Home exercise 4.2.2: Wire and rectangular loop

$B_1=\frac{{\mu }_0{I}_1}{2\pi \rho }{a}_{\varphi }$ (Problem 4.2.1), normal to the plane of the loop and of the same sign over the whole loop.

Cut the loop into strips of width $d\rho$ and length $b$ at the distance $\rho$ from the wire, $d\le \rho \le d+a$. Each strip carries the flux $B_1(\rho )bd\rho$:

For $d\gg a$, $\ln (1+a/d)\approx a/d$ and $L_{12}\approx {\mu }_0{N}_{2}ab/(2\pi d)\to 0$: the loop area $ab$ times the (almost uniform) field per ampere at the distance $d$, times the turns (Figure 4.2.22).

The result is independent of $I_1$, as every inductance must be, and by reciprocity it is also the flux the loop would send through the wire's circuit per ampere in the loop, which would be much harder to compute directly. Units: ${\mu }_{0}b$ is (H/m)$\cdot$m $=$ H. With $a=b=10\mathrm{cm}$, $d=5\mathrm{cm}$ and $N_2=1$: $L_{12}=2\times 10^{-7}\times 0.1\times \ln 3=22\mathrm{nH}$.