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(C) The motional electromotive force $(EMF)$ induced in a rod of length $l$ rotating about one end with angular velocity $\omega$ in a uniform magneti

Vedclass · Accepted Answer

$V_Q - V_O = 2B\omega a^2$

Vedclass · Answer

(C) The motional electromotive force $(EMF)$ induced in a rod of length $l$ rotating about one end with angular velocity $\omega$ in a uniform magnetic field $B$ is given by $\varepsilon = \frac{1}{2} B \omega l^2$.Here,the ring rotates about point $O$. Any chord of length $l$ passing through $O$ acts as a rotating rod.The distance $OP = \sqrt{a^2 + a^2} = \sqrt{2}a$. Thus,$V_P - V_O = \frac{1}{2} B \omega (\sqrt{2}a)^2 = B \omega a^2$.The distance $OQ = 2a$. Thus,$V_Q - V_O = \frac{1}{2} B \omega (2a)^2 = 2 B \omega a^2$.Comparing these with the options,option $C$ is correct.

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