Magnetic Force On A Moving Charge

As stated in NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), a charge $q$ moving with velocity $v$ in a magnetic field $B$ experiences the Lorentz force $F=qvB\sin \theta$, where $\theta$ is the angle between velocity and field. This force is always perpendicular to both the velocity and the field, so it changes direction of motion without doing any work on the particle. Here the proton enters at right angles, so $\theta =90^{\circ }$ and $\sin \theta =1$, giving maximum force. Substituting the proton charge $q=1.6\times 10^{-19}$ C, $v=2\times 10^6$ m/s and $B=0.5$ T: $F=(1.6\times 10^{-19})(2\times 10^6)(0.5)=1.6\times 10^{-13}$ N. The option $1.6\times 10^{-19}$ N is wrong because it is merely the electronic charge and ignores $v$ and $B$. The value $0.8\times 10^{-13}$ N is half the correct answer, arising from wrongly using $\sin 30^{\circ }$. The value $3.2\times 10^{-13}$ N doubles the charge incorrectly. Plausibility check: the units work out as $\text{C}\cdot \text{m/s}\cdot \text{T}=\text{N}$, and the magnitude ($\sim 10^{-13}$ N) is reasonable for a subatomic particle in a moderate field.