JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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 $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$ . After travelling a distance such that $x$ -coordinate has changed from $x=0$ to $x=L$ , the change in the kinetic energy is _______.

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]$
B
$\left(\dfrac{v_0 q B_0}{2\lambda}\right)[2 - e^{-2\lambda L}]$
C
$\dfrac{qE_0}{\lambda}[1 + e^{-\lambda L}]$
D
$q\left(\dfrac{E_0 + v_0 B_0}{\lambda}\right)[1 - e^{-\lambda L/2}]$

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Solution

According to the work-energy theorem, the change in kinetic energy of a particle is equal to the total work done by all the forces acting on it.

The forces acting on the particle are the electric force and the magnetic force. The magnetic force is given by $\vec{F}_B = q(\vec{v} \times \vec{B})$ . Since the magnetic force is always perpendicular to the velocity vector of the particle, the work done by the magnetic field is zero.

The change in kinetic energy is equal to the work done by the electric field alone. The electric force is $\vec{F}_E = q\vec{E} = qE_0 e^{-\lambda x}\hat{x}$ .

The work done by the electric field as the particle moves from $x = 0$ to $x = L$ is given by

$W_E = \int_{0}^{L} \vec{F}_E \cdot d\vec{r} = \int_{0}^{L} qE_0 e^{-\lambda x} dx$

Evaluating the integral, we get

$W_E = qE_0 \left[ \dfrac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}$

$W_E = -\dfrac{qE_0}{\lambda} (e^{-\lambda L} - e^{0})$

$W_E = \dfrac{qE_0}{\lambda} [1 - e^{-\lambda L}]$

Thus, the change in kinetic energy is $\dfrac{qE_0}{\lambda} [1 - e^{-\lambda L}]$ .

Answer: $\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]$

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Options

$\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]$

$\left(\dfrac{v_0 q B_0}{2\lambda}\right)[2 - e^{-2\lambda L}]$

$\dfrac{qE_0}{\lambda}[1 + e^{-\lambda L}]$

$q\left(\dfrac{E_0 + v_0 B_0}{\lambda}\right)[1 - e^{-\lambda L/2}]$

Solution

According to the work-energy theorem, the change in kinetic energy of a particle is equal to the total work done by all the forces acting on it.

The change in kinetic energy is equal to the work done by the electric field alone. The electric force is $\vec{F}_E = q\vec{E} = qE_0 e^{-\lambda x}\hat{x}$ .

The work done by the electric field as the particle moves from $x = 0$ to $x = L$ is given by

$W_E = \int_{0}^{L} \vec{F}_E \cdot d\vec{r} = \int_{0}^{L} qE_0 e^{-\lambda x} dx$

Evaluating the integral, we get

$W_E = qE_0 \left[ \dfrac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}$

$W_E = -\dfrac{qE_0}{\lambda} (e^{-\lambda L} - e^{0})$

$W_E = \dfrac{qE_0}{\lambda} [1 - e^{-\lambda L}]$

Thus, the change in kinetic energy is $\dfrac{qE_0}{\lambda} [1 - e^{-\lambda L}]$ .

Answer: $\dfrac{qE_0}{\lambda}[1 - e^{-\lambda L}]$

JEE Main 2026 — Physics Electromagnetism

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism