Mutual inductance of a straight wire and square coil ($\mathcal{M}=\frac{{\mu }_{0}Na}{2\pi }\ln \frac{s+a}{s}$)

An N-turn square coil of side a lies in the plane of a long straight wire, with nearest edge at perpendicular distance s and parallel to the wire. Integrating $B={\mu }_{0}I/(2\pi r)$ across the coil gives flux-linkage magnitude $\Lambda ={\mu }_{0}NaI\ln [(s+a)/s]/(2\pi )$. Its coefficient of I is the mutual inductance magnitude; its sign depends on the selected loop normal and current directions. Flux through one turn and N-turn flux linkage must be distinguished.

In a fixed square coil beside a long wire, neglect self-inductance compared with resistance. A changing wire current induces coil current proportional to $dI/dt$. Forces on the two perpendicular sides cancel, and the unequal parallel-side fields give magnitude ${\mu }_{0}Na|Ii|[1/s-1/(s+a)]/(2\pi )$. When $I=I_0\sin \omega t$, the signed force alternates at twice the drive frequency and has zero cycle mean, while Joule dissipation is positive with mean ${\mathcal{M}}^2{\omega }^2{I}_0^2/(2R)$. A fixed coil has no mechanical work despite its time-dependent force.