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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 magnitude 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/πm
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Two particles each of mass m and charge q, are attached to the two end...
Current i = (frequency) (charge)

magnetic moment M = (i) (A)

M = qωR2
L = 2lω = 2(mR2)ω Angular momentum
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Most Upvoted Answer
Two particles each of mass m and charge q, are attached to the two end...
The magnetic moment of a particle with charge q and mass m rotating around an axis with angular momentum L is given by:

μ = (q/2m)L

In this case, we have two particles attached to the ends of the rod, so the total angular momentum is the sum of the angular momenta of each particle:

L_total = L_particle1 + L_particle2

Since the rod is light and rigid, its angular momentum is negligible compared to the angular momentum of the particles. Therefore, the total angular momentum is approximately equal to the sum of the angular momenta of the particles.

L_total ≈ 2L_particle

The ratio of the magnitude of the magnetic moment of the system (μ_total) to the total angular momentum (L_total) is:

μ_total / L_total = (q/2m)L_total / L_total = q/2m

Therefore, the correct answer is:

a) q/2m
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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 magnitude of the magnetic moment of the system and its angular momentum about the centre of the rod isa)q/2mb)q/mc)2q/md)q/πmCorrect answer is option 'A'. Can you explain this answer?
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