# Ferromagnetic Hysteresis

1. Apr 15, 2007

### Tomsk

1. The problem statement, all variables and given/known data

It's a past paper question, but I just don't understand it. I must have missed something here...

2. Relevant equations

Maxwell's equations, $$\frac{B}{\mu_{0}} = H + M$$, $$M=\chi H$$ change in stored energy = H.dB

3. The attempt at a solution

(a) $$\nabla\times H = J_{free}$$(because$$\epsilon_{0} \partial D/\partial t = 0$$)
$$\int H.dl = \int J.dA$$
$$H.2\pi R = NI$$
$$H=\frac{NI}{2\pi R}$$
$$B=\mu_{0}(1+\chi)H$$
$$B=\mu_{0}(1+\chi)\frac{NI}{2\pi R}$$
Is that OK?

(b)This is where I get stuck. I'm just not sure how it works. Is that curve parameterised by time, or not? I.e, does the system move round the curve automatically once you switch on the current (and have moved it from its equilibrium position) because of the magnetization (if so how?), in the way that a pendulum tries to get to it's equilibrium position, or do you have to adjust the current (or something else?) to change the applied magnetic field, and when you do you find that it moves around that curve? And how do I get from there to the energy required?

Last edited: Apr 15, 2007
2. Apr 15, 2007

### Meir Achuz

You can't use the susceptiblity in ferromagnetism.
The energy is given by \integral H.dB, the area under the hysteresis curve.

3. Apr 16, 2007

### Tomsk

Thanks for the reply. Why can I not use susceptibility? I can't think why they would have given it to me otherwise... And would I want to find an equation for B in terms of H, and integrate that, or do I just work out the area approximately by looking at the graph? I would have thought that wouldn't be very accurate...

4. Apr 16, 2007

### Meir Achuz

The ratio B/H is different for every point on the curve.
The susceptibility was given only for part a.
There is no equation for the curve since it is different for different materials.
You do have to just find the appropriate area.