JEE Main 2026 — Physics Electromagnetism

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

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

Substituting the given values into the force equation:

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

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

$$\vec{F} = 10^{-9} [0.4\hat{j} - 4 \times 10^{-3} v_x \hat{j} + 4 \times 10^{-3} v_y \hat{i}]$$F=10−9[0.4j^​−4×10−3vx​j^​+4×10−3vy​i^]

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

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

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

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

For the $\hat{i}$i^ component:

$$4 \times 10^{-3} v_y = 0.4$$4×10−3vy​=0.4

$v_y = \dfrac{0.4}{4 \times 10^{-3}} = 100$vy​=4×10−30.4​=100 m/s

For the $\hat{j}$j^​ component:

$$0.4 - 4 \times 10^{-3} v_x = 0.2$$0.4−4×10−3vx​=0.2

$$4 \times 10^{-3} v_x = 0.2$$4×10−3vx​=0.2

$v_x = \dfrac{0.2}{4 \times 10^{-3}} = 50$vx​=4×10−30.2​=50 m/s

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

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