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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

Starting from the Lorentz force law, in the form of Eq. 5.16, show that the torque on any steady current distribution (not just a square loop) in a uniform field B is m脳B.

Notice the following parallel:

{鈭�路D=0鈭�脳E=0,蔚0E=D-P(Nofreecharge)鈭�路B=0鈭�脳H=0,渭0H=B-渭0M(Nofreecharge)

Thus, the transcription D鈫�B,E鈫�H,P鈫�渭0M,蔚0鈫�渭0,, turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive.

(a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16);

(b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18);

(c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93).

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 forL鈮�a.) Compare this bar magnet with the bar electret of Prob. 4.11.

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.

(a)Show that the energy of a magnetic dipole in a magnetic field B is

U=鈭�m鈰�B.

[Assume that the magnitude of the dipole moment is fixed, and all you have to do is move it into place and rotate it into its final orientation. The energy required to keep the current flowing is a different problem, which we will confront in Chapter 7.] Compare Eq. 4.6.

Figure 6.30

(b) Show that the interaction energy of two magnetic dipoles separated by a displacement r is given by

U=渭04蟺1r3[m1鈰�m2鈭�3(m1鈰�r^)(m2鈰�r^)]

Compare Eq. 4.7.

(c) Express your answer to (b) in terms of the angles 胃1 and 胃2 in Fig. 6.30, and use the result to find the stable configuration two dipoles would adopt if held a fixed distance apart, but left free to rotate.

(d) Suppose you had a large collection of compass needles, mounted on pins at regular intervals along a straight line. How would they point (assuming the earth's magnetic field can be neglected)? [A rectangular array of compass needles aligns itself spontaneously, and this is sometimes used as a demonstration of "ferromagnetic" behaviour on a large scale. It's a bit of a fraud, however, since the mechanism here is purely classical, and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism. 13]
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