Voltage Induced by slowing varying magnetic moment

[keniwas]

Homework Statement

A small magnetic moment $$\vec{m}(t)$$ fixed at position $$\vec{r}$$ varies in a complicated way as a function of time.

(a) Show that if the time variation is "slow enough", the voltage generated around an arbitrary loop of wire near the origin is

$$E(t)=\alpha \frac{\vec{B}_{test}(\vec{r})}{I_{test}}\cdot \frac{d\vec{m}}{dt}$$

and determine the coefficient $$\alpha$$. Here, $$\vec{B}_{test}(\vec{r})$$ is the magnetic field due to a "fictitious test current" $$I_{test}$$ flowing in the wire. This field is evaluated at the position of the magnetic moment. State your result in terms of the voltage generated by integrating around the loop in the direction of the test current.

(b) Define what is meant by "slow enough"

2. The attempt at a solution

I have tried figuring out how to go about calculating the magnetic flux through an "arbitrary loop". But I am a little lost on how to incorporate the test current and test magnetic field into it.

Any input on this would be much appreciated.

 

[nickjer]

Not sure about the "slow enough" part, but (a) I can help with. I have a guess for (b), but we will get to that after (a).

First start by writing out the formula for emf in terms of flux. Keep the flux in integral form. The trick is to use Stoke's theorem for the magnetic vector potential. You will start to see it look similar to the final answer after that.

What I mean by the final answer, is write out B_test(r) using the Biot-Savart law in integral form. And you will see the similarities.

 

[keniwas]

Great! thanks for the reply. I will try that and see how it goes.

I am curious though what gave you the idea to do this? Why would one use a test current in this fashion?

 

[keniwas]

So I have tried what you suggested but I seem to be missing a vital step...

I started with Faradays law and reduced it via stokes theorem to the following

$$\int_c \vec{E}\cdot d\vec{l} = -\frac{d}{dt}\int\int_s \vec{B}\cdot d\vec{a}$$

which is the EMF and the magnetic flux (in integral form). I don't however see how putting in the fact that $$\vec{B}=\nabla \times\vec{A}$$ is going to help me get something closer to the biot-savart law..?? It seems to give me a line integral over $$A$$...

Thanks again for your time on this, I really appreciate it.

 

[nickjer]

If you don't see it yet, then start with the solution you are given. What is $$\vec{B}_{test}(\vec{r})$$ equal to using the Biot-Savart law for a closed loop.

It is best to get rid of these test functions and set it up in an integral form. That way you can play with the integrals until they are in a similar form as your solution.

 

[keniwas]

Ok, so here is what I have now.

From Biot-Savart Law I find the Btest(r) to be

$$\vec{B}_{test}(r)=\frac{1}{c}\int_c I_{test} \frac{d\vec{l}\times\vec{r}}{|r|^3}$$

Which allows me to place the integral form into the equation given in the problem, hence

$$E(t)=\alpha\frac{1}{c}\int_c \frac{d\vec{l}\times\vec{r}}{|r|^3}\cdot \frac{d\vec{m}}{dt}=\alpha\frac{1}{c}\frac{d}{dt}\int_c d\vec{l}\cdot \frac{\vec{r}\times \vec{m}}{|r|^3}=-\alpha\frac{1}{c}\frac{d}{dt}\int_c d\vec{l}\cdot \frac{\vec{m}\times \vec{r}}{|r|^3}$$

Which effectively eliminates the test fields.

I also have the emf in terms of the flux given by

$$\int_c \vec{E}\cdot d\vec{l} = -\frac{1}{c}\frac{d}{dt}\int\int_S \vec{B}\cdot d\vec{a}=-\frac{1}{c}\frac{d}{dt}\int_c \vec{A}\cdot d\vec{l}$$

the latter equality comes from stokes theorem and $$\vec{B}=\nabla\times\vec{A}$$.

Finally, since the vector potential due to a magnetic dipole moment is given by
$$\vec{A}(\vec{r})=\frac{\vec{r}\times\vec{m}}{|r|^3}$$

we see the relation (for $$\alpha=1$$)

$$E(t)=\int_c \vec{E}\cdot d\vec{l}$$

so indeed the induced voltage E(t) can take the given form.

I think this argument works, but if you see anything that looks funny please let me know. I am a little lost on what sort of physical insight this is suppose to provide? When solving it, all I was concerned with was matching up the math, but the physics eludes me... for that reason it makes me feel there is something off in the solution.

Anyway, thank you again for your help on this.

 

[nickjer]

It looks right.

You are able to calculate the emf caused by a changing dipole only by initially measuring the magnetic field at the point caused by running a current through your loop. So no matter what the loop looks like (or even the direction it is facing in), as long as you know the magnetic field at a certain point, then you will know how a dipole at that point will affect the loop. Measuring magnetic fields are easier than solving integrals for some strange looking wire.

Now you need to solve (b). The only thing I could think of is to use the fact that the fields are retarded. So a change in the dipole's magnetic field will travel at the speed of light. But I could be wrong on this.

 

[keniwas]

Excellent, it makes more sense now thanks.


As for part B, I said slow enough would imply that the field varies slowly such that the current can be modeled as a steady current at any given point in time (hence why using the biot-savart law was valid). I think that even if the source were very far away, if it is rapidly varying it would change the analysis regardless of whether or not it was retarded in time.

 

[nickjer]

Looks good.