Question

The wire of arbitrary shape, as shown below, moves with a constant velocity $v$ in a uniform magnetic field $B$ directed into the plane of paper. The potential difference between ends $P$ and $Q$ is -

• A. $\frac{BLv}{2}$ with $P$ at lower potential
• B. $\frac{BLv}{2}$ with $P$ at higher potential
• C. $BLv$ with $Q$ at higher potential
• D. $BLv$ with $Q$ at lower potential

Answer 1

The correct option is B $\frac{BLv}{2}$ with $P$ at higher potential


We know that for arbitrary shaped wire, we take the effective length between the ends.

Here, the effective length is $PQ=L/2$ which is moving perpendicular to the magnetic field.

So, induced emf across the ends, $E=Bvl=Bv\left(L/2\right)=\frac{BLv}{2}$

Also, from right-hand rule, the end $P$ is at higher potential.

Hence, option $\left(B\right)$ is the correct answer.