Question

A rod $PQ$ of length $2l$ is rotating in the horizontal plane about one end $P$ in a uniform vertical magnetic field $B$. Point $M$ is the midpoint of the rod. Find the induced emf between $M$ and $Q$ if that between $P$ and $Q$ is $100\text{V}$.

• A. $75\text{V}$
• B. $50\text{V}$
• C. $25\text{V}$
• D. $15\text{V}$

Answer 1

The correct option is A $75\text{V}$


Potential difference across $P$ and $Q$,

$V_{PQ}=\frac{B\omega L^2}{2}=2B\omega l^2=100\text{V}[\because L=2l]$

$\Rightarrow B\omega l^2=50\text{V}$

Similarly,

$V_{PM}=\frac{B\omega l^2}{2}=\frac{50}{2}=25\text{V}$

Assuming, $V_Q=0$,

$V_P=100\text{V}$ and $V_M=75\text{V}$

So, $V_{MQ}=V_M-V_Q=75-0=75\text{V}$

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