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uniformly charged disc of radius R and total charge Q is rotating about its axis passing through the centre of diae and perpendicular to the plane of dise, with an angular velocity omega.Calculate its magnetic dipole moment. Also find the ratio of angular momentum to that with the calculated magnetic moment of the system. |
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Answer» Solution :Consider a disc of radius R having centre at O. Given that, Q is the total charge on disc. Consider a small element of thickness dx at a distance x from the centre. DQ = Charge on this small element ` = Q/(pi R^(2)) xx 2pi x dx `  Magnetic moment `(d mu)` of mall element ` = 1/2 d q omegax^(2)` ` = 1/2 (Q/(piR^(2)) * 2pi xdx) omegax^(2)` ` dmu = (Qomega)/R^(2) x^(3) dx ` The total magnetic moment can be obtained by integrating between the limits 0 to R. `mu = intdmu = UNDERSET(0)overset(R)int (Qomega)/R^(2) x^(3) dx ` ` = (Qomega)/R^(2) [x^(4)/4]_(0)^(R) ` ` = 1/4 (Qomega)/R^(2) R^(4)` ` = 1/4 Q omegaR^(2)` ANGULAR momentum , ` L = I omega = (MR^(2))/(2) omega` ` "Magnetic moment"/"Angular momentum" = 1/4 (QomegaR^(2))/(MR^(2) omega) xx 2 ` ` = Q/(2M)` |
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