A particle carrying a charge of 1.0 × 10^ -9 C travels in the x – y...

The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.

Since the particle travels 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:

$\vec{F} = 1.0 \times 10^{-9} [ (v_x \hat{i} + v_y \hat{j}) \times (4.0 \times 10^{-3} \hat{k}) ]$

$\vec{F} = 4.0 \times 10^{-12} [ v_x (\hat{i} \times \hat{k}) + v_y (\hat{j} \times \hat{k}) ]$

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

$\vec{F} = 4.0 \times 10^{-12} (v_y \hat{i} - v_x \hat{j}) \text{ N}$

Equating this to the given force $\vec{F} = (4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-10} \text{ N}$:

$4.0 \times 10^{-12} v_y = 4.0 \times 10^{-10} \Rightarrow v_y = 100 \text{ m s}^{-1}$

$-4.0 \times 10^{-12} v_x = 3.0 \times 10^{-10} \Rightarrow v_x = -75 \text{ m s}^{-1}$

Therefore, the velocity of the particle is $\vec{v} = (-75 \hat{i} + 100 \hat{j}) \text{ m s}^{-1}$.