Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 5 April, Evening Shift — Question 13

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\bar{v} = \left(\frac{v_0}{\sqrt{2}}\hat{x} + \frac{v_0}{\sqrt{2}}\hat{y}\right)$ . There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}$ 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

Option A:
$\frac{qE_0}{\lambda}[1-e^{-\lambda L}]$
Correct
Option B:
$\frac{v_0 q B_0}{2\lambda}[2-e^{-\lambda L}]$
Option C:
$\frac{qE_0}{\lambda}[1+e^{-\lambda L}]$
Option D:
$\frac{q[E_0+v_0B_0]}{\lambda}[1-e^{-\lambda L/2}]$

Answer: A

Step-by-step solution

Magnetic field does no work. Change in KE = work done by electric field = $\int qE dx = qE_0\int_0^L e^{-\lambda x}dx = \frac{qE_0}{\lambda}(1-e^{-\lambda L})$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in magnetic Fields