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

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.


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).




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)+By0.0.z

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

lf Jf=0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W:

H=W

According to Eq. 6.23, then,

2W=(M)

So Wobeys Poisson's equation, with M as the "source." This opens up all the machinery of Chapter 3. As an example, find the field inside a uniformly magnetized sphere (Ex. 6.1) by separation of variables.

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.

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