An electron (mass \( 9 \times 10^{-31}\,\mathrm{kg} \) and charge \( 1.6 \times 10^{-19}\,\mathrm{C} \)) moving with a speed of \( 5.0\times10^{7}\,\mathrm{m/s} \) enters a region of uniform magnetic field of strength 0.04 T which is perpendicular to its velocity. The radius of the circular path it follows is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 7.03 \times 10^{-3}\,\mathrm{m} \)
A charge moving perpendicular to a uniform magnetic field moves on a circle, because the magnetic force supplies the centripetal force.
\( \displaystyle qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \)
\( \displaystyle r = \frac{9\times10^{-31}\times5\times10^{7}}{1.6\times10^{-19}\times0.04} = \frac{4.5\times10^{-23}}{6.4\times10^{-21}} = 7.03\times10^{-3}\,\mathrm{m} \)
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 – …
