GATE 2020 EE Q.55: solution and answer

Official answer: 138.10 to 139.20Source: Official answer key, GATE 2020 EE (IIT Delhi)

1. Field of the wire (Ampere's circuital law): at perpendicular distance $\rho$ from an infinitely long wire carrying current $I$, $$\mathbf{B}=\frac{{\mu }_{0}I}{2\pi \rho }{\hat{a}}_{\varphi },{\mu }_0=4\pi \times 10^{-7}\text{H/m}$$ Over the loop, which lies in the same plane as the wire, ${\hat{a}}_{\varphi }$ is normal to that plane, so $\mathbf{B}$ is perpendicular to the loop's surface everywhere.

2. Flux through the loop: split the loop into strips parallel to the wire, each of height $a$ and width $d\rho$ at distance $\rho$. $B$ depends only on $\rho$, so it is constant over each strip: $${\Phi }_{21}={\int }_d^{d+a}\frac{{\mu }_{0}I}{2\pi \rho }ad\rho =\frac{{\mu }_{0}Ia}{2\pi }\ln \left(\frac{d+a}{d}\right)$$

3. Mutual inductance $M={\Phi }_{21}/I$: the current cancels, so the given $I=2\text{A}$ does not affect the answer: $$M=\frac{{\mu }_{0}a}{2\pi }\ln \left(\frac{d+a}{d}\right)$$

4. Substitute $a=1\text{m}$, $d=1\text{m}$: $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}\text{H/m}$, so $$M=2\times 10^{-7}\times 1\times \ln 2=2\times 10^{-7}\times 0.693147=1.3863\times 10^{-7}\text{H}$$

5. Convert to nH: $M=138.63\text{nH}$ (rounded to 2 decimal places).