Question

A conductor of length $l$ is rotating with constant angular velocity $\omega$ about an axis $O{O}^{\prime }$ in a uniform magnetic field $B$ as shown in figure. The emf induced between ends $P$ and $Q$ will be

• A. $\frac{1}{4}B\omega l^2$
• B. $\frac{5}{10}B\omega l^2$
• C. Zero
• D. $\frac{1}{2}B\omega l^2$

Answer 1

The correct option is A $\frac{1}{4}B\omega l^2$


$E_1$ is emf generated in the rod from axis to corner $P$
$E_1=\frac{1}{2}B(\frac{L}{4}{)}^2\omega$
$E_2$ is emf generated in the rod from axis to corner $Q$
$E_2=\frac{1}{2}B(\frac{3L}{4}{)}^2\omega$

Potential difference between $P$ and $Q$ is
(From Kirchoffs Voltage Law)
$V_P-E_1+E_2=V_Q$
$\Rightarrow V_P-V_Q=E_1-E_2$
​​​​​​​ $\Rightarrow V_Q-V_P=E_2-E_1$
$\Rightarrow \frac{1}{2}B\omega L^2(\frac{9}{16}-\frac{1}{16})$
$\Rightarrow \frac{B\omega L^2}{4}$


Or


$ϵ={\int }_{\frac{-l}{4}}^{\frac{3l}{4}}B\omega xdx=\frac{1}{4}B\omega l^2$