JEE Main 2026 — Magnetic Effects of Current Question with Solution

The Lorentz force experienced by a charged particle is given by $\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$.

Given that the particle moves in the $x$-$y$ plane, its velocity can be written as $\vec{v}=v_x\hat{i}+v_y\hat{j}$.

Substituting the given values into the force equation:

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

Using the cross product rules $\hat{i}\times \hat{k}=-\hat{j}$ and $\hat{j}\times \hat{k}=\hat{i}$:

$\vec{F}=10^{-9}[0.4\hat{j}-4\times 10^{-3}{v}_x\hat{j}+4\times 10^{-3}{v}_y\hat{i}]$

$\vec{F}=10^{-9}[4\times 10^{-3}{v}_y\hat{i}+(0.4-4\times 10^{-3}{v}_x)\hat{j}]$

The given force is $\vec{F}=(4\hat{i}+2\hat{j})\times 10^{-10}$ N, which can be rewritten as:

$\vec{F}=10^{-9}(0.4\hat{i}+0.2\hat{j})$ N

Comparing the coefficients of $\hat{i}$ and $\hat{j}$ from both expressions:

For the $\hat{i}$ component:
$4\times 10^{-3}{v}_y=0.4$
$v_y=\frac{0.4}{4\times 10^{-3}}=100$ m/s

For the $\hat{j}$ component:
$0.4-4\times 10^{-3}{v}_x=0.2$
$4\times 10^{-3}{v}_x=0.2$
$v_x=\frac{0.2}{4\times 10^{-3}}=50$ m/s

Therefore, the velocity of the particle is $\vec{v}=50\hat{i}+100\hat{j}$ m/s.

Answer: $50\hat{i}+100\hat{j}$