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Question

(C) When a conducting rod of length $r$ rotates with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation,

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) When a conducting rod of length $r$ rotates with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation,the induced electromotive force $(EMF)$ across its ends is given by $\varepsilon = \frac{1}{2} B \omega r^2$.For the segment $AO$ of length $l$:The potential difference between $O$ and $A$ is $V_O - V_A = \frac{1}{2} B \omega l^2$.For the segment $OC$ of length $3l$:The potential difference between $O$ and $C$ is $V_O - V_C = \frac{1}{2} B \omega (3l)^2 = \frac{9}{2} B \omega l^2$.Now,calculating $V_A - V_C$:$V_A - V_C = (V_O - V_C) - (V_O - V_A) = \frac{9}{2} B \omega l^2 - \frac{1}{2} B \omega l^2 = \frac{8}{2} B \omega l^2 = 4 B \omega l^2$.Thus,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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