A charged particle is moving in a cyclotron. What is the effect on time for one revolution of this charged particle, when its radius of the orbit is doubled?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It will remains unchanged
In the magnetic field the force \( qvB \) supplies the centripetal force \( \frac{mv^{2}}{r} \), so \( r = \frac{mv}{qB} \). 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 v cancels, so T depends only on m, q and B and not on the radius. Doubling the radius (the particle moves faster) leaves the time of one revolution unchanged.
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 and is given by:
\( \displaystyle \nu_c = \frac{qB}{2\pi m} \)
This frequency is independent of the particle’s speed and radius. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles.”