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

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 ?

A familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass md and dipole moment m.

(a) If you put two back-to-hack magnets on the rod, the upper one will "float"-the magnetic force upward balancing the gravitational force downward. At what height (z) does it float?

(b) If you now add a third magnet (parallel to the bottom one), what is the ratio of the two heights? (Determine the actual number, to three significant digits.) [Answer:(a)3μ0m2(b)0.8501]


In Prob. 6.4, you calculated the force on a dipole by "brute force." Here's a more elegant approach. First writeB(r)as a Taylor expansion about the center of the loop:

,B(r)B(r0)+[(rr0)0]B(r0)

Wherer0the position of the dipole and 0is denotes differentiation with respect tor0. Put this into the Lorentz force law (Eq. 5.16) to obtain

.F=IdI×[(r0)B(r0)]

Or, numbering the Cartesian coordinates from 1 to 3:

Fi=Ij,k,l=13εijk{rldlj}[0lBk(r0)],

Where εijk is the Levi-Civita symbol (+1ifijk=123,231, or312; 1ifijk=132, 213, or 321;0otherwise), in terms of which the cross-product can be written (A×B)i=j,k=13εijkAjBk. Use Eq. 1.108 to evaluate the integral. Note that

j=13εijkεljm=δilδkmδimδkl

Whereoil is the Kronecker delta (Prob. 3.52).




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.

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×B.

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