An electron of mass \( 9.1\times10^{-31}\,\mathrm{kg} \) carrying charge \( 1.6\times10^{-19}\,\mathrm{C} \), is moving horizontally with a speed of \( 4\times10^{6}\,\mathrm{m/s} \) in a uniform magnetic field of magnitude \( 8\times10^{-4}\,\mathrm{T} \) acting vertically upward. The radius of its circular path is,
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.84 cm
The velocity is perpendicular to the field, so the magnetic force provides the centripetal force.
\( \displaystyle evB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{eB} \)
\( \displaystyle r = \frac{9.1\times10^{-31} \times 4\times10^{6}}{1.6\times10^{-19} \times 8\times10^{-4}} = \frac{3.64\times10^{-24}}{1.28\times10^{-22}}\,\mathrm{m} \)
\( \displaystyle r = 2.84\times10^{-2}\,\mathrm{m} = 2.84\,\mathrm{cm} \)
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 – …
