A conducting rod of length L rotates with ang | vedclass

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(A) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The induced emf $de$ across this small element is given by $de

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$\frac{B L^2 \omega}{4}$

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(A) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The induced emf $de$ across this small element is given by $de = B v dr$,where $v = r \omega$.So,$de = B (r \omega) dr$.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,the emf induced in each half of the rod is $\frac{B \omega L^2}{8}$.Because the two halves are connected in parallel with respect to the center,the total emf between the two ends is the sum of the emf induced in each half: $e_{total} = \frac{B \omega L^2}{8} + \frac{B \omega L^2}{8} = \frac{B \omega L^2}{4}$.

$A$ 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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