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Question 5.59: (a) Prove that the average magnetic field, over a sphere of ...

(a) Prove that the average magnetic field, over a sphere of radius R, due to steady currents inside the sphere, is

Bave=μ04π2mR3B _{ ave }=\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}} ,                    (5.93)

where m is the total dipole moment of the sphere. Contrast the electrostatic result, Eq. 3.105. [This is tough, so I’ll give you a start:

Bave=143πR3∫BdτB _{ ave }=\frac{1}{\frac{4}{3} \pi R^{3}} \int B d \tau .

Eave=−14πϵ0pR3E _{ ave }=-\frac{1}{4 \pi \epsilon_{0}} \frac{ p }{R^{3}}                         (3.105)

Write B as (∇ × A), and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the surface integral first, showing that

∫1ᴫda=43πr′\int \frac{1}{ᴫ} d a =\frac{4}{3} \pi r ^{\prime} .

A(r)=μ04π∫J(r′)ᴫdτ′A ( r )=\frac{\mu_{0}}{4 \pi} \int \frac{ J \left( r ^{\prime}\right)}{ᴫ} d \tau^{\prime}                      (5.65)

(see Fig. 5.65). Use Eq. 5.90, if you like.]

m=12∫(r×J)dτm =\frac{1}{2} \int( r \times J ) d \tau                   (5.90)

(b) Show that the average magnetic field due to steady currents outside the sphere is the same as the field they produce at the center.

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(a)  Bave=1(3/4)πR3∫Bdτ=34πR3∫(∇×A)dτ=−34πR3∮A×da=−34πR3μ04π∮{∫Jᴫdτ′}×da=−3μ0(4π)2R3∫J×{∮1ᴫda}dτ′B _{ ave }=\frac{1}{(3 / 4) \pi R^{3}} \int B d \tau=\frac{3}{4 \pi R^{3}} \int( \nabla \times A ) d \tau=-\frac{3}{4 \pi R^{3}} \oint A \times d a =-\frac{3}{4 \pi R^{3}} \frac{\mu_{0}}{4 \pi} \oint\left\{\int \frac{ J }{ᴫ} d \tau^{\prime}\right\} \times d a =-\frac{3 \mu_{0}}{(4 \pi)^{2} R^{3}} \int J \times\left\{\oint \frac{1}{ᴫ} d a \right\} d \tau^{\prime} .

Note that J depends on the source point r′r ^{\prime}, not on the field point r. To do the surface integral, choose the (x, y, z) coordinates so that  r′r ^{\prime} lies on the z axis (see diagram).

Then ᴫ=R2+(z′)2−2Rz′cos⁡θᴫ=\sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime} \cos \theta} , while da=R2sin⁡θdθdϕr^d a =R^{2} \sin \theta d \theta d \phi \hat{ r } .

By symmetry, the x and y components must integrate to zero; since the z component of r^\hat{ r } is cos⁡θ\cos \theta, we have

∮1ᴫda=z^∫cos⁡θR2+(z′)2−2Rz′cos⁡θR2sin⁡θdθdϕ=2πR2z^∫0πcos⁡θsin⁡θR2+(z′)2−2Rz′cos⁡θdθ\oint \frac{1}{ᴫ} d a =\hat{ z } \int \frac{\cos \theta}{\sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime} \cos \theta}} R^{2} \sin \theta d \theta d \phi=2 \pi R^{2} \hat{ z } \int_{0}^{\pi} \frac{\cos \theta \sin \theta}{\sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime} \cos \theta}} d \theta .

 Let u≡cos⁡θ, so du=−sin⁡θdθ\text { Let } u \equiv \cos \theta, \text { so } d u=-\sin \theta d \theta .

=2πR2z^∫−11uR2+(z′)2−2Rz′udu=2 \pi R^{2} \hat{ z } \int_{-1}^{1} \frac{u}{\sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime} u}} d u

 

=2πR2z^{−2[2(R2+(z′)2)+2Rz′u]3(2Rz′)2R2+(z′)2−2Rz′u}∣−11=\left.2 \pi R^{2} \hat{ z }\left\{-\frac{2\left[2\left(R^{2}+\left(z^{\prime}\right)^{2}\right)+2 R z^{\prime} u\right]}{3\left(2 R z^{\prime}\right)^{2}} \sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime} u}\right\}\right|_{-1} ^{1}

 

=−2πR2z^3(Rz′)2{[R2+(z′)2+Rz′]R2+(z′)2−2Rz′−[R2+(z′)2−Rz′]R2+(z′)2+2Rz′}=-\frac{2 \pi R^{2} \hat{ z }}{3\left(R z^{\prime}\right)^{2}}\left\{\left[R^{2}+\left(z^{\prime}\right)^{2}+R z^{\prime}\right] \sqrt{R^{2}+\left(z^{\prime}\right)^{2}-2 R z^{\prime}}-\left[R^{2}+\left(z^{\prime}\right)^{2}-R z^{\prime}\right] \sqrt{R^{2}+\left(z^{\prime}\right)^{2}+2 R z^{\prime}}\right\}

 

=−[2π3(z′)2z^]{[R2+(z′)2+Rz′]∣R−z′∣−[R2+(z′)2−Rz′](R+z′)}=-\left[\frac{2 \pi}{3\left(z^{\prime}\right)^{2}} \hat{ z }\right]\left\{\left[R^{2}+\left(z^{\prime}\right)^{2}+R z^{\prime}\right]\left|R-z^{\prime}\right|-\left[R^{2}+\left(z^{\prime}\right)^{2}-R z^{\prime}\right]\left(R+z^{\prime}\right)\right\}

 

={4π3z′z^=4π3r′,(r′<R);4πR33(z′)2z^=4π3R3(r′)3r′,(r′>R).}=\left\{\begin{array}{l}\frac{4 \pi}{3} z^{\prime} \hat{ z }=\frac{4 \pi}{3} r ^{\prime}, \quad\left(r^{\prime}<R\right); \\\frac{4 \pi R^{3}}{3\left(z^{\prime}\right)^{2}} \hat{ z }=\frac{4 \pi}{3}\frac{R^{3}}{\left(r^{\prime}\right)^{3}} r ^{\prime},\left(r^{\prime}>R\right) .\end{array}\right\}

For now we want r′<R, so Bave=−3μ0(4π)2R34π3∫(J×r′)dτ′=−μ04πR3∫(J×r′)dτ′. Now m=12∫(r×J)dτr^{\prime}<R, \text { so } B _{ ave }=-\frac{3 \mu_{0}}{(4 \pi)^{2} R^{3}} \frac{4 \pi}{3} \int\left( J \times r ^{\prime}\right) d \tau^{\prime}=-\frac{\mu_{0}}{4 \pi R^{3}} \int\left( J \times r ^{\prime}\right) d \tau^{\prime} . \text { Now } m =\frac{1}{2} \int( r \times J ) d \tau

(Eq. 5.90), so Bave =μ04π2mR3B _{\text {ave }}=\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}} . qed

(b) This time  r′>R, so Bave=−3μ0(4π)2R34π3R3∫(J×r′(r′)3)dτ′=μ04π∫J×ᴫ^ᴫ2dτ′r^{\prime}>R, \text { so } B _{ ave }=-\frac{3 \mu_{0}}{(4 \pi)^{2} R^{3}} \frac{4 \pi}{3} R^{3} \int\left( J \times \frac{ r ^{\prime}}{\left(r^{\prime}\right)^{3}}\right) d \tau^{\prime}=\frac{\mu_{0}}{4 \pi} \int \frac{ J \times \hat{ᴫ} }{ᴫ^{2}} d \tau^{\prime} , where ᴫ now goes from the source point to the center (ᴫ=−r′)\left( ᴫ =- r ^{\prime}\right) .

Thus  Bave =Bcen B _{\text {ave }}= B _{\text {cen }} . qed

5.59

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