Exercise 5.3.2: Rectangular loop beside a long wire

The field of the wire at distance $x$ is $B={\mu }_{0}I/(2\pi x)$, into the page on the side of the loop (for $I$ upward). Taking $dS$ into the page,

$M=2\times 10^{-7}\times 0.2\times \ln 3=4.39\times 10^{-8}\mathrm{H}$ and $\omega I_0=314.2\times 10=3142A/s$, so $\hat{e}=M\omega I_0=1.38\times 10^{-4}\mathrm{V}=0.138\mathrm{mV}$.

While $I$ increases, the flux into the page grows. The induced current circulates so as to produce flux out of the page inside the loop: counter-clockwise. Equivalently, the side nearest the wire carries a current antiparallel to $I$ and is repelled, so the loop is pushed away from the wire.

Units: ${\mu }_{0}a$ is $\mathrm{H}$, and $\mathrm{H}\cdot A/s=\mathrm{V}$. $0.14\mathrm{mV}$ from a $10\mathrm{A}$ conductor $5\mathrm{cm}$ away is the order of magnitude of the mains hum picked up by a wiring loop. It grows with $a$, with the logarithm of the aspect ratio, and linearly with frequency, which is why the same loop near a switching converter at $100\mathrm{kHz}$ picks up $2000$ times more.