A charge of $$4.0 \;\mu$$C is moving with a velocity of $$4.0 \times 10^6 \text{ m s}^{-1}$$ along the positive $$y$$-axis under a magnetic field $$B$$ of strength $$\left(2\hat{k}\right)$$ T. The force acting on the charge is $$x\hat{i}$$ N. The value of $$x$$ is ______.

A charge $$q = 4.0\;\mu\text{C} = 4.0 \times 10^{-6}$$ C moves with velocity $$\vec{v} = 4.0 \times 10^6\;\hat{j}$$ m/s in a magnetic field $$\vec{B} = 2\hat{k}$$ T. The magnetic force on a moving charge is given by $$\vec{F} = q(\vec{v} \times \vec{B}).$$

To compute the cross product $$\vec{v} \times \vec{B},$$ observe that $$\vec{v} \times \vec{B} = (4.0 \times 10^6\;\hat{j}) \times (2\hat{k}) = 4.0 \times 10^6 \times 2\;(\hat{j} \times \hat{k}).$$ Using the rule $$\hat{j} \times \hat{k} = \hat{i},$$ it follows that $$\vec{v} \times \vec{B} = 8.0 \times 10^6\;\hat{i}.$$

Multiplying by the charge yields $$\vec{F} = q(\vec{v} \times \vec{B}) = 4.0 \times 10^{-6} \times 8.0 \times 10^6\;\hat{i},$$ so $$\vec{F} = 32\;\hat{i}\text{ N}.$$

The answer is $$x = 32$$.