JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Electromagnetic Induction

A square loop of side 20 cm and resistance

1Ω

is moved towards right with a constant speed vo.The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4. What should be the value of vo so that a steady current of 2 mA flows in the loop

(A) $1 m / s$
(B) $1 0^{2} m / s$
(C) $1 0^{- 2} cm / s$
(D) $1 cm / s$ ✓ Correct

Answer: (D)

Solution

The motional EMF of the right arm is

ε = B ℓ v_{0}

with

B = 5 T

and

ℓ = 0.20 m

. The four

4 Ω

resistors form two series pairs of

8 Ω

joined in parallel, giving

4 Ω

, which adds to the loop's

1 Ω

for a total of

5 Ω

. From

I = \frac{B ℓ v _{0}}{R}

2 \times 1 0^{- 3} = \frac{5 \times 0.20 \times v _{0}}{5}

, so

v_{0} = 1 0^{- 2} m/s = 1 cm/s

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Electromagnetic Induction →

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Electromagnetic Induction

Solution

More JEE Main questions on Electromagnetic Induction