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(A) Consider a thin ring element of the spherical shell at an angle $	heta$ with the axis of rotation,having a width $Rd	heta$.The radius of this ri

Vedclass · Accepted Answer

$\frac{1}{3}Q\omega R^2$

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(A) Consider a thin ring element of the spherical shell at an angle $	heta$ with the axis of rotation,having a width $Rd	heta$.The radius of this ring is $r = R \sin 	heta$.The area of this ring element is $dA = (2\pi r)(Rd	heta) = 2\pi R^2 \sin 	heta d	heta$.The total surface area of the shell is $A = 4\pi R^2$.The charge on this ring element is $dq = Q \cdot \frac{dA}{A} = Q \cdot \frac{2\pi R^2 \sin 	heta d	heta}{4\pi R^2} = \frac{Q}{2} \sin 	heta d	heta$.The current $dI$ produced by this rotating ring is $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2\pi} = \frac{Q \omega}{4\pi} \sin 	heta d	heta$.The magnetic moment $d\mu$ of this ring is $d\mu = dI \cdot 	ext{Area of ring} = dI \cdot (\pi r^2) = \left( \frac{Q \omega}{4\pi} \sin 	heta d	heta ight) (\pi R^2 \sin^2 	heta) = \frac{Q \omega R^2}{4} \sin^3 	heta d	heta$.To find the total magnetic moment $\mu$,integrate from $	heta = 0$ to $\pi$:$\mu = \int_0^{\pi} \frac{Q \omega R^2}{4} \sin^3 	heta d	heta = \frac{Q \omega R^2}{4} \int_0^{\pi} \sin^3 	heta d	heta$.Using the identity $\int_0^{\pi} \sin^3 	heta d	heta = \int_0^{\pi} \sin 	heta (1 - \cos^2 	heta) d	heta = [-\cos 	heta + \frac{\cos^3 	heta}{3}]_0^{\pi} = (1 - 1/3) - (-1 + 1/3) = 2/3 + 2/3 = 4/3$.Thus,$\mu = \frac{Q \omega R^2}{4} \cdot \frac{4}{3} = \frac{1}{3} Q \omega R^2$.

$A$ spherical shell of radius $R$ carries a uniformly distributed charge $Q$ and is rotated about its diameter with angular speed $\omega$. Find its magnetic moment.

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