= Solution
<Faraday's law of induction> states that the induced <electromotive force> around an oriented closed circuit equals the negative rate of change of linked <magnetic flux>. For a stationary circuit,
$$
\boxed{\mathcal E=\oint\mathbf E\cdot d\boldsymbol\ell=-\frac{d\Phi_B}{dt},\qquad \Phi_B=\int_A\mathbf B\cdot d\mathbf A.}
$$
Here $\mathbf E$ is <electric field>, $\mathbf B$ is <magnetic flux density>, and the oriented surface $A$ has the circuit as its boundary. For $N$ identical series turns, replace $\Phi_B$ by the total <flux linkage> $N\Phi_B$. <Lenz's law> supplies the sign: an induced <electric current> opposes the \b[change] of linked <magnetic flux>, in agreement with <conservation of energy>.
If the wire moves with local <velocity> $\mathbf u$, the more general <electromotive force> includes the magnetic <Lorentz force> per unit <electric charge>:
$$
\mathcal E=\oint(\mathbf E+\mathbf u\times\mathbf B)\cdot d\boldsymbol\ell=-\frac{d}{dt}\int_{A(t)}\mathbf B\cdot d\mathbf A.
$$
The boundary and surface orientations in both expressions are related by the <right-hand rule>.
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