# Work done by doubling the magnetization of a rod

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

Picture a rod of length L with N turns of wire around it

What is the work done when one doubles the magnetization? We’ll assume the coil has negligible resistance.

2. Relevant equations

$\mathcal M$ is total magnetization of the rod.
M is magnetic moment
$C_c$ is curie constant
$\mathcal H$ is magnetic field strength
T is temperature
$B = \mu_0 (\mathcal H + \mathcal M)$
$M = \mu_0 V \mathcal M$
$M = \frac{C_c \mathcal H}{T}$

3. The attempt at a solution

OK, so I want to find the work. The power put into the system is P = (emf) * I where emf is electromotive force and I is current. So the work in time dt will be

$d’W = \mathcal E I dt$

Since $\mathcal H = \frac{NI}{L}$
and $\mathcal E = -NA \frac{dB}{dt}$ (where A is the cross-sectional area of the rod) we can substitute in and get

$d’W = V \mathcal H dB$

from earlier $B = \mu_0 (\mathcal H + \mathcal M)$ so $dB = \mu_0 d \mathcal H + \mu_0 d \mathcal M$ so:

$d’W = \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H d \mathcal M$

Integrating this I should end up with:

$\int d’W = \int \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H d \mathcal M = \int \mu_0 V \mathcal H d \mathcal H + \int^{\mathcal 2M}_{\mathcal M} \mu_0 V \mathcal H d \mathcal M$

$= \mu_0 V ( \int \mathcal H d \mathcal H + \int^{\mathcal 2M}_{\mathcal M} \mathcal H d \mathcal M) = \mu_0 V ( \frac{\mathcal H^2}{2} + \mathcal H \mathcal M)\left. \right|_{\mathcal M}^{\mathcal 2M}$

And putting it all together I should have:
$= \mu_0 V ( \frac{\mathcal H^2}{2} + \mathcal H \mathcal 2M) – \mu_0 V ( \frac{\mathcal H^2}{2} + \mathcal H \mathcal M) = \mu_0 V (H \mathcal M) = \mu_0 V (\mathcal H \frac {M}{\mu_0 V}) = \mu_0 V \frac{C_c \mathcal H^2}{T \mu_0 V} = \frac{C_c \mathcal H^2}{T}$

But that’s wrong. The solution I found says it is $-3\frac{\mathcal H^2}{2} ( \mu_0 V + \frac{C_c}{T} )$

So I think I missed a negative sign or made an error in the integral somewhere. Or I just need to add the factor that expresses the work done in the absence of the rod, which seems to have somehow got lost. I suspect it is that one of the integrals I get should be negative, since we’re talking about work done on the rod by the field.

If anyone can tell me where I lost the plot I’d be most appreciative. Thanks.

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