$A$ magnetic field $\vec{B} = B_0 \sin(\omega t) \hat{k}$ covers a large region where a wire $AB$ slides smoothly over two parallel conductors separated by a distance $d$ as shown in the figure. The wires are in the $xy$-plane. The wire $AB$ (of length $d$) has resistance $R$ and the parallel wires have negligible resistance. If $AB$ is moving with velocity $v$,what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

(N/A) Let the wire $AB$ be at position $x$ at time $t$. The area of the loop is $A = x \cdot d$.
The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (B_0 \sin(\omega t)) \cdot (x d)$.
According to Faraday's law,the induced electromotive force (emf) $\varepsilon$ is:
$\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} [B_0 d x \sin(\omega t)]$
Using the product rule:
$\varepsilon = -B_0 d [\frac{dx}{dt} \sin(\omega t) + x \frac{d}{dt} \sin(\omega t)]$
Since $v = \frac{dx}{dt}$,we have:
$\varepsilon = -B_0 d [v \sin(\omega t) + x \omega \cos(\omega t)]$
The magnitude of the induced current $I$ is:
$I = \frac{|\varepsilon|}{R} = \frac{B_0 d}{R} [v \sin(\omega t) + x \omega \cos(\omega t)]$
The magnetic force $F_m$ on the wire $AB$ is $F_m = I L B = I d (B_0 \sin(\omega t))$.
To keep the wire moving at a constant velocity,an external force $F_{ext}$ must be applied such that $F_{ext} = F_m$ (in magnitude).
$F_{ext} = \frac{B_0^2 d^2}{R} [v \sin(\omega t) + x \omega \cos(\omega t)] \sin(\omega t)$.