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Question

(D) The motional electromotive force $(EMF)$ induced in a rod of length $r$ rotating with angular velocity $\omega$ in a uniform magnetic field $B$ is

Vedclass · Accepted Answer

$V_C - V_O = \frac{9}{2}B\omega l^2$

Vedclass · Answer

(D) The motional electromotive force $(EMF)$ induced in a rod of length $r$ rotating with angular velocity $\omega$ in a uniform magnetic field $B$ is given by $e = \frac{1}{2} B r^2 \omega$.For the segment $AO$ of length $l$:The potential difference is $V_A - V_O = \frac{1}{2} B l^2 \omega$.For the segment $OC$ of length $3l$:The potential difference is $V_C - V_O = \frac{1}{2} B (3l)^2 \omega = \frac{9}{2} B l^2 \omega$.Now,calculating $V_A - V_C$:$V_A - V_C = (V_A - V_O) - (V_C - V_O) = \frac{1}{2} B l^2 \omega - \frac{9}{2} B l^2 \omega = -4 B l^2 \omega$.Thus,$V_C - V_A = 4 B l^2 \omega$. Comparing with the given options,option $D$ is correct as $V_C - V_O = \frac{9}{2} B \omega l^2$.

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

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