6
$\begingroup$

I learnt that, when a ferromagnetic material is heated, it will become paramagnetic at a certain temperature. This temperature is called Curie point or Curie temperature. For example, Curie temperature for iron is given to be $1043\ \text{K}$. So as per my understanding at $1042\ \text K$ the material is ferromagnetic and at $1044 \ \text K$ it becomes paramagnetic. However, I don't understand what happens during this transition.

Earlier, I've learnt that both ferromagnetic and paramagnetic substances have atoms with permanent magnetic moment. A ferromagnetic substance is strongly attracted in the presence of a magnetic field, and a paramagnetic substance is weakly attracted. But "strongly" and "weakly" are relative terms. Or in other words, I think a weak ferromagnet is a strong paramagnet. If so, how could an experimenter say that a substance at a particular temperature is ferromagnetic or paramagnetic? What properties does a ferromagnetic substance loose when it gets converted to a paramagnetic material?

Is the boundary between ferromagnetism and paramagnetism well-defined? Alternatively, is the Curie point actually a single point on the temperature scale or is it a range of temperature?

$\endgroup$
  • 1
    $\begingroup$ Yes, there is a well defined boundary between them. A paramagnet has zero magnetisation at zero external magnetic field. But a ferromagnet has non zero magnetisation at zero external magnetic field. $\endgroup$ – Superfast Jellyfish Mar 2 at 14:08
  • 1
    $\begingroup$ @FellowTraveller Slight nitpick: it's still technically possible that a ferromagnet can have zero magnetization at zero external field, just extremely unlikely. $\endgroup$ – probably_someone Mar 2 at 14:11
  • $\begingroup$ @probably_someone well yes. In the same way entropy of a system can reduce right? Or is it more probable? $\endgroup$ – Superfast Jellyfish Mar 2 at 14:13
  • $\begingroup$ @FellowTraveller If you're talking about an isolated system moving between equilibrium states, then yes, it's more probable than the entropy of that system decreasing (because that's impossible, whereas a ferromagnet having zero magnetization at zero applied field is still possible). Now, if the system is not in equilibrium, then the entropy of the system can fluctuate, and therefore can decrease for short periods of time (see the fluctuation theorem: en.wikipedia.org/wiki/Fluctuation_theorem). $\endgroup$ – probably_someone Mar 2 at 14:16
5
$\begingroup$

It turns out that if you work out the physics behind the magnetism of a material, the magnetisation $M$ obeys a transcendental equation of the form $$M=\tanh\left(\frac{\alpha M+\gamma H}{T}\right)$$ where $\alpha$ and $\gamma$ are some constants, $H$ is the magnitude of the external magnetic field and $T$ is the temperature. What this says is that at a given temperature and external field, the magnetisation of the material will be given by the value which is a solution to the above equation. All well and good but how do we solve such an equation?

One way to do so is numerically (graphically) as can be seen below. In this simplified picture, I am considering no external field and only temperature $T$ varying. The solution to the equation will be given by the points of intersection between the hyperbolic tan curve and the straight line.

enter image description here

As you can see, the only solution for $T>1$ is at $M=0$ as that is the only intersection point. This is the behaviour of a paramagnet. However for $T<1$, we see that there are two more solutions emerging. What does this mean? This means that we have $|M|>0$, for zero external field. This is the behaviour of a ferromagnet.

So there is a clear distinction between a paramagnet and a ferromagnet.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer. As per your comment "A paramagnet has zero magnetisation at zero external magnetic field. But a ferromagnet has non zero magnetisation at zero external magnetic field." can we say that this kind of distinction is due to lack of domains in paramagnetic materials? Or is that because the thermal randomisation is larger in paramagnets attributing to their low retentivity? $\endgroup$ – Guru Vishnu Mar 3 at 6:36
  • $\begingroup$ can we say that this kind of distinction is due to lack of domains in paramagnetic materials? Or is that because the thermal randomisation is larger in paramagnets attributing to their low retentivity?” turns out they are both the same thing! The lowest energy state for materials is when they form domains. But thermal randomisation above the critical temperature gives the material enough energy to get out of the ground state which results in the disappearance of domains $\endgroup$ – Superfast Jellyfish Mar 3 at 6:46
  • $\begingroup$ Ok thanks. In the graph in your answer, I can see that the $x$ axis represents the intensity of magnetisation $M$. But could you tell what does the $y$ axis denote? $\endgroup$ – Guru Vishnu Mar 3 at 6:51
  • $\begingroup$ The y axis represents $f(M)$ which for the green curve is straight line and for the red curve is tanh $\endgroup$ – Superfast Jellyfish Mar 3 at 6:53
  • 1
    $\begingroup$ Thank you. For those who wish to play with the graph: Desmos $\endgroup$ – Guru Vishnu Mar 3 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.