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Chapter 6: Q20P (page 291)

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 ?

Short Answer

Expert verified

Therefore, the permanent magnet can demagnetise by heating the material above the Curie temperature.

Step by step solution

01

define the Curie temperature.

The Curie temperature is the temperature at which magnetic material changes its magnetic properties, or certain material loses their permanent magnetic properties. It is also known as the Curie point.The magnetism of the material is obtained by the dipole produced by the electron's momentum and spin.

02

Demagnetizing the permanent magnet.

When the permanent magnetic is heated above the curie temperature or the curie point, it loses its magnetism permanently and demagnetized. Then it goes to its original state. Where, M = 0, I = O when the magnetic material is placed between the electromagnet and magnetized, reverse the field direction, which reduces the field strength, and the material gets demagnetized.

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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/ฮผ0)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.]

Derive Eq. 6.3. [Here's one way to do it: Assume the dipole is an infinitesimal square, of side E (if it's not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate F = I J (dl x B) along each of the four sides. Expand B in a Taylor series-on the right side, for instance,

B=B(0,โˆˆ,z)โ‰…B(0,0,Z)+โˆˆโˆ‚Bโˆ‚y0.0.z

For a more sophisticated method, see Prob. 6.22.]

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=Mฯ•^if they are Gilbert-type, the magnetization is equivalent to surface monopole densities ฯƒb=ยฑMat 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?

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

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?
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