A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = \left(\dfrac{v_0}{\sqrt{2}}\hat{x} + \dfrac{v_0}{\sqrt{2}}\hat{y}\right)$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0e^{-\lambda x}\hat{x}$ within the region $0 \le x \le L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is ______.
Answer: (A) $\dfrac{qE_0}{\lambda}\left[1 - e^{-\lambda L}\right]$
Concept. The magnetic force does no work, so the change in kinetic energy is the work done by $\vec E$ alone. $\vec E$ is along $x$, so only the change in $x$ matters: $W=\displaystyle\int_0^LqE_0e^{-\lambda x}dx$.
Step 1. Integrate. $W=\dfrac{qE_0}{\lambda}\left[-e^{-\lambda x}\right]_0^L=\dfrac{qE_0}{\lambda}\left(1-e^{-\lambda L}\right)$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
