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 :

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 \(r = L\):

\(E = \int_0^L B_0 \omega r e^{-\lambda r} dr\)

\(E = B_0 \omega \int_0^L r e^{-\lambda r} dr\)

Using integration by parts:

\(\int r e^{-\lambda r} dr = r \left( \dfrac{e^{-\lambda r}}{-\lambda} \right) - \int 1 \cdot \left( \dfrac{e^{-\lambda r}}{-\lambda} \right) dr\)

\(= -\dfrac{r}{\lambda} e^{-\lambda r} - \dfrac{1}{\lambda^2} e^{-\lambda r}\)

Evaluating the definite integral from \(0\) to \(L\):

\(\int_0^L r e^{-\lambda r} dr = \left[ -\dfrac{L}{\lambda} e^{-\lambda L} - \dfrac{1}{\lambda^2} e^{-\lambda L} \right] - \left[ 0 - \dfrac{1}{\lambda^2} e^0 \right]\)

\(= \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right)\)

Therefore, the total induced emf is:

\(E = B_0 \omega \left[ \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right) \right]\)

Answer: \(B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]\)