Question

A conducting rod AC of length $4l$ is rotated about a point O in a uniform magnetic field directed into the paper. AO $=l$ and OC $=3l$. Then

• A. $V_A-V_O=\frac{B\omega l^2}{2}$
• B. $V_O-V_C=\frac{9}{2}B\omega l^2$
• C. $V_A-V_C=4B\omega l^2$
• D. $V_C-V_O=\frac{9}{2}B\omega l^2$

Answer 1

Answer: (B), (C)

$V_A-V_C=4B\omega l^2$

We know that, emf induced across a rotating rod
$\in =\frac{1}{2}B\omega l^2$
where $l$ is the distance from the axis of rotation.

$\Rightarrow V_{OC}=V_O-V_C=\frac{1}{2}B\omega (3l{)}^2=\frac{9}{2}B\omega l^2---(1)$
[$V_O>V_C$]

Similarly, $V_{OA}=V_O-V_A=\frac{1}{2}B\omega (l{)}^2=\frac{B\omega l^2}{2}---(2)$

Subtracting equation (2) from (1),
$V_A-V_C=4B\omega l^2$

Hence , options (b) and (c) are correct.