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Question

(C) 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

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

Vedclass · Answer

(C) 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 $e = \frac{1}{2}B\omega r^2$.For the segment $AO$ of length $l$,the induced $EMF$ is $V_O - V_A = \frac{1}{2}B\omega l^2$. Thus,$V_A - V_O = -\frac{1}{2}B\omega l^2$.For the segment $OC$ of length $3l$,the induced $EMF$ is $V_O - V_C = \frac{1}{2}B\omega (3l)^2 = \frac{9}{2}B\omega l^2$.To find $V_A - V_C$,we can write: $V_A - V_C = (V_A - V_O) + (V_O - V_C) = -\frac{1}{2}B\omega l^2 + \frac{9}{2}B\omega l^2 = \frac{8}{2}B\omega l^2 = 4B\omega l^2$.Therefore,the correct option is $C$.

$A$ 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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