JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Electromagnetic Induction

A 12 V battery connected to a coil of resistance

6Ω

through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is :

(A)5 mH
(B)12 mH
(C)8 mH
(D)10 mH✓ Correct

Answer: (D)

Solution

The steady current before the switch opens is

I = \frac{V}{R} = \frac{12}{6} = 2 A

. On opening, this current falls to zero in

1 ms

, and the induced emf is

ε = L \frac{Δ I}{Δ t}

. Hence

20 = L \times \frac{2}{1 0 ^{- 3}}

, giving

L = 10 mH

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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