Class 12 Physics · Original EZT practice
Electromagnetic Induction (Class 12 Chapter 6): summary and practice
Electromagnetic Induction is where changing magnetism makes electricity: Faraday's flux rule, Lenz's law for the direction, motional emf, mutual and self inductance, and the AC generator that turns rotation into voltage. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Flux and Faraday's law
Faraday and Henry's experiments showed that a changing magnetic environment induces an emf. Magnetic flux phi = B A cos(theta) counts the field through a loop, and Faraday's law says the induced emf equals the rate of change of flux linkage: emf = -N d(phi)/dt.
2. Lenz's law and energy conservation
The minus sign is Lenz's law: the induced current always opposes the change that creates it. This is not arbitrary - if the induced current helped the change instead, you would get energy for free, so Lenz's law is conservation of energy applied to induction.
3. Motional emf
A conductor of length l moving at speed v across a field B drives charge along itself, giving emf = B l v when the three are mutually perpendicular. A rod rotating about one end sweeps flux at emf = (1/2) B omega l^2.
4. Mutual and self inductance
A changing current in one coil induces emf in a neighbour through mutual inductance M, and in itself through self-inductance L: emf = -L dI/dt. An inductor carrying current I stores energy (1/2) L I^2 in its magnetic field.
5. The AC generator
A coil of N turns and area A rotating at angular speed omega in a field B produces emf = N B A omega sin(omega t) - an alternating voltage with peak N B A omega. This is the chapter's payoff: mechanical rotation becomes the mains supply.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Apply Faraday's law
Choose the method: Average emf = N x (change in flux) / (time taken).
EZT practice question
The flux through a 100-turn coil falls from 0.05 Wb to 0 in 0.1 s. Find the average induced emf.
Show the worked solution
- emf = N x delta phi / delta t.
- = 100 x 0.05 / 0.1.
Answer: 50 V.
Check: The faster the flux changes, the bigger the emf - halving the time would double the voltage.
Watch for this: Do not use the flux VALUE alone - the emf depends on how fast the flux CHANGES.
Now try it yourself
A 50-turn coil, flux changing by 0.02 Wb in 0.05 s. Average emf?
Check your answer
20 V.
50 x 0.02 / 0.05 = 50 x 0.4 = 20 V.
Find the induced current direction
Choose the method: The induced current opposes the change that creates it.
EZT practice question
The north pole of a magnet moves toward a circular loop. Seen from the magnet's side, which way does the induced current flow?
Show the worked solution
- The approaching north pole increases flux toward the loop.
- The loop fights back by making its own north pole face the magnet.
- A north-facing face needs an anticlockwise current.
Answer: Anticlockwise as seen from the magnet's side.
Check: If the current ran the other way, the loop would attract the magnet in - accelerating it and creating energy from nothing.
Watch for this: Decide WHICH side you are viewing from before saying clockwise or anticlockwise.
Now try it yourself
The same north pole is pulled AWAY from the loop. Direction seen from the magnet side?
Check your answer
Clockwise.
Now the loop attracts the retreating pole to oppose the loss of flux - a south-facing face, so clockwise current.
Compute motional emf
Choose the method: emf = B l v for mutually perpendicular B, l and v.
EZT practice question
A 0.5 m rod moves at 4 m/s at right angles to a 0.2 T field. Find the emf across its ends.
Show the worked solution
- emf = B l v.
- = 0.2 x 0.5 x 4.
Answer: 0.4 V.
Check: Double the speed or the field and the emf doubles - the relation is linear in each factor.
Watch for this: The rod must CUT field lines - motion parallel to B induces nothing.
Now try it yourself
A 1 m rod at 2 m/s in a 0.1 T field. emf?
Check your answer
0.2 V.
0.1 x 1 x 2 = 0.2 V.
Emf of a rotating rod
Choose the method: emf = (1/2) B omega l^2 about one end.
EZT practice question
A 1 m rod rotates at omega = 2 rad/s in a 0.5 T field, about one end. Find the emf between its ends.
Show the worked solution
- emf = (1/2) B omega l^2.
- = (1/2) x 0.5 x 2 x 1.
Answer: 0.5 V.
Check: The emf grows with the SQUARE of the length - a 2 m rod in the same setting gives 2 V, four times as much.
Watch for this: Do not use B l v with the tip speed alone - every point of the rod moves at a different speed, which is where the 1/2 comes from.
Now try it yourself
A 2 m rod at the same omega and field. emf?
Check your answer
2 V.
(1/2) x 0.5 x 2 x 4 = 2 V.
Use mutual inductance
Choose the method: emf in the second coil = M x (rate of change of current in the first).
EZT practice question
Two adjacent coils have mutual inductance 1.5 H. The current in the first changes at 20 A/s. Find the emf induced in the second.
Show the worked solution
- emf = M x dI/dt.
- = 1.5 x 20.
Answer: 30 V.
Check: Self-inductance is the same idea turned inward: a coil opposes changes in its OWN current through L.
Watch for this: A steady current induces nothing - only the CHANGE in current matters.
Now try it yourself
Same pair, current changing at 4 A/s. Induced emf?
Check your answer
6 V.
1.5 x 4 = 6 V.
Peak emf of an AC generator
Choose the method: Peak emf = N B A omega; the output is emf = N B A omega sin(omega t).
EZT practice question
A generator coil has 100 turns, area 0.1 m^2, field 0.5 T, and spins at omega = 100 pi rad/s. Find the peak emf.
Show the worked solution
- Peak emf = N B A omega.
- = 100 x 0.5 x 0.1 x 100 pi.
- = 500 pi.
Answer: Peak emf = 500 pi V, about 1570.8 V.
Check: The output swings between +1570.8 V and -1570.8 V sinusoidally; the rms value is the peak divided by root 2.
Watch for this: Peak and rms are different numbers - multiplying or dividing by root 2 converts between them.
Now try it yourself
Same coil but 50 turns. Peak emf?
Check your answer
About 785.4 V.
Halving N halves the peak: 250 pi = 785.4 V.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Faraday's law?
Show answer and reason
emf = -N d(phi)/dt.
The emf follows the rate of change of flux linkage.
2. Lenz's law?
Show answer and reason
The induced current opposes the change producing it.
It is conservation of energy for induction.
3. Motional emf?
Show answer and reason
emf = B l v.
B, l and v must be mutually perpendicular.
4. AC generator output?
Show answer and reason
emf = N B A omega sin(omega t).
Peak value is N B A omega.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A 50-turn coil sees its flux change by 0.02 Wb in 0.05 s. Find the average emf.
- A 1 m rod moves at 2 m/s across a 0.1 T field. Find the motional emf.
- A 50-turn generator coil with A = 0.1 m^2, B = 0.5 T and omega = 100 pi rad/s. Find the peak emf.
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Questions about this practice page
- What is magnetic flux?
- Flux phi = B A cos(theta) measures how much magnetic field passes through a loop - the field strength times the area, adjusted for the tilt angle. Induction happens whenever this quantity changes, however the change is produced.
- How does Lenz's law follow from energy conservation?
- If the induced current aided the change instead of opposing it, a small push would run away into ever-growing current and motion - free energy. Opposing the change means you always have to do work to induce current, which is exactly what conservation of energy demands.
- What is the difference between self and mutual inductance?
- Mutual inductance M links two coils: a changing current in one induces emf in the other. Self-inductance L is a coil acting on itself: its own changing current induces an emf that opposes the change, and the inductor stores energy (1/2) L I^2.
- Why does a rotating rod get an emf of (1/2) B omega l^2?
- Each point of the rod moves at a different speed, from zero at the pivot to omega l at the tip. Averaging the motional emf along the rod gives half of B l v_tip, which works out to (1/2) B omega l^2.
- Are these the NCERT exercise questions for Electromagnetic Induction (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.