Question

Sol. [Key Concept: $e = \int_{0}^{L}B(r)\ \omega\ r\ dr$ with $B = B_{0}e^{- \lambda\ r}$; integrate by parts.]

$${e = \int_{0}^{L}\mspace{2mu} B(\omega r)dr = B_{0}\omega\int_{0}^{L}\mspace{2mu} e^{- \lambda r} \cdot rdr }{= B_{0}\omega\left\lbrack r\int e^{- \lambda r}dr - \int 1 \cdot \frac{e^{- \lambda r}}{- \lambda}dr \right\rbrack_{0}^{L} }{= B_{0}\omega\left\lbrack \frac{{re}^{- \lambda r}}{- \lambda} - \frac{e^{- \lambda r}}{\lambda^{2}} \right\rbrack_{0}^{L} }{= B_{0}\omega\left\lbrack \left( 0 + \frac{1}{\lambda^{2}} \right) - \left( \frac{{Le}^{- \lambda L}}{\lambda} + \frac{e^{- \lambda L}}{\lambda^{2}} \right) \right\rbrack }{e = B_{0}\omega\left\lbrack \frac{1}{\lambda^{2}} - e^{- \lambda L}\left( \frac{1}{\lambda^{2}} + \frac{L}{\lambda} \right) \right\rbrack }$$