Question

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = \left( \frac{v_o}{\sqrt{2}} \hat{x} + \frac{v_o}{\sqrt{2}} \hat{y} \right)$. There exists a uniform magnetic field $\vec{B} = B_o \hat{z}$ and a space varying electric field $\vec{E} = E_o e^{-\lambda x} \hat{x}$ 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 \dots

• A. $\frac{q E_o}{\lambda} [ 1 - e^{-\lambda L} ]$
• B. $\left( \frac{v_o q B_o}{2 \lambda} \right) [ 2 - e^{-2\lambda L} ]$
• C. $\frac{q E_o}{\lambda} [ 1 + e^{-\lambda L} ]$
• D. $q \left( \frac{E_o + v_o B_o}{\lambda} \right) [ 1 - e^{-\lambda L/2} ]$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The work-energy theorem states that the work done by all forces equals the change in kinetic energy ($\Delta K = W_{total}$).
Magnetic force ($\vec{F}_m = q(\vec{v} \times \vec{B})$) is always perpendicular to velocity, so it does zero work.
Only the electric field does work.
Step 2: Detailed Explanation:
Work done by electric force $W_e = \int \vec{F}_e \cdot d\vec{r} = \int (q \vec{E}) \cdot d\vec{r}$.
Since $\vec{E}$ is along $\hat{x}$, $d\vec{r} \cdot \hat{x} = dx$.
\[ \Delta K = W_e = \int_0^L q E_o e^{-\lambda x} dx \]
\[ \Delta K = q E_o \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_0^L \]
\[ \Delta K = -\frac{q E_o}{\lambda} ( e^{-\lambda L} - e^0 ) \]
\[ \Delta K = \frac{q E_o}{\lambda} ( 1 - e^{-\lambda L} ) \]
Step 3: Final Answer:
The change in kinetic energy is $\frac{q E_o}{\lambda} [ 1 - e^{-\lambda L} ]$.