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(C) The induced $EMF$ in a rotating rod of length $l$ in a magnetic field $B$ with angular velocity $\omega$ is given by $\varepsilon = \frac{1}{2} B

Vedclass · Accepted Answer

$\frac{3}{2} B \omega L^2$

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(C) The induced $EMF$ in a rotating rod of length $l$ in a magnetic field $B$ with angular velocity $\omega$ is given by $\varepsilon = \frac{1}{2} B \omega l^2$.For the horizontal part of length $2L$,the induced $EMF$ is $\varepsilon_1 = \frac{1}{2} B \omega (2L)^2 = 2 B \omega L^2$.The potential increases towards the outer end (away from the pivot). Thus,$V_A$ is higher than the potential at the pivot.For the vertical part of length $L$,the induced $EMF$ is $\varepsilon_2 = \frac{1}{2} B \omega L^2$.The potential increases towards the outer end $B$. Thus,$V_B$ is higher than the potential at the pivot.Let $V_P$ be the potential at the pivot. Then $V_A = V_P + 2 B \omega L^2$ and $V_B = V_P + \frac{1}{2} B \omega L^2$.Subtracting the two equations: $V_A - V_B = (V_P + 2 B \omega L^2) - (V_P + \frac{1}{2} B \omega L^2) = \frac{3}{2} B \omega L^2$.

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