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Chapter 6: Q6P (page 274)

Question: Of the following materials, which would you expect to be paramagnetic and which diamagnetic: aluminum, copper, copper chloride (Cucl2), carbon, lead, nitrogen (N2), salt (Nacl ), sodium, sulfur, water? (Actually, copper is slightly diamagnetic; otherwise, they're all what you'd expect.)

Short Answer

Expert verified

Carbon, lead nitrogen, sodium chloride, sulfur and water are diamagnetic.

Copper, copper chloride, aluminium and sodium are paramagnetic.

Step by step solution

01

What is paramagnetic and diamagnetic material.

The diamagnetic material creates a field opposite to the external field and, in the absence of the external magnetic field, does not retain the magnetism.

These materials have weak and negative magnetic material susceptibility. These materials contain an even number of electrons.

The paramagnetic material produces a field in the direction of the external magnetic field. This material gets weakly magnetized in the external field and has positive but small susceptibility. These materials contain an even number of electrons.

02

Find the paramagnetic and diamagnetic material.

Carbon, lead nitrogen, sodium chloride, sulphur and water contain an even number of electrons; therefore, there are diamagnetic.

Copper, copper chloride, aluminium and sodium contain an odd number of electrons; therefore, there are paramagnetic.

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Most popular questions from this chapter

(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]

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.

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,鈭�,z)鈮�B(0,0,Z)+鈭�鈭�B鈭�y0.0.z

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

Suppose the field inside a large piece of magnetic material is B0, so that H0=(1/渭0)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.]

A magnetic dipole m is imbedded at the center of a sphere (radius R) of linear magnetic material (permeability 渭). Show that the magnetic field inside the sphere 0<r鈮�R is

渭4蟺{1r3[3(m.r^r^-m)]+2(渭0-渭)m(2渭0+渭)R3}

What is the field outside the sphere?

See all solutions

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