JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Electromagnetic Induction

A uniform magnetic field

B

exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate

\frac{d B}{d t} = 0.032 Ts^{- 1}

. The induced current in the loop is close to (Resistivity of the metal wire is

1.23 \times 1 0^{- 8} Ω m

)

(A)0.61 A✓ Correct
(B)0.43 A
(C)0.53 A
(D)0.34 A

Answer: (A)

Solution

By Faraday's law the induced emf is the rate of change of flux through the loop,

ε = A \frac{d B}{d t}

, where

A

is the area the square encloses. The wire of total length

30 cm

makes a square of side

a = 0.30/4 = 0.075 m

, so

A = a^{2} = 5.625 \times 1 0^{- 3} m^{2}

. The loop's resistance comes from the wire itself,

R = ρ L / A_{c}

, with cross-section

A_{c} = π r^{2} = π (2 \times 1 0^{- 3})^{2} = 4 π \times 1 0^{- 6} m^{2}

, giving

R = \frac{1.23 \times 1 0 ^{- 8} \times 0.30}{4 π \times 1 0 ^{- 6}} \approx 2.9 \times 1 0^{- 4} Ω

. The induced current is then

i = \frac{A ( d B / d t )}{R} = \frac{5.625 \times 1 0 ^{- 3} \times 0.032}{2.9 \times 1 0 ^{- 4}} \approx 0.61 A

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Electromagnetic Induction

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