A particle having charge $10^{-9} \mathrm{C}$ moving in $x-y$ plane in fields of $0.4 \hat{\mathrm{j}} \mathrm{N} / \mathrm{C}$ 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 ____ $\mathrm{m} / \mathrm{s}$.

(C) $-50 \hat{i}+100 \hat{j}$

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Magnetic Effects of Current

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

Answer: (A)

The force on the charge is

F = q E + q (v \times B)

with

v = v_{x} \hat{i} + v_{y} \hat{j}

and

B = 4 \times 1 0^{- 3} \hat{k}

. The cross product is

v \times B = 4 \times 1 0^{- 3} (v_{y} \hat{i} - v_{x} \hat{j})

, so

F = q [4 \times 1 0^{- 3} v_{y} \hat{i} + (0.4 - 4 \times 1 0^{- 3} v_{x}) \hat{j}]

. Matching the

\hat{i}

component,

4 \times 1 0^{- 10} = 1 0^{- 9} \times 4 \times 1 0^{- 3} v_{y}

gives

v_{y} = 100 m/s

; matching the

\hat{j}

component,

2 \times 1 0^{- 10} = 1 0^{- 9} (0.4 - 4 \times 1 0^{- 3} v_{x})

gives

v_{x} = 50 m/s

. The velocity of the particle is

50 \hat{i} + 100 \hat{j} m/s

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option