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

Question: (a) Find the density ρof mobile charges in a piece of copper, assuming each atom contributes one free electron. [Look up the necessary physical constants.]

(b) Calculate the average electron velocity in a copper wire 1 mm in diameter, carrying a current of 1 A. [Note:This is literally a snail'space. How, then, can you carry on a long distance telephone conversation?]

(c) What is the force of attraction between two such wires, 1 em apart?

(d) If you could somehow remove the stationary positive charges, what would the electrical repulsion force be? How many times greater than the magnetic force is it?

(a) one way to fill in the "missing link" in Fig. 5.48 is to exploit the analogy between the defining equations for A(viz-A=0,×A=B)and Maxwell's equations forB(viz.B=0×B=μ0J).Evidently A depends on B in exactly the same way that B depends onμ0J(to wit: the Biot-Savart law). Use this observation to write down the formula for A in terms of B.

(b) The electrical analog to your result in (a) is

localid="1658557463395" V(r)=-14πE(r')-r^r2dτ'

Derive it, by exploiting the appropriate analogy.

A current flows to the right through a rectangular bar of conducting material, in the presence of a uniform magnetic fieldBpointing out of the page (Fig. 5.56).

(a) If the moving charges are positive, in which direction are they deflected by the magnetic field? This deflection results in an accumulation of charge on the upper and lower surfaces of the bar, which in turn produces an electric force to counteract the magnetic one. Equilibrium occurs when the two exactly cancel. (This phenomenon is known as the Hall effect.)

(b) Find the resulting potential difference (the Hall voltage) between the top and bottom of the bar, in terms ofB,v(the speed of the charges), and the relevant dimensions of the bar.23

(c) How would your analysis change if the moving charges were negative? [The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material.]

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.

(a) Check that Eq. 5.65 is consistent with Eq. 5.63, by applying the divergence.

(b) Check that Eq. 5.65 is consistent with Eq. 5.47, by applying the curl.

(c) Check that Eq. 5.65 is consistent with Eq. 5.64, by applying the Laplacian.

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