Velocity of Charged Particle in Electromagnetic Field

The total force acting on a charged particle moving in combined electric and magnetic fields (Lorentz force) is given by: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$

Given data:

• Charge, $q = 10^{-9} \text{ C}$
• Electric field, $\vec{E} = 0.4\hat{j} \text{ N/C}$
• Magnetic field, $\vec{B} = 4 \times 10^{-3}\hat{k} \text{ T}$
• Net force, $\vec{F} = (4\hat{i} + 2\hat{j}) \times 10^{-10} \text{ N}$

Since the particle is moving in the $x\text{-}y$ plane, its velocity vector can be represented as: $\vec{v} = v_x \hat{i} + v_y \hat{j}$

First, compute the magnetic force term per unit charge: $\vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3}\hat{k})$

Using the cross product relations $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$: $\vec{v} \times \vec{B} = 4 \times 10^{-3} v_y \hat{i} - 4 \times 10^{-3} v_x \hat{j}$

Now, substitute $\vec{E}$ and $\vec{v} \times \vec{B}$ into the Lorentz force equation: $\frac{\vec{F}}{q} = \vec{E} + \vec{v} \times \vec{B}$

Substitute the given values: $\frac{(4\hat{i} + 2\hat{j}) \times 10^{-10}}{10^{-9}} = 0.4\hat{j} + \left(4 \times 10^{-3} v_y \hat{i} - 4 \times 10^{-3} v_x \hat{j}\right)$ $0.4\hat{i} + 0.2\hat{j} = \left(4 \times 10^{-3} v_y\right)\hat{i} + \left(0.4 - 4 \times 10^{-3} v_x\right)\hat{j}$

Equating the $\hat{i}$ and $\hat{j}$ components on both sides:

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

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

Thus, the velocity of the particle at that instant is: $\vec{v} = 50\hat{i} + 100\hat{j} \text{ m/s}$

This corresponds to Option A.