= Solution
<Electromagnetic induction> produces an <electromotive force> when <magnetic flux linkage> changes. For an oriented moving loop,
$$
\mathcal E=\oint(\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell
=-\frac{d\Phi_B}{dt}.
$$
For a stationary loop this reduces to <Faraday's law of induction>, $\nabla\times\mathbf E=-\partial_t\mathbf B$. Motion in a static field can instead supply the motional term. <Lenz's law> gives the minus sign: the induced <electric current> opposes the flux change, with the required electrical <energy> supplied by mechanical <work> or the field source.
For linear fixed-geometry circuits, flux linkages satisfy $\Lambda_1=L_1I_1+MI_2$ and $\Lambda_2=MI_1+L_2I_2$. <Self-inductance> gives $\mathcal E_1=-L_1\dot I_1$; <mutual inductance> gives the additional term $-M\dot I_2$. Reciprocity gives the same mutual coefficient for both circuits, with its sign determined by winding orientation. The stored field <energy> is
$$
W_B=\frac12L_1I_1^2+\frac12L_2I_2^2+MI_1I_2,
\qquad |M|\le\sqrt{L_1L_2}.
$$
The bound follows from nonnegative <energy> for arbitrary <electric currents>. In a linear medium the equivalent field expression is $W_B=\tfrac12\int\mathbf B\cdot\mathbf H\,dV$.
An ideal <electrical transformer> links both windings with one changing core flux, giving $V_1/V_2=N_1/N_2$ for corresponding voltage magnitudes. A loaded secondary drives an opposing magnetomotive force and increases primary <electric current>; ideally $I_1/I_2=N_2/N_1$ and input and output powers agree. \b[A <electrical transformer> exchanges voltage for <electric current>, not for extra <power>.] Real devices have winding resistance, leakage flux, eddy-current and <hysteresis> losses, and finite magnetizing <electric current>. A steady DC input gives no continuing secondary voltage after the switching transient and may cause excessive primary <electric current> or core saturation.
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