JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Electromagnetic Induction

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A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of

1 mm / s

. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be
____

μ V

Answer: 50

Solution

The flux through the loop is

ϕ = B A = B π r^{2}

, and the induced emf is its rate of change:

ε = \frac{d ϕ}{d t} = B \frac{d A}{d t} = B (2 π r) \frac{d r}{d t}

. Substituting

B = 0.4 T

r = 2 cm = 0.02 m

and

\frac{d r}{d t} = 1 mm s^{- 1} = 1 0^{- 3} m s^{- 1}

gives

ε = 0.4 \times 2 π \times 0.02 \times 1 0^{- 3} = 16 π \times 1 0^{- 6} V

. The magnitude of the induced emf is about

50 μ V

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer

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