# Work done by doubling the magnetization of a rod

• Emspak
In summary, the work done when doubling the magnetization of a rod of length L with N turns of wire around it is given by the equation W = -3(\frac{\mathcal H^2}{2}(\mu_0V + \frac{C_c}{T})) and can be calculated by integrating the equations \int \mu_0 V \mathcal H d \mathcal H + \mu_0 V \mathcal H(\frac{C_c}{T \mu_0 V}) d \mathcal H and subtracting the result from the work done in the absence of the rod.
Emspak

## Homework Statement

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.

## Homework 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
$B = \mu_0 (\mathcal H + \mathcal M)$
$M = \mu_0 V \mathcal M$
$M = \frac{C_c \mathcal H}{T}$

## 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.

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

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?

TSny said:
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?

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...

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.

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...

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.

1 person
TSny said:
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.

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!

Not dumb. Just an oversight.

## 1. What is the definition of work in terms of magnetization?

Work is defined as the energy required to move a magnetized object against an external force, such as the force of gravity or an opposing magnetic field.

## 2. How is the work done affected by doubling the magnetization of a rod?

Doubling the magnetization of a rod will result in an increase in the amount of work required to move the rod against an external force. This is because the stronger magnetic field of the rod will create a greater force, requiring more energy to overcome.

## 3. Can the work done by doubling the magnetization of a rod be negative?

Yes, the work done can be negative if the rod is moving in the opposite direction of the external force. In this case, the rod's increased magnetization may create a repulsive force, resulting in a decrease in the amount of work required to move the rod.

## 4. What factors besides magnetization can affect the work done?

The strength of the external force and the distance the rod is moved can also affect the work done. A stronger external force will require more work, while a longer distance will also result in more work being done.

## 5. How is the work done by doubling the magnetization of a rod related to energy?

The work done by doubling the magnetization of a rod is directly related to the energy required to create the stronger magnetic field. This means that the more work is done, the more energy is being used to increase the magnetization of the rod.

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