A particle having charge 10^ -9 C moving in x - y plane in fields...

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:
$4 \times 10^{-3} v_y = 0.4$
$v_y = \dfrac{0.4}{4 \times 10^{-3}} = 100$ m/s

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

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

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