A particle having charge 10 -9 C moving in x-y plane in fields of 0.4 hat j N/C and 4 × 10 -3 hat k T experiences a force of (4hat i +2hat j )× 10 -10 N . The velocity of the particle at that instant is m/s .

Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 2 April, Evening Shift — Question 4

A particle having charge $10^{-9}$ C moving in $x-y$ plane in fields of $0.4 \hat{j} \mathrm{N/C}$ and $4 \times 10^{-3} \hat{k} \mathrm{T}$ experiences a force of $(4\hat{i}+2\hat{j})\times 10^{-10} \mathrm{N}$ . The velocity of the particle at that instant is $\mathrm{m/s}$ .

Option A:
$50\hat{i}+100\hat{j}$
Option B:
$100\hat{i}+50\hat{j}$
Option C:
$-50\hat{i}+100\hat{j}$
Correct
Option D:
$50\hat{i}-100\hat{j}$

Answer: C

Step-by-step solution

$\mathrm{Energy} = \frac{\mathrm{hc}}{\lambda} = \frac{\mathrm{GM}_{1}\mathrm{M}_{2}}{\mathrm{r}}$ $\left[\mathrm{ML}^{-2}\mathrm{T}^{-2}\right] = \left[\mathrm{h}\right]\left[\mathrm{T}^{-1}\right] = \left[\mathrm{G}\right]\left[\mathrm{M}^{2}\mathrm{L}^{-2}\right]$ $\left[\mathrm{G}\right] = \left[\mathrm{M}^{-2}\mathrm{L}^{-1}\mathrm{T}^{-1}\mathrm{h}^{-1}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in Combined Electric and Magnetic Fields