JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

A particle having charge $1 0^{- 9}$ C moving in $x$ - $y$ plane in fields of $0.4 \hat{j}$ N/C and $4 \times 1 0^{- 3} \hat{k}$ T experiences a force of $(4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10}$ N. The velocity of the particle at that instant is______m/s.

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

Solution

$F = q E + q (V \times B)$

$(4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10}$

$=10^{-9} \left[ 0.4\,\hat{j} + (v_x\hat{i}+v_y\hat{j}) \times (4\times10^{-3})\hat{k} \right]$

$(4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10}$

$=10^{-9} \left( 0.4\,\hat{j} - 4\times10^{-3}v_x\,\hat{j} + 4\times10^{-3}v_y\,\hat{i} \right)$

On comparing

$v_{y} = 100, v_{x} = 50$

$V = 50 \hat{i} + 100 \hat{j}$

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →