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An infinitely long circular cylinder carries a uniform magnetization Mparallel to its axis. Find the magnetic field (due toM) inside and outside the cylinder.

Short Answer

Expert verified

The value of magnetic field inside and outside the cylinder isBins=μ0Mz^ andBout=0 .

Step by step solution

01

Write the given data from the question.

Consideran infinitely long circular cylinder carries a uniform magnetizationM parallel to its axis.

02

Determine the formula of magnetic field inside the cylinder.

Write the formula of magnetic field inside the cylinder.

B=μ0K …… (1)

Here,μ0 is permeability andK is surface current.

03

Determine the value of magnetic field inside and outside the cylinder.

Thereisnovolumeboundcurrentsincethecylinder'smagnetizationishomogeneous;instead,thereissurfaceboundcurrent.

Determine the surface bound current.

Kb=M×n=Mz^×s^=Mϕ

This is the field of an infinite solenoid with surface current,nI=K=M therefore the outside field is and the inner field is:Bout=0

Determine the magnetic field inside the cylinder.

SubstituteMz^ forK into equation (1).

Bins=μ0Mz^=μ0M

Therefore, the value of magnetic field inside and outside the cylinder is Bins=μ0Mz^ and Bout=0.

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Most popular questions from this chapter

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.]

A sphere of linear magnetic material is placed in an otherwise uniform magnetic field B0. Find the new field inside the sphere.

Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ,P , and Mare uniform, the same integral is involved in all three:

r^r2dτ'

Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized object, and the vector potential of a uniformly magnetized object, of the same shape. Use this observation to obtain Vinside and outside a uniformly polarized sphere (Ex. 4.2), andA inside and outside a uniformly magnetized sphere (Ex. 6.1).

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?

A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility χm. A currentI flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field.

Figure 6.24

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