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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 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?

(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]

A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility 蠂m. A currentI flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field.

Figure 6.24

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.

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.]

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