An electron (mass \( 9.1\times10^{-31} \) kg and charge \( 1.6\times10^{-19} \) C) is moving in a circular path with a speed of \( 3\times10^{7} \) m/s in a magnetic field of \( 6\times10^{-4} \) T perpendicular to it. Its energy (in keV) is
(Take \( 1 \) eV \( = 1.6\times10^{-19} \) J)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sim 2.56 \)
A magnetic force is always perpendicular to the velocity, so it does no work: it only bends the path into a circle and leaves the speed and the kinetic energy unchanged. The energy asked for is therefore just \( \frac{1}{2}mv^{2} \); the value of the field is not needed.
\( \displaystyle K = \frac{1}{2}mv^{2} = \frac{1}{2}(9.1\times10^{-31})(3\times10^{7})^{2} \)
\( K = 0.5\times9.1\times10^{-31}\times9\times10^{14} = 4.095\times10^{-16}\,\mathrm{J} \)
\( \displaystyle K = \frac{4.095\times10^{-16}}{1.6\times10^{-19}}\,\mathrm{eV} \approx 2.56\times10^{3}\,\mathrm{eV} \approx 2.56\,\mathrm{keV} \)
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 – …
