Under the influence of a uniform magnetic field, a charged particle moves with a constant speed v in a circle of radius r. The time period of the revolution of the particle:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → is independent of both v and r
The magnetic force \( qvB \) on the particle acts as the centripetal force for its circular motion:
\( \displaystyle qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \)
The time for one revolution is the circumference divided by the speed:
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB} \)
The speed cancels: a faster particle moves on a proportionally larger circle and takes the same time. So the period depends only on \( m \), \( q \) and \( B \), and is independent of both v and r (this is the principle of the cyclotron).
NCERT (2026-27):
Motion in a Magnetic Field – “… If ω is the angular frequency, then v = ωr. So,
\( \displaystyle \omega = 2\pi\nu = qB/m \qquad [4.6(a)] \)
which is independent of the velocity or energy. Here ν is the frequency of rotation. The independence of ν from energy has important application in the design of a cyclotron. The time taken for one revolution is
\( \displaystyle T = 2\pi/\omega \equiv 1/\nu \)”