A particle of charge q and mass m moves in a circular orbit of radius…

The correct answer is Option (2) → \( m \) and \( q \) only

A revolving charge is a current loop, so its magnetic moment is current times area, and its angular momentum is \( L = m r^2\omega \).

    \( \displaystyle T = \frac{2\pi}{\omega} \qquad I = \frac{q}{T} = \frac{q\omega}{2\pi} \)

    \( \displaystyle \mu = I\,A = \frac{q\omega}{2\pi}\,\pi r^{2} = \frac{q\omega r^{2}}{2} \)

    \( \displaystyle \frac{\mu}{L} = \frac{q\omega r^{2}/2}{m r^{2}\omega} = \frac{q}{2m} \)

The ratio depends on \( q \) and \( m \) only; \( \omega \) and \( r \) cancel.

Note: It seems that there is a mistake in the original NTA paper, which uses the symbol \( m \) for both the magnetic moment and the mass of the particle. The options mean the mass of the particle. Here \( \mu \) is used for the magnetic moment.

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 – “… There will be a magnetic moment, usually denoted by \( \mu_l \), associated with this circulating current. From Eq. (4.28) its magnitude is, \( \mu_l = I\pi r^2 = evr/2 \). … Multiplying and dividing the right-hand side of the above expression by the electron mass \( m_e \), we have,

    \( \displaystyle \mu_l = \frac{e}{2m_e}\left(m_e v r\right) = \frac{e}{2m_e}\,l \qquad [4.34(a)] \)

Here, l is the magnitude of the angular momentum of the electron about the central nucleus (“orbital” angular momentum). … The ratio

    \( \displaystyle \frac{\mu_l}{l} = \frac{e}{2m_e} \qquad (4.35) \)

is called the gyromagnetic ratio and is a constant.”