Along straight wire carries a current $I$. A rectangular loop of sides $a$ and $b$ is placed in the same plane such that the side of length $b$ is parallel to the wire at a distance $d$ from it. The mutual inductance of the system is:

Solution

1. The magnetic field at a distance $r$ from the wire is $B=\frac{{\mu }_{0}I}{2\pi r}$.
2. The flux through a strip of width $dr$ at distance $r$ is $$\begin{array}{rl}d\Phi & =B\cdot b\cdot dr\\ & =\frac{{\mu }_{0}Ib}{2\pi r}dr\end{array}$$.
3. Integrating from $r=d$ to $r=d+a$, we get $\Phi =\frac{{\mu }_{0}Ib}{2\pi }{\int }_d^{d+a}\frac{1}{r}dr$.
4. The total flux is $\Phi =\frac{{\mu }_{0}Ib}{2\pi }\ln (\frac{d+a}{d})$.
5. Since $M=\frac{\Phi }{I}$, the mutual inductance is $\frac{{\mu }_{0}b}{2\pi }\ln (\frac{d+a}{d})$. Hence the answer is (A).