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An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the axis

M=ksz^,

Where is a constant and is the distance from the axis; there is no free current anywhere. Find the magnetic field inside and outside the cylinder by two different methods: (a) As in Sect. 6.2, locate all the bound currents, and calculate the field they produce. (b) Use Ampere's law (in the form of Eq. 6.20) to find, and then get from Eq. 6.18. (Notice that the second method is much faster, and avoids any explicit reference to the bound currents.)

Short Answer

Expert verified

(a)

The values of all bound currents areKb=kRϕ^ and Jb=-kϕ^.

The value of magnetic field they produce is B=μ0ksz^.

(b) The value of magnetic field for any loop there is no enclosed free current, is B=π0ksz^.

Step by step solution

01

Write the given data from the question.

Consider an infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the axis.

02

Determine the formula of all bound currents and of magnetic field they produce.

Write the formula ofall bound currents.

Kb=M×s^ …… (1)

Here, Mis frozen in magnetization and s^is distance from the axis.

Write the formula of all bound currents.

Jb=×M …… (2)

Here, Mis frozen in magnetization.

Write the formula ofmagnetic field they produce.

BI=μ0KbI-μ0IsRkbs …… (3)

Here,μ0 is permeability, k is constant, s^ is distance from the axis and I is current passing area.

Write the formula ofmagnetic field for any loop there is no enclosed free current.

B=μ0M …… (4)

Here, μ0is permeability andM is frozen in magnetization.

03

(a) Determine the value of all bound currents and of magnetic field they produce.

Determine the bound currents first:

Substitute k for Mand Rϕ^for s^into equation (1).

kb=kRϕ^

Determine the bound currents second:

Substitute -k for Mand ϕ^for into equation (2).

Jb=-kϕ^

Although the magnetic field will be in the z-direction, it will be zero outside since Ienc = 0. We have the following inside, utilising an Amperian rectangle that spans the surface:

BI=kri-k(R-s)B=μ0ksz^

Therefore, the value of magnetic field they produce is B=μ0ksz^.

In the gray-shaded areas in the image below, the total current is seen passing through the RHS.

Figure 1

04

(b) Determine the value of magnetic field for any loop there is no enclosed free current.

for any loop there is no enclosed free current, so we have:

H = 0

Determine the magnetic field.

H=Bμ0-M0=Bμ0-MB=μ0M=μ0ksz^

Therefore, the value of magnetic field for any loop there is no enclosed free current, is B=μ0ksz^.

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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?

How would you go about demagnetizing a permanent magnet (such as the wrench we have been discussing, at point in the hysteresis loop)? That is, how could you restore it to its original state, with M = 0 at / = 0 ?

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.

lf Jf=0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W:

H=W

According to Eq. 6.23, then,

2W=(M)

So Wobeys Poisson's equation, with M as the "source." This opens up all the machinery of Chapter 3. As an example, find the field inside a uniformly magnetized sphere (Ex. 6.1) by separation of variables.

In Sect, 6.2.1, we began with the potential of a perfect dipole (Eq. 6.10), whereas in fact we are dealing with physical dipoles. Show, by the method of Sect. 4.2.3, that we nonetheless get the correct macroscopic field.

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