# Work done by doubling the magnetization of a rod

1. Feb 23, 2014

### Emspak

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.

2. Feb 23, 2014

### TSny

In the second integral it looks like you assumed $\mathcal H$ remains constant as $\mathcal M$ changes.

I'm not sure why the overall sign of the answer should be negative. Could it have to do with the difference between work done on a system and work done by a system?

3. Feb 23, 2014

### Emspak

Yes that is what i was thinking -- work done by is positive work done on a system is negative.

As to the integral you still end up with

$\mu_0 V (\frac{\mathcal H^2}{2} + \frac{\mathcal H^2}{2}\mathcal M)$

and that doesn't get me a factor of 3 anyplace...

4. Feb 24, 2014

### TSny

Can you show more detail of how you did the second integral? You need to write the integrand in terms of just one variable. When you evaluate at the limits you will see how the 3 appears.

5. Feb 24, 2014

### Emspak

Thanks I think I just got a chunk f this. I realized that I was trying a technique that only works with a different kind of gas-pressure equation where you can integrate by parts. So, I tried this:

$\int \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H d \mathcal M$

But $d \mathcal M$ can be written in terms of other things to get it in terms of $\mathcal H$. I can do the following:

$M = \mu_0 V \mathcal M \rightarrow \mathcal M = \frac{M}{\mu_0 V}$ and $M= \frac {\mathcal H C_c}{T}$ so plugging that in I should have $\mathcal M = \frac{C_c \mathcal H}{T \mu_0 V}$ which turns the integral into:

$\int \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H d \left( \frac{C_c \mathcal H}{T \mu_0 V} \right) = \int \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H \left( \frac{C_c}{T \mu_0 V} \right) d \mathcal H = \mu_0 V \int (\mathcal H + \frac{C_c \mathcal H}{T \mu_0 V} ) d \mathcal H$

which integrated gets me

$\mu_0 \left( \frac{\mathcal H^2}{2}V + \frac{\mathcal H^2}{2} \frac{C_c}{T \mu_0 }\right)$

Remembering that $M= \frac {\mathcal H C_c}{T}$ we can see that

$W = \mu_0 \left( \frac{\mathcal H^2}{2}V + \frac{\mathcal H}{2} \frac{M}{\mu_0 }\right)$ which would get us, when subtracting,

$W = \mu_0 \left( \frac{\mathcal H^2}{2}V + \frac{\mathcal H}{2} \frac{2M}{\mu_0 }\right) - \mu_0 \left( \frac{\mathcal H^2}{2}V + \frac{\mathcal H}{2} \frac{M}{\mu_0 }\right) = \frac{\mathcal H M}{2 \mu_0}$

That still doesn't quite get me there but let me know if this makes any more sense.

EDIT: I know that $\mu_0 \left( \frac{\mathcal H^2}{2}V + \frac{\mathcal H^2}{2} \frac{C_c}{T \mu_0 }\right)$ is close. A little algebra gets me closer to the given answer, but I am thinking that the reason it is -3 is that the work in one case has t be negative and you are subtracting a positive, but that still doesn't quite make sense to ma from a physical standpoint. Almost, I think...

6. Feb 24, 2014

### TSny

To see how a factor of $\frac{3}{2}$ comes about, what do you get when you evaluate $\int^{2H}_{H} xdx$? Be careful with how you handle the limits.

7. Feb 24, 2014

### Emspak

Hm. That should give me:

$\int^{2H}_{H} xdx = \frac{H^2}{2} \left. \right|^{2H}_H = (\frac{4H^2}{2} - \frac{H^2}{2})$

Ah, I see. That was sort of dumb of me. Thanks!

8. Feb 24, 2014

### TSny

Not dumb. Just an oversight.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted