A proton moves with the velocity (3.0 î + 4.0 ĵ) × 10⁵ m/s in a uniform field B = 0.60 k̂ T…

Step 1. The cross product. $\vec{v} \times \vec{B} = v_yB_z\,\hat{i} - v_xB_z\,\hat{j} = (4.0 \times 10^{5} \times 0.60)\,\hat{i} - (3.0 \times 10^{5} \times 0.60)\,\hat{j} = (2.4\,\hat{i} - 1.8\,\hat{j}) \times 10^{5}\ \text{T m/s}$.

Step 2. The force. $\vec{F} = e\,(\vec{v} \times \vec{B}) = 1.6 \times 10^{-19}\ \text{C} \times (2.4\,\hat{i} - 1.8\,\hat{j}) \times 10^{5}\ \text{T m/s}$. So $F_x = 3.84 \times 10^{-14}$ N and $F_y = -2.88 \times 10^{-14}$ N.

Step 3. The size. $F = \sqrt{3.84^2 + 2.88^2} \times 10^{-14}\ \text{N} = 4.8 \times 10^{-14}\ \text{N}$. Check with $evB$: $v = 5.0 \times 10^{5}$ m/s, so $evB = 1.6 \times 10^{-19} \times 5.0 \times 10^{5} \times 0.60 = 4.8 \times 10^{-14}$ N.

Step 4. At right angles. $\vec{F} \cdot \vec{v} \propto 3.84 \times 3.0 + (-2.88) \times 4.0 = 11.52 - 11.52 = 0$.