Exercise 4.1.7: Flux through a rectangle near a long wire

In the plane of the wire $B={\mu }_{0}I/(2\pi \rho )a_{\varphi }$ is normal to the loop (into the page in the figure). Take strips of width $d\rho$ and length $l$:

$\ln (0.25/0.05)=\ln 5=1.609$, so $\Phi =2\times 10^{-7}\times 100\times 1\times 1.609=3.22\times 10^{-5}$ Wb $=32$ μWb.

$A$ has only a $z$-component, so only the two sides parallel to the wire contribute to ${\oint }_{C}A\cdot dl$. Going round the loop so that its normal is $a_{\varphi }$ (right-hand rule), the near side at $\rho =d$ is traversed along $+z$ and the far side at $\rho =d+w$ along $-z$:

the same flux. The far side, where $A_z$ is smaller, subtracts less than the near side adds, and the flux is their difference. The ends of the loop are perpendicular to $A$ and contribute nothing.

Units: ${\mu }_{0}Il$ is (H/m)(A)(m) $=$ Wb. This flux divided by $I$ is the mutual inductance between the wire and the loop, $M=\frac{{\mu }_{0}l}{2\pi }\ln \frac{d+w}{d}=0.32$ μH, a notion of the next chapter. That $A$ is determined only up to a gradient (the constant $\ln {\rho }_0$ one may add) does not matter: it drops out of the closed line integral. Exercise 4.1.9 derives this $A$.