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Physics Question on Electric charges and fields

Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is

A

q/2m

B

q/m

C

2q/m

D

q/πmq/ \pi m

Answer

q/2m

Explanation

Solution

Current, i = (frequency) (charge) = (ω2π)(2q)=qωπ \bigg( \frac{ \omega}{2 \pi }\bigg) (2 q) = \frac{q \omega }{\pi } Magnetic moment, M=(i)(A)=(qωπ)(πR2)=(qωR2)M = (i) (A) = \bigg( \frac{ q \omega}{ \pi } \bigg) (\pi R^2 ) = (q \omega R^2 ) Angular momentum, L=2Iω=2(mR2)ωL = 2 I \omega = 2 (m R^2) \omega ∴         ML=qωR22(mR2)ω=q2m \therefore \, \, \, \, \, \, \, \, \, \frac{M}{L} = \frac{q \omega R^2}{2 (mR^2) \omega } = \frac{ q}{2m}