Skip to content
Question 88 of 91

Q.A non conducting charged ring carrying a charge of q, mass m and radius r is rotated about its axis with constant angular speed ω\omega. The ratio of its magnetic moment with angular momentum is :

(a) 2qm\dfrac{2q}{m}
(b) q2m\dfrac{q}{2m}
(c) q4m\dfrac{q}{4m}
(d) qm\dfrac{q}{m}
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026MCQ· 1mImportance★★★★★
97% · 88/91 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Computing the equivalent current and hence magnetic moment of the rotating charged ring, and dividing by its angular momentum mr2ωmr^2\omega, gives the classic gyromagnetic ratio q/(2m)q/(2m).

Working

1. Equivalent current. A ring of charge qq rotating at angular speed ω\omega passes a point ω/2π\omega/2\pi times per second, so it is equivalent to a current

I=qω2πI = \dfrac{q\omega}{2\pi}

2. Magnetic moment.

μ=IA=qω2π×πr2=qωr22\mu = IA = \dfrac{q\omega}{2\pi}\times\pi r^2 = \dfrac{q\omega r^2}{2}

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.