# HiI know that for a short solenoid (L<R) the magnetic field at the

1. Feb 3, 2012

### Niles

Hi

I know that for a short solenoid (L<R) the magnetic field at the axis is (standard EM)
$$B(z) = \frac{1}{2}\mu_0 \frac{N}{L}I(\frac{z+\frac{L}{2}}{\sqrt{(z+L/2)^2+R^2}} - \frac{z-\frac{L}{2}}{\sqrt{(z-L/2)^2+R^2}})$$
where R is the radius of the solenoid, N the number of turns along the axis and L the length. In this system, each vertical plane consists of a single turn, but say I am looking at a solenoid, where each vertical plane consists of e.g. 2 turns. First I thought about using the above equation twice, but that is wrong since it is not 2 independent solenoids.

Is it correct to regard the system simply as a collection of coils with 2 turns each? I'm not quite sure how this would work out, since this way I can't take into account the widths of each individual coil. If my description is confusing, please let me know.

Best,
Niles.

Last edited: Feb 3, 2012
2. Feb 3, 2012

### clem

Re: Coils

Your formula is for a solenoid of any length, not just L<R.
I don't understand what you mean be 'two turns'.
Why would it make a difference?

3. Feb 3, 2012

### Niles

Re: Coils

Hi, thanks for replying. I really appreciate it. What I mean by two turns is the following:

OOOOOO
OOOOOO
------------ (axis of solenoid)
OOOOOO
OOOOOO

There I have a solenoid with 12 turns in total, i.e. a 6-turn solenoid on top of another (larger radii) 6-turn solenoid. How would one calculate the B-field for that? In principle I could just add the B-field for each one of the 12 turns, right? The reason why I am asking is because I am looking for the most general way to calculate this, because I would also have to look at e.g.

OOOO
OOOOOO
OOOOOOOO
---------------------- (axis)
OOOOOOOO
OOOOOO
OOOO