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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

At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that tanθ2/tanθ1=μ2/μ1 assuming there is no free current at the boundary. Compare Eq. 4.68.

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).

On the basis of the naïve model presented in Sect. 6.1.3, estimate the magnetic susceptibility of a diamagnetic metal such as copper. Compare your answer with the empirical value in Table 6.1, and comment on any discrepancy.

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.)

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,∈,z)≅B(0,0,Z)+∈∂B∂y0.0.z

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

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