What is the magnetic force acting on a proton (charge \( +1.6\times10^{-19}\,\mathrm{C} \)) moving at an angle of \( 30^\circ \) across a magnetic field of \( 5.3\times10^{-3}\,\mathrm{T} \) at a speed of \( 3.0\times10^{5}\,\mathrm{m\,s^{-1}} \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.3\times10^{-16}\,\mathrm{N} \)
The magnetic (Lorentz) force on a charge moving in a magnetic field is \( F = qvB\sin\theta \), where \( \theta \) is the angle between the velocity and the field.
\( \displaystyle F = 1.6\times10^{-19} \times 3.0\times10^{5} \times 5.3\times10^{-3} \times \sin 30^\circ \)
\( \displaystyle F = 1.6\times10^{-19} \times 3.0\times10^{5} \times 5.3\times10^{-3} \times 0.5 = 1.272\times10^{-16}\,\mathrm{N} \approx 1.3\times10^{-16}\,\mathrm{N} \)