Question

A conducting rod $PQ$ rotates with a constant angular velocity $\omega$ about the axis which passes through the point $O$ and is perpendicular to its length. A uniform magnetic field $B$ exists in this region, as shown in the figure. The potential difference between the two ends of the rod is :

• A. $6B\omega l^2$
• B. $5B\omega l^2$
• C. $4B\omega l^2$
• D. Zero

Answer 1

The correct option is A $6B\omega l^2$


EMF developed across $OP$,

$E_1=\frac{B\omega (2l{)}^2}{2}=2B\omega l^2$

EMF developed across $OQ$,

$E_2=\frac{B\omega (4l{)}^2}{2}=8B\omega l^2$

The equivalent circuit is shown in the below figure.

The direction of higher potential of each cell has been found using right-hand rule.

Using, KVL,

$V_P-E_1+E_2=V_Q$

$\Rightarrow V_Q-V_P=E_2-E_1=8B\omega l^2-2B\omega l^2=6B\omega l^2$