A conducting solid sphere of radius R and mas | vedclass

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Question

(A) Consider a thin ring element of the sphere at an angle $	heta$ with the axis of rotation,having width $R d	heta$ and radius $r = R \sin 	heta$.

Vedclass · Accepted Answer

$(1.66$ to $1.67)$

Vedclass · Answer

(A) Consider a thin ring element of the sphere at an angle $	heta$ with the axis of rotation,having width $R d	heta$ and radius $r = R \sin 	heta$.The surface charge density is $\sigma = \frac{Q}{4 \pi R^2}$.The charge on this ring is $dq = \sigma (2 \pi r) (R d	heta) = \sigma (2 \pi R \sin 	heta) (R d	heta) = 2 \pi \sigma R^2 \sin 	heta d	heta$.The current produced by this rotating ring is $dI = \frac{dq}{T} = \frac{dq \omega}{2 \pi} = \sigma R^2 \omega \sin 	heta d	heta$.The magnetic dipole moment of this ring is $d\mu = dI \cdot A = (\sigma R^2 \omega \sin 	heta d	heta) (\pi r^2) = \sigma R^2 \omega \sin 	heta d	heta (\pi R^2 \sin^2 	heta) = \pi \sigma R^4 \omega \sin^3 	heta d	heta$.The total magnetic dipole moment is $\mu = \int_0^{\pi} \pi \sigma R^4 \omega \sin^3 	heta d	heta = \pi \sigma R^4 \omega \int_0^{\pi} \sin^3 	heta d	heta$.Using $\int_0^{\pi} \sin^3 	heta d	heta = \frac{4}{3}$,we get $\mu = \pi \left( \frac{Q}{4 \pi R^2} ight) R^4 \omega \left( \frac{4}{3} ight) = \frac{Q R^2 \omega}{3}$.The angular momentum of the solid sphere is $L = I \omega = (\frac{2}{5} M R^2) \omega$.The ratio is $\frac{\mu}{L} = \frac{Q R^2 \omega / 3}{(2/5) M R^2 \omega} = \frac{Q}{2M} \left( \frac{5}{3} ight)$.Thus,$\alpha = \frac{5}{3} \approx 1.67$.

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