JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Electromagnetic Induction

A square loop of area

25 cm^{2}

has a resistance of

10 Ω

. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be

(A) $2.5 \times 1 0^{- 3} J$
(B) $1.0 \times 1 0^{- 3} J$ ✓ Correct
(C) $1.0 \times 1 0^{- 4} J$
(D) $5 \times 1 0^{- 3} J$

Answer: (B)

Solution

The side of the loop is

ℓ = 25 \times 1 0^{- 4} = 0.05

m, and pulling it fully out in

1.0

s means a steady speed

v = \frac{0.05}{1.0} = 0.05

m/s. The motional emf is

ε = B ℓ v = 40 \times 0.05 \times 0.05 = 0.1

V, driving a current

i = \frac{ε}{R} = \frac{0.1}{10} = 0.01

A. The opposing force is

F = B i ℓ = 40 \times 0.01 \times 0.05 = 0.02

N, and acting over the distance

ℓ = 0.05

m the external agent does

W = F ℓ = 0.02 \times 0.05 = 1.0 \times 1 0^{- 3}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Electromagnetic Induction

Solution

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