Case: Read the passage and answer the question.
A charged particle of mass \( 19.9\times10^{-27}\,\mathrm{kg} \) and charge \( 1.6\times10^{-19}\,\mathrm{C} \) moves with a speed of \( 3\times10^{5}\,\mathrm{m/s} \) at right angles to a uniform magnetic field of 0.75 T.
What is the magnitude of the magnetic force acting on the particle?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.6\times10^{-14}\,\mathrm{N} \)
The magnetic force on a charge \( q \) moving with velocity \( v \) in a field \( B \) has magnitude \( F=qvB\sin\theta \). Here \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \). The particle moves at right angles to the field, so \( \theta=90^\circ \) and \( \sin\theta=1 \).
\( \begin{aligned} F &= qvB \\ &= 1.6\times10^{-19}\,\mathrm{C}\times3\times10^{5}\,\mathrm{m\,s^{-1}}\times0.75\,\mathrm{T} \\ &= 3.6\times10^{-14}\,\mathrm{N} \end{aligned} \)