A rigid square loop of side 'a' and carrying current I2 is lying on a… | JEE Main 2019 (9 Apr, Shift 1)

Q48JEE Main 2019 · 9 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

A rigid square loop of side 'a' and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be:

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Answer: (a)

Forces on the top and bottom sides cancel.
Near side (distance $a$ ) carries current opposite to $I_1$ : repulsion $\frac{\mu_0 I_1 I_2 a}{2\pi a} = \frac{\mu_0 I_1 I_2}{2\pi}$ .
Far side (distance $2a$ ) carries current along $I_1$ : attraction $\frac{\mu_0 I_1 I_2 a}{2\pi (2a)} = \frac{\mu_0 I_1 I_2}{4\pi}$ .
Net $= \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi} = \frac{\mu_0 I_1 I_2}{4\pi}$ , repulsive.
So the answer is (a).

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