JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Electromagnetic Induction

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10Ω, 20 mH

coil carrying constant current is connected to a battery of 20 V through a switch is opened current becomes zero in

100 μ s

. The average emf induced in the coil is ____ V.

Answer: 400

Solution

The average induced emf is given by

⟨ ε ⟩ = \frac{L Δ i}{Δ t}

.

First, find the initial steady-state current:

i_{0} = \frac{V}{R} = \frac{20}{10} = 2 A

.

When the switch is opened, the current drops from 2 A to 0 A in

Δ t = 100 μ s = 100 \times 1 0^{- 6} s

.

Using

L = 20 mH = 20 \times 1 0^{- 3} H

⟨ ε ⟩ = \frac{20 \times 1 0 ^{- 3} \times ( 2 - 0 )}{100 \times 1 0 ^{- 6}} = \frac{40 \times 1 0 ^{- 3}}{100 \times 1 0 ^{- 6}} = \frac{40 \times 1 0 ^{- 3}}{1 0 ^{- 4}} = 400 V

Therefore, the average induced emf is 400 V.

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer

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