A current carrying rectangular loop PQRS is made of uniform wire. The… | JEE Main 2023 (30 Jan, Shift 2)

Q36JEE Main 2023 · 30 JAN · SHIFT 2Medium

Force on Current-Carrying Conductors

A current carrying rectangular loop PQRS is made of uniform wire. The length $PR = QS = 5\text{ cm}$ and $PQ = RS = 100\text{ cm}$ . If ammeter current reading changes from I to $2I$ , the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively $\left(f_{PQ}^{I} : f_{PQ}^{2I}\right)$ is:

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Answer: (a)

PQ and RS are identical branches in parallel, so each carries half the total current.
Force per unit length $f = \frac{\mu_0 (I/2)(I/2)}{2\pi d} \propto I^2$ .
$\frac{f^{I}}{f^{2I}} = \frac{I^2}{(2I)^2} = \frac{1}{4}$
So the answer is (a) 1 : 4.

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