An electron of charge 'e' moves in a circular orbit of radius r around a nucleus with a frequency \( \nu \). The magnetic moment associated with the orbital motion of the electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \pi\,\nu\,e\,r^{2} \)
The revolving electron is a current loop, and the magnetic moment of a loop is current × area. The electron passes any point of the orbit \( \nu \) times per second, so the charge crossing it per second is
\( I = e\nu \)
\( \displaystyle \mu = IA = e\nu \times \pi r^{2} = \pi\,\nu\,e\,r^{2} \)
Shortcut: Dimensional check: magnetic moment has the unit A m2, that is C s−1 m2. Only \( \pi\nu e r^{2} \) (C × s−1 × m2) has these units; options 2, 3 and 4 do not.
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,
\( \displaystyle I = \frac{e}{T} \qquad (4.32) \)
and T is the time period of revolution. … There will be a magnetic moment, usually denoted by μl, associated with this circulating current. From Eq. (4.28) its magnitude is,
\( \displaystyle \mu_l = I\pi r^2 = evr/2 \)”