# 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**

[itex]\mathcal M[/itex] is total magnetization of the rod.

M is magnetic moment

[itex]C_c[/itex] is curie constant

[itex]\mathcal H[/itex] is magnetic field strength

T is temperature

[itex]B = \mu_0 (\mathcal H + \mathcal M)[/itex]

[itex]M = \mu_0 V \mathcal M[/itex]

[itex]M = \frac{C_c \mathcal H}{T}[/itex]

**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

[itex] d’W = \mathcal E I dt [/itex]

Since [itex]\mathcal H = \frac{NI}{L}[/itex]

and [itex]\mathcal E = -NA \frac{dB}{dt}[/itex] (where A is the cross-sectional area of the rod) we can substitute in and get

[itex] d’W = V \mathcal H dB [/itex]

from earlier [itex]B = \mu_0 (\mathcal H + \mathcal M)[/itex] so [itex]dB = \mu_0 d \mathcal H + \mu_0 d \mathcal M[/itex] so:

[itex] d’W = \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H d \mathcal M[/itex]

Integrating this I should end up with:

[itex] \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[/itex]

[itex] = \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} [/itex]

And putting it all together I should have:

[itex]= \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}[/itex]

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

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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