A uniform charged disc whose total charge has magnitude q and whose radius is r rotates with constant angular velocity of magnitude `omega`. What is the magnetic dipole moment?
Text Solution
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The surface charge density is `q//pir^(2)`. Hence the charge within a ring of radius R and width dR is `dq = (q)/(pi r^(2)) (2 pi RdR) = (2q)/(r^(3)) (RdR)` The current carried by this ring is its charge divided by the rotation period, `di = (dq)/(2 pi // omega) = (q omega)/(pi r^(2)) [R.dR]` The magnetic moment contributed by this ring has the magnitude dM = a |di|, where a is the area of ring. `dM = pi R^(@) |di| = (q omega)/(r^(2)) R^(3) dR` `M = int dM = int_(R - 0)^(r ) q. (omega)/(r^(2)) (R^(3) dR) = q omega (r^(2))/(4)`
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