In a vacuum chamber, a particle of charge $1 \mu \mathrm{C}$ and mass 1 mg is projected with a velocity $(\hat{i}+2 \hat{j}) \mathrm{ms}^{-1}$ from the $XZ$ plane at time $t=0$ in an electric field of $1 \hat{\imath} \mathrm{Vm}^{-1}$. At $t=0.2 \mathrm{~s}$, the electric field is switched off and a magnetic field of $6 \hat{j} \mathrm{~T}$ is switched on. The acceleration due to gravity is $-10 \hat{j} \mathrm{~ms}^{-2}$. Correct option(s) is/are:

(A) The vertical distance of the particle from the $XZ$ plane at $t=0.3 \mathrm{~s}$ is 15 cm.; (C) The radius of the trajectory of the particle for $t>0.2 \mathrm{~s}$ is 20 cm.

(B) The vertical distance of the particle from the $XZ$ plane at $t=0.4 \mathrm{~s}$ is 10 cm.

(D) The particle will be in the $XZ$ plane at $t=0.35 \mathrm{~s}$.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Magnetic Effects of Current

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In a vacuum chamber, a particle of charge

1 μ C

and mass 1 mg is projected with a velocity

(\hat{i} + 2 \hat{j}) ms^{- 1}

from the

X Z

plane at time

t = 0

in an electric field of

1 \overset{}{^} Vm^{- 1}

. At

t = 0.2 s

, the electric field is switched off and a magnetic field of

6 \hat{j} T

is switched on. The acceleration due to gravity is

- 10 \hat{j} ms^{- 2}

. Correct option(s) is/are:

(A)The vertical distance of the particle from the $X Z$ plane at $t = 0.3 s$ is 15 cm.✓ Correct
(B)The vertical distance of the particle from the $X Z$ plane at $t = 0.4 s$ is 10 cm.
(C)The radius of the trajectory of the particle for $t > 0.2 s$ is 20 cm.✓ Correct
(D)The particle will be in the $X Z$ plane at $t = 0.35 s$ .

Answer: (A), (C)

Solution

For

0 \leq t \leq 0.2

s: acceleration

a_{1} = \frac{q E}{m} + g = \hat{i} - 10 \hat{j}

m/s². At

t = 0.2

v_{(0.2)} = (\hat{i} + 2 \hat{j}) + (\hat{i} - 10 \hat{j}) (0.2) = 1.2 \hat{i}

m/s. Vertical position:

y = 2 (0.2) - \frac{1}{2} (10) (0.2)^{2} = 0.2

m = 20 cm.

For

t > 0.2

s: magnetic force causes circular motion in XZ plane. Vertical motion:

y = 0.2 - 5 τ^{2}

where

τ = t - 0.2

.

(A) At

t = 0.3

y = 0.2 - 5 (0.1)^{2} = 0.15

m = 15 cm. Correct.

(B) At

t = 0.4

y = 0.2 - 5 (0.2)^{2} = 0

cm. Incorrect.

(C) Radius:

R = \frac{m v _{⊥}}{q B} = \frac{1 0 ^{- 6} \times 1.2}{1 0 ^{- 6} \times 6} = 0.2

m = 20 cm. Correct.

(D) Particle reaches XZ plane when

y = 0

, i.e., at

t = 0.4

s, not 0.35 s. Incorrect.

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Magnetic Effects of Current

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