# Quasi-stastic solenoid magnetic field

1. Mar 6, 2010

### Pengwuino

The construction of and excitation of an infinite, straight, right circular solenoid of radius a, with N turns per unit length, are such taht tis current I(t) is the same everywehre along its length and is changed very slowly in time. Show that the fields far from the solenoid are approximately:

$$E \approx \frac{{ - \mu _0 Na^2 }}{{2\rho }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi$$

Attempt:

We know from previous work that for a localized charge distribution far from the origin for slowly changing currents with no charge density and only dipole terms is:

$$E = \frac{{\mu _0 }}{{4\pi r^2 }}\hat r \times \frac{{\partial m(t - \frac{r}{c})}}{{\partial t}}$$

So it should be a simple matter of computing the dipole moment per unit length. Now I've been second guessing myself continually on the charge density but I've kinda come to the conclusion that it the multipole moment should be

$$\begin{array}{l} m = \frac{1}{2}\int {I(t - \frac{\rho }{c})N\delta (\rho ' - a)\rho '^2 d\rho 'd\phi '(\hat \rho ' \times \hat \phi ')} \\ = 2\pi a^2 I(t - \frac{\rho }{c})N\hat z \\ \end{array}$$

However, simply plugging this making the assumption that due to the symmetry, r becomes $$\rho$$, gives:

$$E = \frac{{ - \mu _0 a^2 N}}{{2\rho ^2 }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi$$

which is off by that $$\rho^2$$ instead of $$\rho$$. Is there something obvious im missing here?

2. Mar 7, 2010

### nickjer

It is an infinitely long solenoid. So you will need to integrate along the z-axis (since dm is proportional to dz), your A-vector will have the integral in it. Once you solve it, you will see the 1/r and not 1/r^2.

3. Mar 7, 2010

### Pengwuino

Integrating along dz will blow the integral up. I can't find a way to do the z-integration without that happening.

4. Mar 8, 2010

### nickjer

Hmm... So you have:

$$dA = \frac{\mu_0}{4\pi}\frac{dm\times r}{r^2}\hat{r}$$

Next you know from a single current loop:

$$m_{loop} = IA \hat{z}$$

So for an infinitesmal cross-section of our cylinder, we have:

$$dm = NIAdz \hat{z}$$

Substituting this into our first equation gives:

$$dA =\frac{\mu_0 N I a^2}{4} \frac{dz}{r} \hat{z}\times\hat{r}$$

Now you need to find an analytical form for $$\hat{z}\times\hat{r}$$ using only $$z$$ and $$r$$. You know it will be $$sin(\theta)$$ where $$\theta$$ is the angle formed between $$\hat{z}$$ and $$\hat{r}$$.

So you will want to draw a picture of an infinite cylinder. Then draw a point far out on the x-axis, lets call $$\rho$$. Then somewhere along the z-axis you will draw a vector out to this point and call it $$r$$. So that:

$$\rho^2+z^2=r^2$$

I set up the differential form for you in its most basic form. Now you will need to find this cross product and integrate it from -infinity to infinity. It won't blow up on you since it didn't blow up in Maple for me.

5. Mar 9, 2010

### Pengwuino

I absolutely do not want to use the vector potential in this problem though. To find the B-field, i'd have to do $$\nabla \times \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over A$$ which would result in eventually having terms like $$\nabla I(t - \frac{\rho }{c})$$ eventually which I can't operate on. Then again using Jefimenko's equations, I might be able to get somewhere with using the moment.

Actually I don't even think your method is valid. Your dA is with respect to source points but you are doing your integration using field points (or maybe not, I'm still thinking about this...)

Last edited: Mar 9, 2010
6. Mar 9, 2010

### nickjer

You are not solving for the magnetic field. You want just the electric field, so you use:

$$E=-\frac{\partial A}{\partial t}$$

I probably left out a few constants.

Also, it is very similar to integrating over charge points to get the full electric field at a point. Except now I am integrating over magnetic dipoles along an infinite wire to get the vector potential.

Last edited: Mar 9, 2010
7. Mar 9, 2010

### Pengwuino

Oh right, what's with me?

However, $$$\hat z \times \hat r = \frac{{ - \rho }}{r}\hat \phi$$$. Doing the integration along z, if valid, returns something funny.

$$$A = \frac{{\mu _0 NIa^2 }}{4}\int_{ - \infty }^\infty {\frac{{ - \rho }}{{(\rho ^2 + z^2 )}}dz = } \frac{{ - \pi \mu _0 NIa^2 }}{4}$$$

Did I miss something?

8. Mar 9, 2010

### nickjer

I made a mistake, I lost a factor of 'r' in my original equation. The correct form is:

$$dA =\frac{\mu_0 N I a^2}{4} \frac{dz}{r^2} \hat{z}\times\hat{r}$$

So you should have a power of 3/2 in your denominator of the integral.

9. Mar 9, 2010

### Pengwuino

Yah I found that at the same time. The one thing I'm just hesitant about is integrating over field points as if they were source points. For example, if the solenoid was simply of length 2L, one would have to have a z-dependence on the electric field, but you would lose it running through the method we have done here I would think...

10. Mar 10, 2010

### nickjer

This is similar to the magnetic field from an infinite wire. If it were a finite wire then the math would be more difficult and there would be a z-dependence on the magnetic field. Since it is infinite, then the magnetic field only depends on radial distance from the wire.