If charge is moving perpendicular to uniform magnetic field then its time period of revolution is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → independent of its radius of circular path
A charge moving perpendicular to the field travels on a circle, with the magnetic force supplying the centripetal force, \( qvB = mv^{2}/r \), so \( r = mv/qB \).
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi}{v}\cdot\frac{mv}{qB} = \frac{2\pi m}{qB} \)
The speed cancels, so T does not depend on the radius or the speed. A faster particle moves on a bigger circle in exactly the same time. T does depend on the mass, the charge and the field B, so the statements in Options (2), (3) and (4) are false.
NCERT (2026-27):
Summary – “… 3. In a uniform magnetic field B, a charge q executes a circular orbit in a plane normal to B. Its frequency of uniform circular motion is called the cyclotron frequency … This frequency is independent of the particle’s speed and radius.”