GATE 2023 IN Q.27: solution and answer

Official answer: 4.9 to 5.1Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Field of one infinitely long conductor carrying current $I_1$, at distance $d$ from it (Ampère's circuital law over a circle of radius $d$): $$B_1=\frac{{\mu }_0{I}_1}{2\pi d},{\mu }_0=4\pi \times 10^{-7}\text{H/m}$$ Its direction is circumferential, so it is perpendicular to the second conductor.

2. Force on a length $L$ of the second conductor, carrying $I_2$ (Lorentz force $d\mathbf{F}=I_{2}d\mathbf{l}\times {\mathbf{B}}_1$, with $d\mathbf{l}\perp {\mathbf{B}}_1$): $$\frac{F}{L}=I_2{B}_1=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$$

3. Only the gap changes; the currents $I_1$, $I_2$ and ${\mu }_0$ stay the same. Therefore $\frac{F}{L}\propto \frac{1}{d}$.

4. Take the ratio of the two cases: $$\frac{(F/L{)}_2}{(F/L{)}_1}=\frac{d_1}{d_2}=\frac{2\text{cm}}{4\text{cm}}=\frac{1}{2}$$

5. Hence $$(F/L{)}_2=\frac{10\mu \text{N/m}}{2}=5.0\mu \text{N/m}$$