A short circular cylinder of radius and length L carries a "frozen-in" uniform magnetization M parallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one forL>>a, one forL<<a, and one forLa.) Compare this bar magnet with the bar electret of Prob. 4.11.

Short Answer

Expert verified

The value of bound current is Kb=Mϕ^.

Draw the magnetic field of the cylinder for L>>a.

Draw the magnetic field of the cylinder for L<<a.

Draw the magnetic field of the cylinder for La.

Step by step solution

01

Write the given data from the question.

Consider a short circular cylinder of radius and length carries a "frozen-in" uniform magnetization parallel to its axis.

02

Determine the formula of bound current.

Write the formula of bound current.

Kb=M×s..........(1)

Here, Mis frozen-in uniform magnetization ands is distance from axis.

03

Determine the value of bound current and draw the magnetic field of the cylinder for L>>a, for L<<a and for L≈a.

Draw the circuit diagram of magnetic field of the cylinder for L>>a.

Figure 1

Draw the circuit diagram of magnetic field of the cylinder for L<<a.

Figure 2

Draw the circuit diagram of magnetic field of the cylinder for La.

Figure 3

Determine the bound current is given by:

Substitute ϕ^for sinto equation (1).

kb=Mϕ^

Thus, the field is made up of a solenoid with the following dimensions: L, a, and for the surface current.

The lines of the magnetic field are drawn above. They resemble the bar electret case in appearance, but inside they are entirely different since the magnetic field does not have discontinuities at the top or bottom like the electric field does.

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Most popular questions from this chapter

You are asked to referee a grant application, which proposes to determine whether the magnetization of iron is due to "Ampere" dipoles (current loops) or "Gilbert" dipoles (separated magnetic monopoles). The experiment will involve a cylinder of iron (radius Rand length L=10R), uniformly magnetized along the direction of its axis. If the dipoles are Ampere-type, the magnetization is equivalent to a surface bound current Kb=Mϕ^if they are Gilbert-type, the magnetization is equivalent to surface monopole densities σb=±Mat the two ends. Unfortunately, these two configurations produce identical magnetic fields, at exterior points. However, the interior fields are radically different-in the first case Bis in the same general direction as M, whereas in the second it is roughly opposite to M. The applicant proposes to measure this internal field by carving out a small cavity and finding the torque on a tiny compass needle placed inside.

Assuming that the obvious technical difficulties can be overcome, and that the question itself is worthy of study, would you advise funding this experiment? If so, what shape cavity would you recommend? If not, what is wrong with the proposal?

Suppose the field inside a large piece of magnetic material is B0, so that H0=(1/μ0)B0-M, where M is a "frozen-in" magnetization.

(a) Now a small spherical cavity is hollowed out of the material (Fig. 6.21). Find the field at the center of the cavity, in terms of B0 and M. Also find H at the center of the cavity, in terms of H0 and M.

(b) Do the same for a long needle-shaped cavity running parallel to M.

(c) Do the same for a thin wafer-shaped cavity perpendicular to M.

Figure 6.21

Assume the cavities are small enough so M, B0, and H0 are essentially constant. Compare Prob. 4.16. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite magnetization.]


In Prob. 6.4, you calculated the force on a dipole by "brute force." Here's a more elegant approach. First writeB(r)as a Taylor expansion about the center of the loop:

,B(r)B(r0)+[(rr0)0]B(r0)

Wherer0the position of the dipole and 0is denotes differentiation with respect tor0. Put this into the Lorentz force law (Eq. 5.16) to obtain

.F=IdI×[(r0)B(r0)]

Or, numbering the Cartesian coordinates from 1 to 3:

Fi=Ij,k,l=13εijk{rldlj}[0lBk(r0)],

Where εijk is the Levi-Civita symbol (+1ifijk=123,231, or312; 1ifijk=132, 213, or 321;0otherwise), in terms of which the cross-product can be written (A×B)i=j,k=13εijkAjBk. Use Eq. 1.108 to evaluate the integral. Note that

j=13εijkεljm=δilδkmδimδkl

Whereoil is the Kronecker delta (Prob. 3.52).




A current Iflows down a long straight wire of radius. If the wire is made of linear material (copper, say, or aluminium) with susceptibility Xm, and the current is distributed uniformly, what is the magnetic field a distances from the axis? Find all the bound currents. What is the net bound current flowing down the wire?

Calculate the torque exerted on the square loop shown in Fig. 6.6, due to the circular loop (assume is much larger than or ). If the square loop is free to rotate, what will its equilibrium orientation be?

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