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.
When the particle has gone round half of its circular path, the magnitude of the change in its momentum is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.2\times10^{-20}\,\mathrm{kg\,m\,s^{-1}} \)
The magnetic force does no work. So the particle keeps the same speed all along its circle, but the direction of its velocity keeps turning. After half a revolution, the particle moves with the same speed in the opposite direction. So its momentum is reversed.
The magnitude of the momentum is:
\( \begin{aligned} p &= mv \\ &= 2.0\times10^{-26}\,\mathrm{kg}\times3\times10^{5}\,\mathrm{m\,s^{-1}} \\ &= 6.0\times10^{-21}\,\mathrm{kg\,m\,s^{-1}} \end{aligned} \)
Take the first direction of motion as positive. The momentum goes from \( +p \) to \( -p \), and the change is:
\( \begin{aligned} |\Delta p| &= |(-p)-(+p)| \\ &= 2p \\ &= 1.2\times10^{-20}\,\mathrm{kg\,m\,s^{-1}} \end{aligned} \)
NCERT (2026-27):
Motion in a Magnetic Field – “… In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).”