Case: Read the passage and answer the question(s).
A charged particle of mass \( 2.0\times10^{-26}\,\mathrm{kg} \) and charge \( 1.6\times10^{-19}\,\mathrm{C} \) moves with a speed of \( 3\times10^{5}\,\mathrm{m\,s^{-1}} \) at right angles to a uniform magnetic field of 0.75 T.
Calculate the centripetal acceleration of the charged particle.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.8\times10^{12}\,\mathrm{m\,s^{-2}} \)
The particle moves at right angles to the field, so the magnetic force on it has magnitude \( qvB \) and acts as the centripetal force. By Newton’s second law, the centripetal acceleration is this force divided by the mass.
In SI units the data are:
\( m=2.0\times10^{-26}\,\mathrm{kg} \)
\( q=1.6\times10^{-19}\,\mathrm{C} \)
\( v=3\times10^{5}\,\mathrm{m\,s^{-1}} \)
\( B=0.75\,\mathrm{T} \)
The magnetic force is:
\( \begin{aligned} F &= qvB \\ &= 1.6\times10^{-19}\,\mathrm{C}\times3\times10^{5}\,\mathrm{m\,s^{-1}}\times0.75\,\mathrm{T} \\ &= 3.6\times10^{-14}\,\mathrm{N} \end{aligned} \)
The centripetal acceleration is:
\( \begin{aligned} a &= \frac{F}{m} \\ &= \frac{3.6\times10^{-14}\,\mathrm{N}}{2.0\times10^{-26}\,\mathrm{kg}} \\ &= 1.8\times10^{12}\,\mathrm{m\,s^{-2}} \end{aligned} \)