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

Conducting electrons experience a magnetic force $F_{m}=evB.$ So they move from P to Q within the rod. The end P of the rod becomes positively charged while end Q becomes negatively charged, hence an electric field is set up within the rod which opposes the further downward movement of electrons, i.e., an equilibrium $F_{e}=F_{m},$ i.e.,

$eE=evB \; or\; E=vB\Rightarrow \text {Induced emf}e=El=Bvl\left [ E=\frac{V}{l} \right ]$

If the rod is moved by making an angle $\theta$ with the direction of the magnetic field or length. Induced emf,

$e=Bvl\; \sin \theta$

Emf induced across PQ due to its motion or change in magnetic flux linked with the loop change due to the change of enclosed area. The induced electric field E along the dotted line CD (perpendicular to both $\vec{V}\; \text {and }\vec{B}$ and along $\vec{V}\times \vec{B}$ ) $=vB$

Therefore, the motional emf along.

$PQ=\left ( \text {length PQ} \right )\times \left ( \text {field along PQ} \right )$

$= \left ( \text {length PQ} \right )\times \left ( \text {vB} \sin \theta \right )$

$= \left ( \frac{d}{\sin \theta } \right )\times \left ( \text {vB} \sin \theta \right )=vBd$

This induced emf make flow of current in closed circuit of resistnce R.

$I=\frac{dvB}{R}$ and independent of $\theta$.