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

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.)

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.

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

For the bar magnet of Problem. 6.9, make careful sketches of M, B, and H, assuming L is about 2a. Compare Problem. 4.17.

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.]

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