A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $B=0.8\,\mathrm{T}$. When released the radius of the loop starts shrinking at a constant rate of $2\,\mathrm{cm\,s^{-1}}$. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be ______ mV.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Electromagnetic Induction

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field

B = 0.8 T

. When released the radius of the loop starts shrinking at a constant rate of

2 cm s^{- 1}

. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be ______ mV.

Answer: 10

Solution

With flux

Φ = B π r^{2}

and a steadily shrinking radius, Faraday's law gives

∣ ε ∣ = \frac{d Φ}{d t} = 2 π B r \frac{d r}{d t}

. Substituting

B = 0.8

r = 0.1

m and

\frac{d r}{d t} = 0.02

m s

^{- 1}

∣ ε ∣ = 2 π (0.8) (0.1) (0.02) \approx 0.01005

V. The induced emf is about

10

mV.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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