The radius of the circular path of an electron moving at a speed of \( 2.4\times10^{7}\ \mathrm{m/s} \) in a uniform magnetic field of \( 6\times10^{-4}\ \mathrm{T} \) perpendicular to its direction, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 22.75\ \mathrm{cm} \)
A charged particle moving perpendicular to a uniform magnetic field moves in a circle, with the magnetic force \( evB \) providing the centripetal force.
\( \displaystyle evB = \frac{mv^{2}}{r} \quad\Rightarrow\quad r = \frac{mv}{eB} \)
Using \( m = 9.1\times10^{-31}\ \mathrm{kg} \) and \( e = 1.6\times10^{-19}\ \mathrm{C} \):
\( \displaystyle r = \frac{(9.1\times10^{-31}\ \mathrm{kg}) \times (2.4\times10^{7}\ \mathrm{m/s})}{(1.6\times10^{-19}\ \mathrm{C}) \times (6\times10^{-4}\ \mathrm{T})} \)
\( \displaystyle r = \frac{2.184\times10^{-23}}{9.6\times10^{-23}}\ \mathrm{m} = 0.2275\ \mathrm{m} = 22.75\ \mathrm{cm} \)
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 – …
