What is the radius of the path of an electron moving with velocity of \( 3\times10^{7}\,\mathrm{m/s} \) in a magnetic field of \( 6\times10^{-4}\,\mathrm{T} \) perpendicular to it?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 28 cm
In a field perpendicular to its velocity, a charged particle moves in a circle because the magnetic force supplies the centripetal force.
\( \displaystyle evB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{eB} \)
With \( m = 9.1\times10^{-31}\,\mathrm{kg} \) and \( e = 1.6\times10^{-19}\,\mathrm{C} \):
\( \displaystyle r = \frac{9.1\times10^{-31}\times3\times10^{7}}{1.6\times10^{-19}\times6\times10^{-4}} = \frac{2.73\times10^{-23}}{9.6\times10^{-23}}\,\mathrm{m} \)
\( r \approx 0.28\,\mathrm{m} = 28\,\mathrm{cm} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Motion in a Magnetic Field – …
