JEE Main 2026 Apr 5 Shift 2, Physics Q13: Electric Potential

JEE Main2026Apr 5, Shift 2Physics

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

A
$\frac{q E _{0}}{λ} [1 - e^{- λ L}]$
B
$(\frac{v _{0} q B _{0}}{2 λ}) [2 - e^{- 2 λ L}]$
C
$\frac{q E _{0}}{λ} [1 + e^{- λ L}]$
D
$q [\frac{E _{0} + v _{0} B _{0}}{λ}] [1 - e^{- λ L /2}]$

Solution

The magnetic force is always perpendicular to velocity, so it does no work. Only the electric force contributes to the change in kinetic energy.

$Δ K = \int_{0}^{L} q E_{0} e^{- λ x} d x = q E_{0} [- \frac{e ^{- λ x}}{λ}]_{0}^{L} = \frac{q E _{0}}{λ} (1 - e^{- λ L})$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q13: Electric Potential

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