Force On A Moving Charge

A charged particle moving through a magnetic field experiences the Lorentz magnetic force $\vec{F}=q\vec{v}\times \vec{B}$, whose magnitude is $F=qvB\sin \theta$. Because the proton moves perpendicular to the field, $\theta =90^{\circ }$ and $\sin \theta =1$, so the full product applies. Substituting $q=1.6\times 10^{-19}\text{C}$, $v=5\times 10^6{\text{m s}}^{-1}$ and $B=0.2\text{T}$ gives $F=(1.6\times 10^{-19})(5\times 10^6)(0.2)=1.6\times 10^{-13}\text{N}$. The option $3.2\times 10^{-13}\text{N}$ doubles the speed incorrectly. The option $1.6\times 10^{-19}\text{N}$ mistakenly reports only the charge, ignoring velocity and field. The option $6.4\times 10^{-13}\text{N}$ uses four times the charge. As stressed in the NCERT treatment of the Lorentz force, this magnetic force is always perpendicular to the velocity and therefore does no work, only bending the path. A unit check confirms coulomb times metre per second times tesla equals newton, validating the answer. It is important to appreciate that this force never changes the proton's speed or kinetic energy; being perpendicular to the velocity at every instant, it only redirects the motion. As a result the proton settles into uniform circular motion within the field, with the magnetic force supplying exactly the centripetal force required. This distinction between changing direction and changing speed is a recurring conceptual theme that separates magnetic forces from the electric forces studied earlier in electrostatics.