Past exam of the physics B course of the University of Cambridge / 2010 / Paper 1 / Section A / 1 / Solution

For a fixed linear circuit, self-inductance relates the magnetic flux linkage produced by its own current to that current: $\Lambda =LI$, with $\Lambda ={\sum }_{\mathrm{turns}}{\Phi }_B$. The self-induced electromotive force is $\mathcal{E}=-d\Lambda /dt=-LdI/dt$ when $L$ is constant. Its unit is the henry, $1\mathrm{H}=1Wb/A$.

mutual inductance relates the linkage in one circuit to current in another: ${\Lambda }_2=M_{21}{I}_1$ and ${\Lambda }_1=M_{12}{I}_2$. For stationary reciprocal linear media, $M_{12}=M_{21}=M$ with consistently oriented turns, and ${\mathcal{E}}_2=-Md{I}_1/dt$. Orientation fixes the sign. The coupled magnetic energy is $W=\frac{1}{2}{L}_1{I}_1^2+\frac{1}{2}{L}_2{I}_2^2+M{I}_1{I}_2$, so positivity requires $M^2\le L_1{L}_2$. Moving coils or nonlinear cores require differentiation of the full linkage rather than treating the inductance as a constant.