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

(A) $\frac{q E_{0}}{\lambda}\left[1-\mathrm{e}^{-\lambda \mathrm{L}}\right]$

(B) $\left(\frac{v_{0} \mathrm{qB}_{0}}{2 \lambda}\right)\left[2-\mathrm{e}^{-2 \lambda \mathrm{~L}}\right]$

(C) $\frac{q E_{0}}{\lambda}\left[1+\mathrm{e}^{-\lambda \mathrm{L}}\right]$

(D) $q\left[\frac{E_{0}+v_{0} B_{0}}{\lambda}\right]\left[1-\mathrm{e}^{-\lambda \mathrm{L} / 2}\right]$

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Magnetic Effects of Current

A particle of charge

q

and mass

m

is projected from 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 x-coordinate has changed from

x = 0

x = L

, the change in the kinetic energy is ______.

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

Answer: (A)

Solution

The magnetic force on a moving charge is always perpendicular to its velocity and so does no work; the whole change in kinetic energy is therefore the work done by the electric field. As the particle moves from

x = 0

x = L

, this work is

Δ K = \int_{0}^{L} q E d x = q E_{0} \int_{0}^{L} e^{- λ x} d x

. Carrying out the integration,

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

. The change in kinetic energy is

\frac{q E _{0}}{λ} (1 - e^{- λ L})

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Magnetic Effects of Current

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