Q 12-04-110JEE MainJEE Main 2023 (30 Jan, Shift 2)Medium
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 $\left(f^I_{PQ}:f^{2I}_{PQ}\right)$ is
Answer: (B) $1:4$
The current enters at side $PR$ and leaves at side $QS$, so it splits equally between $PQ$ and $RS$, each carrying $\dfrac I2$ in the same direction.
$$f=\frac{\mu_0(I/2)(I/2)}{2\pi d}\propto I^2$$
Doubling $I$ multiplies $f$ by $4$: ratio $1:4$.
Solution by Sreeraj P, M.Sc Physics