An electron is shot into the uniform magnetic field, normal to the direction of field. Then the frequency of revolution of the electron in its circular orbit :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → is independent of its speed
For motion perpendicular to a uniform field, the magnetic force provides the centripetal force:
\( \displaystyle e v B = \frac{m v^{2}}{r} \;\Rightarrow\; r = \frac{m v}{e B} \)
The time for one revolution is
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi m}{e B} \)
\( \displaystyle f = \frac{1}{T} = \frac{e B}{2\pi m} \)
The speed cancels: a faster electron moves on a proportionally larger circle and takes the same time per revolution. So the frequency (the cyclotron frequency) is independent of the speed and of the radius.
Option (1) is independent of its speed — Correct. \( f = eB/2\pi m \) contains neither \( v \) nor \( r \).
Option (2) decreases with its speed — Incorrect. The frequency does not change with speed.
Option (3) increases with its speed — Incorrect. A greater speed only gives a larger radius; the frequency stays the same.
Option (4) increases with radius of revolution — Incorrect. The radius grows with the speed, but the frequency stays \( eB/2\pi m \).
NCERT (2026-27):
Motion in a Magnetic Field – “… If ω is the angular frequency, then \( v = \omega\, r \). 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.”