A metal rod of length L rotates about one end at origin with a...

The motional emf $dE$ induced in a small element of length $dr$ at a distance $r$ from the origin is given by:

$dE = B(r) v(r) dr$

Since the rod is rotating with angular velocity $\omega$, the velocity of the element is $v(r) = \omega r$.

$dE = B_0 e^{-\lambda r} (\omega r) dr$

The total emf induced $E is the integral of$dE$from$r = 0$to$r = L$:<br><br>$E = \int_0^L B_0 \omega r e^{-\lambda r} dr$<br><br>$E = B_0 \omega \int_0^L r e^{-\lambda r} dr$<br><br>Using integration by parts:<br><br>$\int r e^{-\lambda r} dr = r \left( \dfrac{e^{-\lambda r}}{-\lambda} \right) - \int 1 \cdot \left( \dfrac{e^{-\lambda r}}{-\lambda} \right) dr$<br><br>$= -\dfrac{r}{\lambda} e^{-\lambda r} - \dfrac{1}{\lambda^2} e^{-\lambda r}$<br><br>Evaluating the definite integral from$0$to$L$:<br><br>$\int_0^L r e^{-\lambda r} dr = \left[ -\dfrac{L}{\lambda} e^{-\lambda L} - \dfrac{1}{\lambda^2} e^{-\lambda L} \right] - \left[ 0 - \dfrac{1}{\lambda^2} e^0 \right]$<br><br>$= \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right)$<br><br>Therefore, the total induced emf is:<br><br>$E = B_0 \omega \left[ \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right) \right]$<br><br>Answer:$B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]