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Physics Question 40 – JEE-MAIN 2026

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A 30 cm long solenoid has 10 turns per cm and area of 5 cm2. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is a×10−5 V. The value a is _______.

Recall Faraday's law of induction and the concept of self-inductance for a solenoid.

Step 1: Calculate the self-inductance of the solenoid✦ Active

First, convert all given values to SI units: length l=30 cm=0.3 m, number of turns per unit length n=10 turns/cm=1000 turns/m, and area A=5 cm2=5×10−4 m2. The permeability of free space is μ0=4π×10−7 T m/A. The self-inductance L of a solenoid is given by:

L=μ0n2Al

Substitute the values:

L=(4π×10−7)×(1000)2×(5×10−4)×(0.3) L=(4π×10−7)×(106)×(5×10−4)×(0.3) L=(4π×5×0.3)×(10−7×106×10−4) L=6π×10−5 H
Step 2: Calculate the magnitude of the induced e.m.f.○ Expand

The current changes from 2 A to 4 A, so the change in current is ΔI=4 A−2 A=2 A. This change occurs over a time interval Δt=3.14 s. According to Faraday's law, the magnitude of the induced e.m.f. |E| is:

|E|=LΔIΔt

Substitute the calculated inductance and given current change values:

|E|=(6π×10−5)×23.14
💡 Teacher's Secret Hint

Remember to use the magnitude of the induced e.m.f. as the question asks for a positive value 'a'.

Step 3: Determine the value of 'a'○ Expand

Given that π≈3.14, we can simplify the expression for |E|. The term π3.14 approximates to 1.

|E|≈(6×3.14×10−5)×23.14 |E|≈6×2×10−5 |E|≈12×10−5 V

The problem states that the induced e.m.f. is a×10−5 V. Comparing this with our result, we find:

a=12
💡 Teacher's Secret Hint

Pay attention to the units and the power of 10 in the final expression for the e.m.f.

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