A square loop of side 20 cm and resistance 1 Ω is moved towards right… | JEE Main 2021 (1 Sep, Shift 2)

Q44JEE Main 2021 · 1 SEP · SHIFT 2Difficult

Motional EMF

A square loop of side 20 cm and resistance 1 $\Omega$ is moved towards right with a constant speed v0. 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 $\Omega$ . What should be the value of v0 so that a steady current of 2 mA flows in the loop?

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Answer: (b)

Only the right arm is in the field: $e = Blv_0 = 5 \times 0.2 \times v_0 = v_0$
Between P and Q there are two paths, each $4 + 4 = 8\ \Omega$ , in parallel: $R_{PQ} = 4\ \Omega$ .
Total resistance $= 1 + 4 = 5\ \Omega$ , so $I = \dfrac{v_0}{5}$ .
$\dfrac{v_0}{5} = 2 \times 10^{-3} \Rightarrow v_0 = 10^{-2}\text{ m/s} = 1\text{ cm/s}$
So the answer is (b) 1 cm/s.

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