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Question: Using Eq. 5.88, calculate the average magnetic field of a dipole over

a sphere of radius Rcentered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the ambiguity at . (If you get stuck, refer to Prob. 3.48.) Evidently the truefield of a magnetic dipole is

Bdip(r)=μ04πr3[3(m·r^)r^-m]+2μ03mδ3(r)Bdip(r)=μ04πr3[3m·r^r^-m]+2μ03mδ3(r)

Compare the electrostatic analog, Eq. 3.106.

Short Answer

Expert verified

Answer:

The average magnetic field of a dipole over a sphere of radius Rcentered at the origin isμ04πr33m·r^r^-m+2μ03mδ3rμ04πr33m·r^r^-m+2μ03mδ3r.

Step by step solution

01

Given data

A dipole having dipole moment m.

02

Magnetic field far from the dipole

The magnetic field outside an infinitesimal sphere centered at a dipole is

Bdip(r)=μ04πr3[3m·r^r^-m]Bdipr=μ04πr33m·r^r^-m ….. (1)

Here, μ0 is the permeability of free space.

03

Magnetic field of a dipole

Inside the sphere, the magnetic field is a delta function

B=Aδ3r

Thus, the average field inside the sphere is

Bave=143πR3Aδ3rdτ=34πR3A

But the average field is also

Bave=μ04π2mR3

Compare the two and get

A=2μ0m3A=2μ0m3

Thus, the field inside the sphere is

role="math" localid="1658231587685" B=2μ0m3δ3r

From equation (1), the total field is

role="math" localid="1658231606913" Bdipr=μ04πr33m·r^r^-m+2μ0m3δ3r

Thus, the field of the dipole is μ04πr33m·r^r^-m+2μ0m3δ3r.

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Most popular questions from this chapter

A thin glass rod of radius Rand length Lcarries a uniform surfacecharge δ .It is set spinning about its axis, at an angular velocity ω.Find the magnetic field at a distances sR from the axis, in the xyplane (Fig. 5.66). [Hint:treat it as a stack of magnetic dipoles.]

The magnetic field on the axis of a circular current loop (Eq. 5.41) is far from uniform (it falls off sharply with increasing z). You can produce a more nearly uniform field by using two such loops a distanced apart (Fig. 5.59).

(a) Find the field (B) as a function of \(z\), and show that \(\frac{\partial \mathbf{B}}{\partial \mathbf{z}}\) is zero at the point midway between them \((z=0)\)

(b) If you pick d just right, the second derivative of \(B\) will also vanish at the midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing relatively uniform fields in the laboratory. Determine \(d\) such that

\(\partial^{2} B / \partial z^{2}=0\) at the midpoint, and find the resulting magnetic field at the center.

\(\frac{A I_{0}}{5 \sqrt{5} R}\)

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.
  2. The current is distributed in such a way that Jis proportional to s,the distance from the axis.

Question: Use Eq. 5.41 to obtain the magnetic field on the axis of the rotating disk in Prob. 5.37(a). Show that the dipole field (Eq. 5.88), with the dipole moment you found in Prob. 5.37, is a good approximation if z>> R.

The magnetic field on the axis of a circular current loop (Eq. 5.41) is far from uniform (it falls off sharply with increasing z). You can produce a more nearly uniform field by using two such loops a distanced apart (Fig. 5.59).

(a) Find the field (B) as a function of z, and show that Bzis zero at the point midway between them (z=0)

(b) If you pick d just right, the second derivative ofBwill also vanish at the midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing relatively uniform fields in the laboratory. Determine dsuch that

2B/z2=0at the midpoint, and find the resulting magnetic field at the center.

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