A spherical shell of radius R, carrying a uniform surface charge σ, is set spinning at angular velocity ω. Find the vector potential it produces at point r (Fig. 5.45).

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Accepted Answer

It might seem natural to set the polar axis along ω, but in fact the integration is easier if we let r lie on the z axis, so that ω is tilted at an angle ψ. We may as well orient the x axis so that ω lies in the xz plane, as shown in Fig. 5.46. According to Eq. 5.66,

[latex]A=\frac{\mu}{4\pi}\int{\frac{I}{\eta }d\acute{l} }=\frac{\mu_0I}{4\pi}\int{\frac{I}{\eta }d\acute{I} }; A= \frac{\mu_0}{4\pi}\int{\frac{K}{\eta } }d\acute{a}. [/latex] (5.66)

[latex]A(r)=\frac{\mu_0}{4\pi}\int{\frac{K(\acute{r})}{\eta } d\acute a,} [/latex]

where [latex]K=\sigma v,\eta =\sqrt{R^2+r^2-2Rr\cos\acute{ heta}}, [/latex] and [latex]d\acute{a}=R^2\sin\acute{ heta}d\acute{ heta}d\acute{\phi}.[/latex] Now the velocity of a point [latex]\acute{r}[/latex] in a rotating rigid body is given by [latex]ω × \acute{r}[/latex]; in this case,

[latex]v = ω ×\acute{r}=\begin{vmatrix} \hat{x} & \hat{y}&\hat{z} \ \omega \sin\psi & 0&\omega \cos\psi \ R\sin\acute{ heta}\cos\acute{\phi}&R\sin\acute{ heta}\sin\acute{\phi}&R\cos\acute{ heta}\end{vmatrix} \=R\omega [-(\cos\psi \sin\acute{ heta}\sin\acute{\phi})\hat{x}+(\cos\psi \sin\acute{ heta}\cos\acute{ heta}-\sin\psi \cos\acute{ heta})\hat{y}\+(sin\psi \sin\acute{ heta}\sin\acute{\phi})\hat{z}].[/latex]

Notice that each of these terms, save one, involves either sin [latex]\acute{\phi}[/latex] or [latex]\cos \acute{\phi}[/latex]. Since

[latex]\int_{0}^{2\pi}{\sin\acute{\phi}d\acute{\phi}}=\int_{0}^{2\pi}{\cos\acute{\phi}d\acute{\phi}}=0, [/latex]

such terms contribute nothing. There remains

[latex]A(r)=-\frac{\mu_0R^3\sigma \omega \sin\psi }{2}\left(\int_{0}^{\pi}{\frac{\cos\acute{ heta}\sin\acute{ heta}}{\sqrt{R^2+r62-2Rr\cos\acute{ heta}} }d\acute{ heta} } ight)\hat{y}. [/latex]

Letting [latex]u\equiv \cos\acute{ heta}[/latex], the integral becomes

[latex]\int_{-1}^{+1}{\frac{u}{\sqrt{R^2+r^2-2Rru} } du}=-\frac{(R^2+r^2+Rru)}{3R^2r^2}\sqrt{R^2+r^2-2Rru}\mid ^{+1}_{-1}\=-\frac{1}{3R^2r^2}[(R^2+r^2+Rr)\left|R-r ight|\-(R^2+r^2-Rr)(R+r) ] . [/latex]

If the point r lies inside the sphere, then R > r , and this expression reduces to [latex](2r/3R^2);[/latex] if r lies outside the sphere, so that R < r , it reduces to [latex](2R/3r^2)[/latex]. Noting that [latex](ω × r) = −ωr \sin ψ \hat{y}[/latex], we have, finally,

[latex] A(r)=\begin{cases}\frac{\mu_0R\sigma }{3}(\omega imes r), for points inside the sphere, \ \frac{\mu_0R^4\sigma }{3r^3}(\omega imes r), for points outside the sphere.\end{cases} [/latex] (5.68)

Having evaluated the integral, I revert to the “natural” coordinates of Fig. 5.45,in which ω coincides with the z axis and the point r is at (r, θ, φ):

[latex]A(r, heta,\phi)=\begin{cases}\frac{\mu_0R\omega \sigma }{3}r\sin heta\acute{\phi}, (r\leq R), \ \frac{\mu_0R^4\omega \sigma }{3}\frac{\sin heta}{r^2} , (r\geq R).\end{cases} [/latex] (5.69)

Curiously, the field inside this spherical shell is uniform:

[latex]B= abla imes A=\frac{2\mu_0R\omega \sigma }{3}(\cos heta\hat{r}-\sin heta \hat{ heta})=\frac{2}{3}\mu_0\sigma R\omega \hat{z}=\frac{2}{3}\mu_0\sigma R\pmb{\omega} . [/latex] (5.70)

Question 5.11: A spherical shell of radius R, carrying a uniform surface ch...

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