Exercises — Electromagnetic Induction

1. 13 marksNCERT Exercises, Chapter 6

   Predict the direction of the induced current in the situations described by Figs. 6.15(a) to (f).

   Hint. Apply Lenz's law in each case: the induced current opposes the change producing it.

   Every part is answered by the same rule, so it is worth stating it once and applying it mechanically.

   Lenz's law states that the induced current flows in whichever direction opposes the change in magnetic flux that produced it. In practice: identify whether the flux through the loop is increasing or decreasing, then choose the current sense whose own field opposes that change.

   If the flux increases, the induced current opposes it, so its own field points opposite to the applied field through the loop.

   If the flux decreases, the induced current tries to maintain it, so its own field points along the applied field.

   Applying this to the six configurations, and reading the sense with the right-hand rule, gives the directions:

   (a) $q$ to $r$, (b) $p$ to $q$ and $y$ to $z$, (c) $y$ to $z$, (d) $z$ to $y$, (e) $x$ to $y$, (f) no induced current, because the field lines lie in the plane of the loop so the flux through it is zero and does not change.

   Part (f) is the instructive one: a field can be present and still induce nothing, provided the flux through the loop never changes.

   ✦ (a) q to r (b) p to q and y to z (c) y to z (d) z to y (e) x to y (f) no current, since the flux through the loop is zero throughout

2. 23 marksNCERT Exercises, Chapter 6

   Use Lenz's law to determine the direction of the induced current in Fig. 6.16: (a) a wire of irregular shape turning into a circular shape; (b) a circular loop being deformed into a narrow straight wire.

   Hint. Ask in each case whether the enclosed area, and hence the flux, is growing or shrinking.

   Both parts turn on how the enclosed area changes, since the field is fixed and flux is $BA$.

   (a) Of all shapes with a given perimeter, the circle encloses the greatest area. So as the irregular loop becomes circular, its area and therefore its flux increase.

   By Lenz's law the induced current opposes this increase, so it flows in the sense whose own field opposes the applied field — that is, adcba in the figure's lettering.

   (b) A circular loop flattened into a narrow straight wire encloses an area approaching zero, so the flux decreases.

   The induced current now acts to maintain the flux, flowing in the sense whose field reinforces the applied one, namely a'd'c'b'a'.

   The contrast between the two parts is the point of the question: the same law gives opposite senses because the flux changes in opposite directions.

   ✦ (a) adcba, opposing the increasing flux as the area grows to a maximum (b) a'd'c'b'a', opposing the decreasing flux as the area shrinks to zero

3. 33 marksNCERT Exercises, Chapter 6

   A long solenoid with 15 turns per cm has a small loop of area 2.0 cm${}^2$ placed inside it, normal to its axis. If the current in the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

   Hint. The loop sits in the solenoid's field, so express B in terms of the solenoid current and differentiate.

   The loop lies inside the solenoid, where the field is $B={\mu }_{0}nI$, so the flux through the small loop is $\Phi =BA={\mu }_{0}nIA$.

   Convert the units first: $n=15$ turns/cm $=1500$ turns/m, and $A=2.0$ cm${}^2=2.0\times 10^{-4}$ m${}^2$.

   Since only the current varies:

   $\epsilon =\frac{d\Phi }{dt}={\mu }_{0}nA\frac{dI}{dt}$

   The rate of change of current is $\frac{4.0-2.0}{0.1}=20$ A/s, so:

   $\epsilon =4\pi \times 10^{-7}\times 1500\times 2.0\times 10^{-4}\times 20$

   $\epsilon \approx 7.5\times 10^{-6}\text{V}$

   The emf lasts only while the current is changing; once it steadies, the flux is constant and the emf falls to zero.

   ✦ emf = 7.5 x 10^-6 V, lasting only while the current changes

4. 44 marksNCERT Exercises, Chapter 6

   A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the loop's velocity is 1 cm s${}^{-1}$ normal to the (a) longer side, (b) shorter side? For how long does the induced voltage last in each case?

   Hint. The motional emf uses the length of the side that cuts the field lines, and the duration is how long the loop takes to leave the field.

   For motional emf, $\epsilon =Blv$, where $l$ is the length of the side perpendicular to the motion — the side sweeping across field lines.

   (a) Moving normal to the longer side, that side of length $0.08$ m does the cutting:

   $\epsilon =0.3\times 0.08\times 0.01=2.4\times 10^{-4}\text{V}$

   The loop must travel its own $2$ cm width to leave the field, so the emf lasts $\frac{0.02}{0.01}=2$ s.

   (b) Moving normal to the shorter side, the cutting length is $0.02$ m:

   $\epsilon =0.3\times 0.02\times 0.01=6\times 10^{-5}\text{V}$

   Now the loop must travel $8$ cm to clear the field, so the emf lasts $\frac{0.08}{0.01}=8$ s.

   Notice the trade-off: the larger emf persists for the shorter time, and the product of emf and duration is the same in both cases, as it must be since the total flux change is identical.

   ✦ (a) 2.4 x 10^-4 V lasting 2 s (b) 6 x 10^-5 V lasting 8 s

5. 53 marksNCERT Exercises, Chapter 6

   A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s${}^{-1}$ about an axis normal to the rod passing through one end. The other end is in contact with a circular metallic ring. A constant uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

   Hint. Different parts of the rod move at different speeds, so integrate along its length or use the standard result.

   A rotating rod is not like a translating one, because each element moves at a different speed — the tip moves fastest and the pivot not at all.

   Consider an element $dx$ at distance $x$ from the pivot, moving with speed $v=\omega x$. It contributes $d\epsilon =B(\omega x)dx$.

   Integrating along the whole rod:

   $\epsilon ={\int }_0^{L}B\omega xdx=\frac{1}{2}B\omega L^2$

   Substituting $B=0.5$ T, $\omega =400$ rad/s and $L=1.0$ m:

   $\epsilon =\frac{1}{2}\times 0.5\times 400\times (1.0{)}^2=100\text{V}$

   The factor of $\frac{1}{2}$ arises directly from the integration and is what distinguishes this from the simple $Blv$ case.

   ✦ emf = 100 V

6. 63 marksNCERT Exercises, Chapter 6

   A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s${}^{-1}$ at right angles to the horizontal component of the earth's magnetic field, $0.30\times 10^{-4}$ Wb m${}^{-2}$. (a) What is the instantaneous emf induced? (b) What is its direction? (c) Which end is at the higher potential?

   Hint. Use the motional emf formula, then apply the force on a positive charge to settle the direction.

   (a) The wire cuts field lines as it falls, giving a motional emf:

   $\epsilon =Blv=0.30\times 10^{-4}\times 10\times 5.0=1.5\times 10^{-3}\text{V}$

   (b) and (c) To fix the direction, consider the force on a positive charge carrier in the wire, $\vec{F}=q\vec{v}\times \vec{B}$.

   The velocity is downward and the horizontal component of the earth's field points north. Taking the cross product, the force on a positive charge is directed towards the east.

   Positive charge therefore accumulates at the eastern end, making it the higher potential terminal, and the induced emf drives conventional current from west to east within the wire.

   The emf is instantaneous in the sense that it persists only while the wire keeps falling.

   The eastern end becomes positive because the downward velocity crossed with the northward field gives an eastward force on positive carriers.

   ✦ (a) 1.5 x 10^-3 V (b) The emf drives current from west to east in the wire (c) The eastern end is at the higher potential

7. 72 marksNCERT Exercises, Chapter 6

   Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, estimate the self-inductance of the circuit.

   Hint. Rearrange the self-induced emf relation for L.

   A changing current in a circuit induces an emf in the circuit itself, given in magnitude by:

   $\epsilon =L\left|\frac{dI}{dt}\right|$

   The rate of change of current is $\frac{5.0-0.0}{0.1}=50$ A/s.

   Rearranging for the self-inductance:

   $L=\frac{\epsilon }{|dI/dt|}=\frac{200}{50}=4\text{H}$

   The large induced emf from a modest current change is why switching off an inductive circuit can produce a damaging voltage spike across the switch contacts.

   ✦ L = 4 H

8. 82 marksNCERT Exercises, Chapter 6

   A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

   Hint. Flux linkage is M times the current, so the change in linkage depends only on the change in current.

   The flux linkage with the second coil is related to the current in the first by $N\Phi =MI$.

   Therefore the change in flux linkage is:

   $\Delta (N\Phi )=M\Delta I=1.5\times (20-0)=30\text{Wb}$

   The time interval of 0.5 s is not needed for this part, because the change in linkage depends only on the change in current, not on how quickly it happened.

   The time would matter only if the question asked for the induced emf, which would be $\frac{30}{0.5}=60$ V.

   ✦ Change of flux linkage = 30 Wb; the 0.5 s is not needed here, though it would give an emf of 60 V

Solutions written by the tuition.in editorial team and checked against leph106.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.