# Magnetization in fluid in Quantum Mechanical Case

Here, we consider the fluid to have N spins up (Nup) and N spins down (Ndown). In the presence of an external field

**B**, the energy is given by where the positive z-axis points upwards.

Again, as in the classical case for magnetization of a fluid (Langevin`s equation), the relative probability of having any angle is, from statistical mechanics, exp(-energy/kT).

So, and . If N is the total number of dipoles, them M is the product of N with the average magnetic moment along the z-axis.

In most normal cases, say, for typical moments, room temperatures, and the fields one can normally get (like 10 000gauss), the ratio represented by u is about 0.02. One must go to very low temperatures to see the saturation. For normal temperatures, tanh(u) can be replaced by u and

The graph of M against u for the quantum mechanical case of magnetization in fluids resembles that for the classical case.

When

**B**gets very large, the hyperbolic tangent approaches 1, and M approaches the limiting value (N x the average magnetic moment along the z-axis). So, at high fields, the magnetization saturates. We can see why that is,; at high enough fields, the moments are all lined up in the same direction . In other words, they are all in the spin-down state, and each atom contains the same moment.

Additionally, when we look at the expressions for magnetization in the fluid at room temperature for both the classical and quantum case, we notice that

**M**is proportional to

**B**/T. From this, we can derive Curie`s law which is

**M**= k

**B**/T, where k is a constant.

i wonder where you got the source for the information on this page, very useful!

ReplyDeleteIt's from my university notes that I prepared. I forgot the exact source.

Delete:) i'm doing physics too btw. anyway, thanks for posting :)

ReplyDelete