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 :

$$d\varepsilon = B(r) \cdot v \cdot dr = B(r) \cdot (r\omega) \cdot dr$$

$$\varepsilon = \int_{0}^{L} B_0 e^{-\lambda r} \cdot (r\omega) \, dr$$

$$\varepsilon = B_0 \omega \int_{0}^{L} r \cdot e^{-\lambda r} \, dr$$

$$\int r e^{-\lambda r} \, dr = r \left( -\frac{1}{\lambda} e^{-\lambda r} \right) - \int \left( -\frac{1}{\lambda} e^{-\lambda r} \right) \, dr$$ (Integration by parts)

$$\int r e^{-\lambda r} \, dr = -\frac{r}{\lambda} e^{-\lambda r} + \frac{1}{\lambda} \int e^{-\lambda r} \, dr$$

$$\int r e^{-\lambda r} \, dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r}$$

$$\int r e^{-\lambda r} \, dr = -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right)$$

$$\varepsilon = B_0 \omega \left[ -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right) \right]_{0}^{L}$$

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

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