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

Bave=μ042m˙4

where mbis 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=14Ï€3R3fBd

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

∫1rd43,

(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.

Short Answer

Expert verified

(a) It is proved that the average magnetic field, over a sphere of radius R, due to steady currents inside the sphere, is Bave=μ04π2mkR3

(b) The average magnetic field due to steady currents outside the sphere is μ04π∫J×r^'r'2dτ'and is same as the field produced at the center.

Step by step solution

01

Given data

There is a sphere of radius R of steady current density J.

02

step 2

The magnetic field as a function of the magnetic vector potential is

B=∇×A^

The magnetic vector potential corresponding to a current density j→is

Here, μ0is the permeability of free space.

The volume integral of the curl of a vector function

The magnetic moment of a current distribution is

m=12∫x×Jd

03

Determine the average magnetic field inside the sphere

(a)

The average magnetic field over a sphere of radius is

Apply equation (1)

Bave=14∫R3∫∇×Adτ

Use equation (3) to get,

Use equation (2) to get,

Bave=-143πR3μ04π∮∫J→rdτ'×da

The point r→''is chosen to be on the zaxis. Therefore,

\begin{aligned}r&=\sqrt{R^{2}+z^{\prime2}-2Rz^{\prime}\cos\theta}\\d^{\lambda}&=R^{2}\sin\thetad\thetad\phi\hat{r}\end{aligned}

The xand yare components of the surface, integration is thus zero. The z component is

∮da^r=∮R2sinθdθdϕz^cosθR2+z'2-2Rz'cosθ

=2πR2z^∫0πsinθcosθdθR2+z'2-2Rz'cosθ

Convert

u=cosθ

du=-sinθdθ

Solve further as,

∮da^r=-2πR2z^∫1-1uduR2+z'2-2Rz'u

=2Ï€R2z^-22R2+z'2+2Rz'u32Rz'2R2+z'2-2Rz'u-11

=2Ï€z^3z'2-R2+z'2+Rz'R-z'+R2+z'2-Rz'R+z'

Inside the sphere, R>z'. Therefore,

\begin{aligned}\prod\frac{dâ}{r}&=\frac{2\pi\hat{z}}{3z^{\prime2}}\left[-\left(R^{2}+z^{\prime2}+Rz^{\prime}\right)\left(R-z^{\prime}\right)+\left(R^{2}+z^{\prime2}-Rz^{\prime}\right)\left(R+z^{\prime}\right)\right]\\&=\frac{2\pi\hat{z}}{3z^{\prime2}}\left[-R^{3}-Rz^{\prime2}-R^{2}z^{\prime}+R^{2}z^{\prime}+z^{\prime3}+Rz^{\prime2}+R^{3}+Rz^{\prime2}-R^{2}z^{\prime}+R^{2}z^{\prime}+z^{\prime3}-Rz^{\prime2}\right]\\&=\frac{2\pi\hat{z}}{3z^{\prime2}}\times2z^{\prime3}\\&=\frac{4\piz^{\prime}\hat{z}}{3}\end{aligned}

Thus, from equation (5),

\begin{aligned}B_{\text{ave}}^{R}&=-\frac{3\mu_{0}}{16\pi^{2}R^{3}}\frac{4\piR^{3}}{3}\intJ\times\frac{\hat{r}^{\prime}}{r^{\prime2}}d\tau^{\prime}\\&=\frac{\mu_{0}}{4\pi}\int\stackrel{D}{J}\times\frac{\hat{r}^{\prime}}{r^{\prime2}}d\tau^{\prime}\end{aligned}

Use equation (4) to get,

Bave=2μ0m~4πR3

Thus, the average field inside the sphere is2μ0mm4πR3

04

Average magnetic field outside the sphere

(b)

Outside the sphere, R<z'. Therefore from equation (6),

\begin{aligned}\dot{y}\frac{da}{r}&=\frac{2\pi\hat{z}}{3z^{\prime2}}\left[-\left(R^{2}+z^{\prime2}+Rz^{\prime}\right)\left(-R+z^{\prime}\right)+\left(R^{2}+z^{\prime2}-Rz^{\prime}\right)\left(R+z^{\prime}\right)\right]\\&=\frac{2\pi\hat{z}}{3z^{\prime2}}\left[R^{3}+Rz^{\prime2}+R^{2}z^{\prime}-R^{2}z^{\prime}-z^{\prime3}-Rz^{\prime2}+R^{3}+Rz^{\prime2}-R^{2}z^{\prime}+R^{2}z^{\prime}+z^{\prime3}-Rz^{\prime2}\right]\\&=\frac{2\pi\hat{z}}{3z^{\prime2}}\times2R^{3}\\&=\frac{4\piR^{3}\hat{z}}{3z^{\prime2}}\end{aligned}

Thus, from equation (5),

\begin{aligned}B_{\text{ave}}^{R}&=-\frac{3\mu_{0}}{16\pi^{2}R^{3}}\frac{4\piR^{3}}{3}\intJ\times\frac{\hat{r}^{\prime}}{r^{\prime2}}d\tau^{\prime}\\&=\frac{\mu_{0}}{4\pi}\int\stackrel{D}{J}\times\frac{\hat{r}^{\prime}}{r^{\prime2}}d\tau^{\prime}\end{aligned}

Thus, the average field outside the sphere is μ04π∫J×r^'r'2dτ'which is also the field at the center..

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Most popular questions from this chapter

Suppose you have two infinite straight line chargesλ, a distance d apart, moving along at a constant speed υ(Fig. 5.26). How great would have tobe in order for the magnetic attraction to balance the electrical repulsion? Work out the actual number. Is this a reasonable sort of speed?

I worked out the multipole expansion for the vector potential of a line current because that's the most common type, and in some respects the easiest to handle. For a volume current J:

(a) Write down the multipole expansion, analogous to Eq. 5.80.

(b) Write down the monopole potential, and prove that it vanishes.

(c) Using Eqs. 1.107 and 5.86, show that the dipole moment can be written

m=12∫(r×J)dτ

(a) By whatever means you can think of (short of looking it up), find the vector potential a distance from an infinite straight wire carrying a current . Check that ∇.A=0and ∇×A=B.

(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.

Is Ampere's law consistent with the general rule (Eq. 1.46) that divergence-of-curl is always zero? Show that Ampere's law cannot be valid, in general, outside magnetostatics. Is there any such "defect" in the other three Maxwell equations?

A steady current Iflows down a long cylindrical wire of radius a(Fig. 5.40). Find the magnetic field, both inside and outside the wire, if

  1. The current is uniformly distributed over the outside surface of the wire.
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