A metal rod of length L rotates about one end | vedclass

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Question

(A) The motional emf $d\varepsilon$ induced in a small element $dr$ at distance $r$ is $d\varepsilon = (v) B(r) dr$, where $v = \omega r$.Thus, $d\var

Vedclass · Accepted Answer

$B_0 \omega [\frac{1}{\lambda^2} - e^{-\lambda L} (\frac{1}{\lambda^2} + \frac{L}{\lambda})]$

Vedclass · Answer

(A) The motional emf $d\varepsilon$ induced in a small element $dr$ at distance $r$ is $d\varepsilon = (v) B(r) dr$, where $v = \omega r$.Thus, $d\varepsilon = (\omega r) (B_0 e^{-\lambda r}) dr$.Integrating from $r=0$ to $L$: $\varepsilon = \int_0^L \omega B_0 r e^{-\lambda r} dr$.Using integration by parts $\int r e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r}$.Evaluating from $0$ to $L$: $\varepsilon = \omega B_0 [(-\frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} e^{-\lambda L}) - (0 - \frac{1}{\lambda^2})]$.Simplifying the expression: $\varepsilon = B_0 \omega [\frac{1}{\lambda^2} - e^{-\lambda L} (\frac{1}{\lambda^2} + \frac{L}{\lambda})]$.

$A$ metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_0 e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

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