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Electromagnetic Induction is short, and JEE Main asks from it almost every year. Faraday for the size of the emf, Lenz for its direction, and then a handful of standard situations: a rod moving in a field, a rotating coil, a solenoid storing energy. Each has one formula, and all of them follow from a single sentence about changing flux. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 2 questions every year, which is 8 marks.
- Motional emf and a rotating coil are the regular ones.
- Self-inductance, stored energy and the LR time constant fill the rest.
What the syllabus says
In the syllabus (Unit 14, Electromagnetic Induction and Alternating Currents): Faraday’s law, induced emf and current, Lenz’s law, eddy currents, self inductance and mutual inductance.
Never in the JEE Main syllabus: LC oscillations. They are in NCERT, so they still turn up in old question banks.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Magnetic flux | Φ = BA cosθ |
| Faraday’s law | e = − N dΦ/dt |
| Motional emf, straight rod | e = Bvl |
| Rotating rod about one end | e = ½Bωl2 |
| Induced charge | q = NΔΦ/R, independent of the time taken |
| Self-inductance | e = −L dI/dt |
| Solenoid | L = μ0n2Al |
| Energy stored | U = ½LI2 |
| LR time constant | τ = L/R |
| Perfectly coupled coils | M = √(L1L2) |
Compare the two motional emf lines. A rod sliding sideways gives Bvl, but a rod turning about one end gives ½Bωl2, because the far end moves faster than the near end and the average speed is half the maximum.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. A coil of 50 turns and area 0.1 m2 lies in a field of 0.2 T and is turned through 90° in 0.1 s. The average emf induced is
- (a) 1 V
- (b) 10 V
- (c) 100 V
- (d) 0.1 V
Q2. The self-inductance of a solenoid of n turns per metre, cross-section A and length l is
- (a) μ0nAl
- (b) μ0n2Al
- (c) μ0n2A/l
- (d) μ0nA
Q3. The total charge that flows round a circuit during a change of flux depends on
- (a) the time the change takes
- (b) the total change in flux and the resistance
- (c) the rate at which the flux changes
- (d) the current already flowing
Q4. An inductor of 2 H carries a current that changes at 3 A/s. The emf induced across it is
- (a) 2 V
- (b) 3 V
- (c) 6 V
- (d) 1.5 V
Q5. The energy stored in an inductor of 0.5 H carrying 4 A is
- (a) 1 J
- (b) 2 J
- (c) 4 J
- (d) 8 J
Q6. The time constant of a circuit containing an inductance L and a resistance R is
- (a) LR
- (b) L/R
- (c) R/L
- (d) 1/LR
Q7. Two perfectly coupled coils have inductances 2 H and 8 H. Their mutual inductance is
- (a) 2 H
- (b) 4 H
- (c) 8 H
- (d) 16 H
Q8. A rod 1 m long turns about one end at 10 rad/s in a field of 0.5 T perpendicular to its plane of rotation. The emf between its ends is
- (a) 2.5 V
- (b) 5 V
- (c) 10 V
- (d) 1.25 V
Q9. Lenz’s law tells you
- (a) the size of the induced emf
- (b) the direction of the induced current
- (c) both the size and the direction
- (d) neither of them
Q10. Eddy current losses in a transformer core are cut down by
- (a) making the core thicker
- (b) laminating the core
- (c) making the core of copper
- (d) raising the current
Answer key, with a hint for each
| Q1 | (b) | ΔΦ through the coil is NBA = 50 × 0.2 × 0.1 = 1 Wb, so e = 1/0.1. |
| Q2 | (b) | L goes as the square of the turns density, which is why winding more turns raises it so sharply. |
| Q3 | (b) | q = NΔΦ/R. Do it quickly or slowly, the same charge flows. |
| Q4 | (c) | e = L dI/dt = 2 × 3. |
| Q5 | (c) | U = ½ × 0.5 × 16. |
| Q6 | (b) | L/R has the unit of time, and after one time constant the current has reached 63% of its final value. |
| Q7 | (b) | M = k√(L1L2) with k = 1, so √16. |
| Q8 | (a) | e = ½Bωl2 = ½ × 0.5 × 10 × 1. |
| Q9 | (b) | Faraday gives the size, Lenz the direction. Together they give the minus sign in the law. |
| Q10 | (b) | Thin sheets separated by insulation break up the circulating current paths, so very little energy is wasted as heat. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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