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 radius of the circular path of the particle?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 49.75\times10^{-3}\,\mathrm{m} \)
When the velocity is perpendicular to the magnetic field, the magnetic force \( qvB \) is perpendicular to the velocity. It acts as the centripetal force, so the particle moves in a circle.
\( \displaystyle \frac{mv^{2}}{r}=qvB \)
\( \begin{aligned} r &= \frac{mv}{qB} \\ &= \frac{19.9\times10^{-27}\,\mathrm{kg}\times3\times10^{5}\,\mathrm{m\,s^{-1}}}{1.6\times10^{-19}\,\mathrm{C}\times0.75\,\mathrm{T}} \\ &= \frac{5.97\times10^{-21}}{1.2\times10^{-19}}\,\mathrm{m} \\ &= 4.975\times10^{-2}\,\mathrm{m} \\ &= 49.75\times10^{-3}\,\mathrm{m} \end{aligned} \)