NCERT Exemplar Solutions Class 12 Physics Electromagnetic Induction — Multiple Choice Questions Q1–10

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

The book numbers these questions 6.1 to 6.10; here they are Questions 1 to 10. Questions 1 to 6 are MCQ I (exactly one option is correct) and Questions 7 to 10 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Flux is a dot product. $\phi = \mathbf{B}\cdot\mathbf{A}$, where the vector area $\mathbf{A}$ points along the normal to the loop. Only the component of $\mathbf{B}$ along that normal counts.
• Lenz’s law. The induced current always opposes the change in flux that causes it, so the sign of the change decides the direction.
• Inductance is geometry. Self and mutual inductance depend on the shape, size and arrangement of the coils, not on the current.
• What produces an emf. An emf appears only when the flux through a circuit changes: by a changing field, a changing area, or a changing orientation.

Key insight. Nearly every option here is settled by asking one question: does the flux through the loop change? If it does not, there is no emf, whatever else is going on.

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

Question 1

A square of side $L$ metres lies in the $x$-$y$ plane in a region where the magnetic field is given by $\mathbf{B} = B_0(2\hat{i} + 3\hat{j} + 4\hat{k})$ T, where $B_0$ is constant. The magnitude of flux passing through the square is

Solution. The square lies in the $x$-$y$ plane, so its normal is along $\hat{k}$ and its vector area is $\mathbf{A} = L^2\hat{k}$. Then

$$\phi = \mathbf{B}\cdot\mathbf{A} = B_0(2\hat{i}+3\hat{j}+4\hat{k})\cdot L^2\hat{k} = 4B_0L^2.$$

The $\hat{i}$ and $\hat{j}$ components lie in the plane of the square and run along its surface without piercing it, so they contribute nothing. Option (d) is the magnitude of $\mathbf{B}$ times $L^2$, which would be the flux only if the field were perpendicular to the square.

Question 2

A loop made of straight edges has six corners at A$(0,0,0)$, B$(L,0,0)$, C$(L,L,0)$, D$(0,L,0)$, E$(0,L,L)$ and F$(0,0,L)$. A magnetic field $\mathbf{B} = B_0(\hat{i}+\hat{k})$ T is present in the region. The flux passing through the loop ABCDEFA (in that order) is

Solution. The loop is not flat, so split it into two flat pieces sharing the edge AD. The square ABCD lies in the $x$-$y$ plane. Going A, B, C, D, the circulation is counterclockwise seen from $+z$, so by the right-hand rule its vector area is $L^2\hat{k}$. The square ADEF lies in the $y$-$z$ plane. Going D, E, F, A, the circulation is counterclockwise seen from $+x$, so its vector area is $L^2\hat{i}$. The shared edge AD is traversed in opposite senses by the two pieces, so it drops out of the loop, leaving exactly ABCDEFA.

The vector area of the whole loop is therefore $\mathbf{A} = L^2(\hat{i}+\hat{k})$, and

$$\phi = \mathbf{B}\cdot\mathbf{A} = B_0(\hat{i}+\hat{k})\cdot L^2(\hat{i}+\hat{k}) = 2B_0L^2.$$

Each piece contributes $B_0L^2$, one from the $\hat{k}$ part of the field through ABCD and one from the $\hat{i}$ part through ADEF.

Question 3

A cylindrical bar magnet is rotated about its axis (Fig. 6.1). A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then

The figure shows a vertical bar magnet with N at the top, spinning with angular speed $\omega$ about its own axis. One end of a wire is attached on the axis at the top, the other touches the curved surface of the magnet, and an ammeter A is in the wire between them.

Solution. This is the homopolar (unipolar) generator. The magnetic field of a bar magnet is symmetric about its axis, so it looks exactly the same at every instant of the rotation. Nothing about the field pattern changes with time, so there is no changing flux in the ordinary sense and no oscillating emf.

Instead, the free charges in the magnet itself move through the field as it spins. The magnetic force $q\mathbf{v}\times\mathbf{B}$ on them is the same at all times and acts along the radius, so it drives them from the axis towards the rim (or the other way). That gives a steady emf between the axis and the curved surface, and the external wire joining them carries a steady current of fixed direction: a direct current. Options (c) and (d) would need a time-varying emf, and (b) ignores the motional emf altogether.

Question 4

There are two coils A and B as shown in Fig. 6.2. A current starts flowing in B as shown, when A is moved towards B, and stops when A stops moving. The current in A is counterclockwise. B is kept stationary when A moves. We can infer that

The figure shows two circular coils side by side in the same plane, A on the left and B on the right, with A being moved towards B. The current in B is drawn counterclockwise.

Solution. Take the facts in turn.

• B carries a current only while A is moving, so a stationary A produces no changing flux in B. Hence the current in A is steady, which rules out (b). It is not zero either, or moving A would do nothing to B, which rules out (c).
• Now the direction. The two coils lie side by side in one plane. By the right-hand rule a counterclockwise current in A gives a field out of the page inside A, and the return field lines pass into the page outside A, which is where B sits. So at B the field of A is into the page.
• As A approaches, that into-the-page flux through B increases. By Lenz’s law, B opposes the increase by setting up a field out of the page inside itself, which needs a counterclockwise current in B. That is exactly what the figure shows.

So a counterclockwise steady current in A is consistent with the given data; a clockwise one would have produced a clockwise current in B instead.

Question 5

Same as Question 4 except the coil A is made to rotate about a vertical axis (Fig. 6.3). No current flows in B if A is at rest. The current in coil A, when the current in B (at $t = 0$) is counterclockwise and the coil A is as shown at this instant, $t = 0$, is

The figure shows coil A on the left, with a dashed vertical diameter as its axis of rotation and an arrow showing angular speed $\omega$, and coil B on the right in the same plane, with a counterclockwise current.

Solution. B has no current when A is at rest, which means A’s own current is steady (if it varied, B would respond even with A at rest). That removes (b) and (c).

At $t = 0$ the two coils are coplanar, and A then starts to turn about its vertical diameter. Its plane tilts away from that of B, so the flux of A’s field through B decreases. B’s current is counterclockwise, which means its induced field is out of the page inside B; by Lenz’s law it is trying to sustain the flux that is being lost, so the lost flux must have been out of the page at B.

Outside a coil, the field is opposite to the field inside it. For the field of A to point out of the page at B (outside A), its field inside A must be into the page. That needs a clockwise current in A.

Compare Question 4: there B’s flux was growing, so B opposed it; here it is falling, so B supports it, which reverses the conclusion.

Question 6

The self inductance $L$ of a solenoid of length $l$ and area of cross-section $A$, with a fixed number of turns $N$ increases as

Solution. For a long solenoid, $B = \mu_0 N I/l$, so the flux linked with all $N$ turns is $N\phi = N\cdot BA = \mu_0N^2AI/l$. Comparing with $N\phi = LI$,

$$L = \frac{\mu_0 N^2 A}{l}.$$

With $N$ fixed, $L$ is directly proportional to $A$ and inversely proportional to $l$. It therefore grows when the area increases and the length decreases: the same turns packed more tightly into a fatter coil link more flux per ampere.

Question 7

A metal plate is getting heated. It can be because

Solution. A metal heats when current flows through it, because the current dissipates energy as $I^2R$.

• (a) A direct current through the plate heats it by Joule heating. True.
• (b) A time-varying field changes the flux through the plate, which induces circulating eddy currents in it, and these heat it. True; this is how induction furnaces work.
• (c) A field that varies from place to place but is constant in time gives no changing flux through a stationary plate, so no emf, no eddy currents and no heating. False.
• (d) Any current, direct or alternating, heats the plate. True.

Question 8

An e.m.f. is produced in a coil, which is not connected to an external voltage source. This can be due to

Solution. An emf is induced whenever the flux through the coil changes.

• (a) A changing field changes the flux directly. True.
• (b) Moving in a changing field certainly changes the flux. True.
• (c) A constant field does not guarantee constant flux: if the coil moves into a region where the field is stronger or weaker (a non-uniform field), or rotates so that its orientation relative to the field changes, the flux changes and an emf appears. True.
• (d) A stationary coil in a field that does not change with time has constant flux, so there is no emf however non-uniform the field is. False.

Question 9

The mutual inductance $M_{12}$ of coil 1 with respect to coil 2

Solution. Mutual inductance is defined by $\phi_{2} = M_{21}I_1$ (flux in one coil per unit current in the other), so it depends only on geometry: sizes, number of turns, separation and relative orientation.

• (a) Bringing the coils closer means more of one coil’s flux threads the other, so $M$ increases. True.
• (b) $M$ is the ratio of flux to current, so the current cancels out. It does not depend on the current. False.
• (c) Rotating a coil about an axis changes the angle between the field of one and the plane of the other; this can reduce the coupling, and in general it need not increase it. It does not necessarily increase $M$. False.
• (d) By the reciprocity theorem $M_{12} = M_{21}$. True.

Question 10

A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because

Solution. No emf means the flux does not change as the coil grows.

• (a) A constant field alone does not help: if it has a component perpendicular to the coil, the flux $B_\perp\pi r^2$ grows with the radius and an emf appears. False.
• (b) If the field lies in the plane of the coil, the flux through the coil is zero at every moment, whatever the radius, so there is no emf. True.
• (c) If the perpendicular component falls at exactly the right rate, the product $B_\perp\pi r^2$ stays constant even though $r$ is increasing, and the flux does not change. True.
• (d) A constant perpendicular field gives flux $B\pi r^2$, which increases as the coil expands, so an emf is produced. False.

Common mistakes

• Using the full magnitude of $\mathbf{B}$ in Question 1. Flux uses only the component along the normal to the surface.
• Forgetting that a bent loop has one vector area in Question 2. Add the vector areas of the flat pieces; the shared edge cancels.
• Expecting an alternating current in Question 3. The field pattern of a spinning bar magnet never changes, so the emf is steady.
• Choosing the wrong direction in Questions 4 and 5. The field of a coil reverses outside it, and Lenz’s law makes B oppose an increase but support a decrease.
• Reading $L \propto 1/l$ the wrong way round in Question 6. More area and a shorter length both raise the inductance.
• Counting a static, non-uniform field as a source of emf in Questions 7 and 8. With a stationary coil and no time variation the flux is constant.
• Letting current into the definition of mutual inductance in Question 9. It is flux per unit current, so it depends on geometry only.
• Saying a constant field gives no emf for an expanding coil in Question 10. The area changes, so the flux does unless the field is in the plane of the coil.

Practise next

• Exemplar Electromagnetic Induction: Very Short, Short and Long Answer Questions — where flux, Lenz’s law, motional emf and inductance are applied in full calculations.
• NCERT textbook Chapter 6 — Electromagnetic Induction for the theory of Faraday’s law, Lenz’s law, eddy currents and inductance.
• Exemplar Magnetism and Matter — the previous unit, on magnetic fields and field lines.
• Exemplar Alternating Current — the next unit, which builds on induced emf in rotating coils.