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A uniform rod of total charge Q is distributed along its entire length 2L. The rod rotates with angular velocity omega about one of its ends. What is the magnetic moment of this rotating charged rod?
- Q*omega*L² / 3
- 2*Q*omega*L² / 3
- 4*Q*omega*L² / 3
- Q*omega*L² / 2
Correct answer: Q*omega*L² / 3
Solution
Each element dx at distance x carries charge dq = (Q/2L)dx; it constitutes a current loop dI = dq * f = dq * omega/(2*pi). Its magnetic moment dM = dI * pi*x². Integrating from 0 to 2L gives M = Q*omega*L²/3... let me re-verify: integral of x² dx from 0 to 2L = (2L)³/3 = 8L³/3; M = (Q/(2L))*(omega/(2*pi))*pi * 8L³/3 = Q*omega*8L³/(12L) = 2*Q*omega*L²/3.
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