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Question

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\overrightarrow{v} = \left( \frac{v_{0}}{\sqrt{2}}\widehat{x} + \frac{v_{0}}{\sqrt{2}}\widehat{y} \right)$. There exists a uniform magnetic field $\overrightarrow{B} = B_{0}\widehat{z}$ and a space varying electric field $\overrightarrow{E} = E_{0}e^{- \lambda x}\widehat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that x -coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is $\_\_\_\_$ .

Options

A.$\frac{qE_{0}}{\lambda}\left\lbrack 1 - e^{- \lambda L} \right\rbrack$
B.$\left( \frac{v_{0}{qB}_{0}}{2\lambda} \right)\left\lbrack 2 - e^{- 2\lambda\text{ }L} \right\rbrack$
C.$\frac{qE_{0}}{\lambda}\left\lbrack 1 + e^{- \lambda L} \right\rbrack$
D.$q\left\lbrack \frac{E_{0} + v_{0}B_{0}}{\lambda} \right\rbrack\left\lbrack 1 - e^{- \lambda L/2} \right\rbrack$

Solution

Sol. [Key Concept: Only the electric force does work: $\Delta\ K = \int_{0}^{L}{qE}_{0}e^{- \lambda\ x}\ dx$.]

$${\Delta K = W_{e} + W_{m} = W_{e} + 0 }{= \int_{0}^{L}\mspace{2mu} qEdx = {qE}_{0}\int_{0}^{L}\mspace{2mu} e^{- \lambda x}dx }{= \frac{{qE}_{0}}{\lambda}\left( 1 - e^{- \lambda L} \right) }$$

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