Find the current in the wire for the configuration shown in the figure. Wire $PQ$ has negligible resistance. $\vec{B}$ is the magnetic field coming out of the paper. $\theta$ is a fixed angle made by $PQ$ as it travels smoothly over two conducting parallel wires separated by a distance $d$.

(D) The induced emf in a moving rod is given by $\varepsilon = B v l \sin \phi$,where $\phi$ is the angle between the velocity vector $\vec{v}$ and the length vector $\vec{l}$.
Alternatively,it can be expressed as $\varepsilon = B v_{\perp} l$,where $v_{\perp}$ is the component of velocity perpendicular to the rod,or $\varepsilon = B v l_{\perp}$,where $l_{\perp}$ is the effective length of the rod perpendicular to the velocity.
In this configuration,the effective length of the rod $PQ$ that cuts the magnetic field lines while moving with velocity $v$ is the perpendicular distance between the two parallel wires,which is $d$.
Thus,the induced emf is $\varepsilon = B v d$.
The induced current $I$ in the circuit with resistance $R$ is given by Ohm's law:
$I = \frac{\varepsilon}{R} = \frac{B v d}{R}$.
According to Lenz's law,the direction of the induced current is such that it opposes the change in magnetic flux,resulting in a clockwise direction in this loop.