An electron having charge \( 1.6\times10^{-19}\,\mathrm{C} \) and mass \( 9\times10^{-31}\,\mathrm{kg} \) is moving with \( 4\times10^{6}\,\mathrm{m/s} \) speed in a magnetic field \( 2\times10^{-1}\,\mathrm{T} \) in a circular orbit. The force experienced by the electron and the radius of the circular orbit, respectively will be :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.28\times10^{-13}\,\mathrm{N},\ 1.1\times10^{-4}\,\mathrm{m} \)
A charge moving perpendicular to a magnetic field feels the magnetic force \( F = qvB \), which acts as the centripetal force and makes it move in a circle of radius \( r = mv/qB \).
Force
\( F = qvB = 1.6\times10^{-19}\times4\times10^{6}\times2\times10^{-1} = 1.28\times10^{-13}\,\mathrm{N} \)
Radius
\( \displaystyle r = \frac{mv}{qB} = \frac{9\times10^{-31}\times4\times10^{6}}{1.6\times10^{-19}\times2\times10^{-1}} = \frac{3.6\times10^{-24}}{3.2\times10^{-20}} \)
\( r = 1.125\times10^{-4}\,\mathrm{m} \approx 1.1\times10^{-4}\,\mathrm{m} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Motion in a Magnetic Field – …
