A charge particle having charge q is moving in a circle of radius R with a uniform speed \( v \). The associated magnetic moment \( \mu \) is given by :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{qvR}{2} \)
A charge going round a circle repeatedly is equivalent to a current loop, and the magnetic moment of a loop is current times area, \( \mu = IA \).
The charge passes any point once per revolution, so the current is the charge divided by the time period.
\( \displaystyle T = \frac{2\pi R}{v} \;\Rightarrow\; I = \frac{q}{T} = \frac{qv}{2\pi R} \)
\( \displaystyle \mu = I\,A = \frac{qv}{2\pi R}\times\pi R^{2} = \frac{qvR}{2} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
The Magnetic Dipole Moment of a Revolving Electron – “… The electron of charge (–e) (e = +1.6 × 10–19 C) performs uniform circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I, where
\( \displaystyle I = \frac{e}{T} \)
and T is the time period of revolution. Let r be the orbital radius of the electron, and v the orbital speed. Then,
\( \displaystyle T = \frac{2\pi r}{v} \)
Substituting in Eq. (4.32), we have I = ev/2πr. There will be a magnetic moment, usually denoted by μl, associated with this circulating current. From Eq. (4.28) its magnitude is, μl = Iπr2 = evr/2.”