The frequency of a charged particle undergoing a circular motion in a uniform magnetic field is independent of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → its speed
When a charge moves perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force.
\( \displaystyle qvB = \frac{mv^{2}}{r} \)
\( \displaystyle r = \frac{mv}{qB} \)
The time for one revolution is
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \)
\( \displaystyle \nu = \frac{1}{T} = \frac{qB}{2\pi m} \)
The speed cancels: a faster particle moves in a proportionally larger circle and takes the same time per revolution. So the cyclotron frequency depends on q, B and m but is independent of the speed (and of the radius).
NCERT (2026-27):
Motion in a Magnetic Field – “… If \( \omega \) is the angular frequency, then \( v = \omega\,r \). So,
\( \displaystyle \omega = 2\pi\,\nu = q B / m \qquad [4.6(a)] \)
which is independent of the velocity or energy. Here \( \nu \) is the frequency of rotation.”