\tan\lambda=\tfrac12$): north of the magnetic equator the dip is $+45^\circ$ and south it is $-45^\circ$.

Printed key error. Minor: the key gives only $\theta=\tan^{-1}2$, the locus for $+45^\circ$, and leaves out the matching $-45^\circ$ circle at $\theta=180^\circ-\tan^{-1}2$.

Question 24

Consider the plane $S$ formed by the dipole axis and the axis of earth. Let $P$ be a point on the magnetic equator and in $S$. Let $Q$ be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at $P$ and $Q$.

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

The magnetic axis is tilted about $11.3^\circ$ from the geographic axis.

At $P$. $P$ lies in the plane $S$ containing both axes, so the geographic meridian and the magnetic meridian through $P$ both lie in $S$ and coincide: the declination is $0$. $P$ is on the magnetic equator, so the dip is $0$.

At $Q$. $Q$ is on the magnetic equator, so the dip is $0$. It is also on the geographic equator, at the point where the two equators cross, and there the magnetic meridian is not the geographic meridian. They differ by the tilt of the axes, so the declination is $11.3^\circ$.

Question 25

There are two current carrying planar coils made each from identical wires of length $L$. $C_1$ is circular (radius $R$) and $C_2$ is square (side $a$). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform $\mathbf B$ and carry the same current. Find $a$ in terms of $R$.

Solution. The frequency of small oscillations of a current loop in a field is given by $\omega^2=\dfrac{mB}{I_{\rm rot}}$, where $m$ is its magnetic moment and $I_{\rm rot}$ its moment of inertia about the axis of rotation. Equal frequencies need equal $m/I_{\rm rot}$.

Moments. The circular coil has $n_1=\dfrac{L}{2\pi R}$ turns, so $$m_1=n_1I\pi R^2=\frac{ILR}{2}.$$ The square coil has $n_2=\dfrac{L}{4a}$ turns, so $$m_2=n_2Ia^2=\frac{ILa}{4}.$$

Moments of inertia. Both coils are made of the same length of the same wire, so they have the same mass $M$. The circular ring about a diameter has $I_1=\tfrac12MR^2$. The square frame is rotated about an in-plane axis through the centre parallel to a side. Two of its sides lie at a distance $a/2$ from the axis and contribute $2\cdot\frac M4\cdot\frac{a^2}{4}=\frac{Ma^2}{8}$. The other two sides are rods turning about their midpoints and contribute $2\cdot\frac{1}{12}\cdot\frac M4\cdot a^2=\frac{Ma^2}{24}$. So $$I_2=\frac{Ma^2}{8}+\frac{Ma^2}{24}=\frac{Ma^2}{6}.$$

Equal frequencies. $$\frac{m_1}{I_1}=\frac{m_2}{I_2}\ \Rightarrow\ \frac{ILR/2}{MR^2/2}=\frac{ILa/4}{Ma^2/6}\ \Rightarrow\ \frac{IL}{MR}=\frac{3IL}{2Ma}\ \Rightarrow\ a=\frac{3R}{2}.$$

Printed key error. The key prints $a=\dfrac{3\pi}{4}R$. It uses $Ma^2/12$ for the square coil, which is the value for a square plate, not a frame, and even with that value its own algebra gives $a=3R$, so the stray factor of $\pi$ is a slip. The correct moment of inertia of the frame is $Ma^2/6$, giving $a=3R/2$.

Common mistakes

• Taking $B_r=\frac{\mu_0}{4\pi}\frac{3m\cos\theta}{r^3}$ in Question 16. The radial component is $2m\cos\theta/r^3$; the $3$ belongs to the vector form $3(\mathbf m\cdot\hat r)\hat r-\mathbf m$.
• Treating the ends of the other two magnets as free in Question 17. The zero-torque condition fixes them, because the three moments must sum to zero.
• Using $I’=I/2$ in Question 19. The length also halves, and $I\propto ML^2$ gives $I/8$; the product with $m’=m/2$ then gives $T/2$.
• Taking the sign of $\mathbf H$ inside a magnet to match $\mathbf B$ in Question 20. Inside, $\mathbf H$ and $\mathbf B$ are opposite; outside they are parallel.
• Dropping the second dip locus in Question 23(iii). A dip of $-45^\circ$ gives another circle, at $116.6^\circ$.
• Using the plate formula $Ma^2/12$ for the square coil in Question 25. A coil is a frame of wire, with $I=Ma^2/6$ about an axis through the centre parallel to a side.

Practise next

• Exemplar Magnetism and Matter: Multiple Choice Questions — the first ten questions of this unit, on para-, dia- and ferromagnetism, shielding and the toroid.
• NCERT textbook Chapter 5 — Magnetism and Matter for the theory of the bar magnet as a solenoid, the dipole field and the three kinds of magnetic material.
• Exemplar Moving Charges and Magnetism — the previous unit, on forces and torques on currents and charges in magnetic fields.
• Exemplar Electromagnetic Induction — the next unit, where changing flux produces emf.