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Question

(B) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The velocity of this element is $v = r\omega$.The induced emf

Vedclass · Accepted Answer

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

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(B) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The velocity of this element is $v = r\omega$.The induced emf $de$ across this small element is given by $de = B v dr = B (r\omega) dr$.To find the total induced emf $e$ between the center and one end of the rod (length $L/2$),we integrate from $r = 0$ to $r = L/2$:$e = \int_{0}^{L/2} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{L/2} = B \omega \frac{(L/2)^2}{2} = \frac{B \omega L^2}{8}$.Since the rod rotates about its center,there are two such segments of length $L/2$ on either side of the axis,but they are connected in parallel. Thus,the potential difference between the two ends is the same as the emf induced in one segment of length $L/2$,which is $\frac{B \omega L^2}{8}$.Wait,if the question implies the rod is rotating about one end,the answer is $\frac{B \omega L^2}{2}$. Given the options and the standard interpretation of a rod of length $L$ rotating about its center,the potential difference between the two ends is $0$ because the emfs induced in the two halves cancel out. However,if the question asks for the emf between the center and one end,it is $\frac{B \omega L^2}{8}$. Looking at the provided options,it is highly likely the question assumes rotation about one end of a rod of length $L$. For a rod of length $L$ rotating about one end,$e = \int_{0}^{L} B \omega r dr = \frac{1}{2} B \omega L^2$.

$A$ conducting rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field of induction $B$ which is perpendicular to its motion. The induced emf developed between the two ends of the rod is

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