# Magnetic Field Outside a Finite Solenoid

1. Nov 4, 2009

### mo09890

I'm trying to write a program to simulate the fields generated by a solenoid but I've hit a bit of a brick wall. There is a vast amount of examples and information on the field generated inside a solenoid however they all assume that the field outside is negligible and as such I have been having trouble finding any equations or information on the field outside of a solenoid.

Any pointers in the right direction would be greatly appreciated.

2. Nov 4, 2009

### Bob S

The field on the axis of a solenoid is given by B = 0.5 μ0 NI (cos β2 – cos β1) where β1 and β2 are the angles of the end of the coils subtended at a point on the solenoid axis. See Smythe “Static and Dynamic Electricity”, 3rd Edition, page 297.

So the field in the center of an infinite solenoid is Bcenter = μ0 NI. The field inside the end of an half-infinite solenoid is Bend = 0.5 μ0 NI

Because the line integral around the coil windings is (Ampere’s Law)

∫B dl = μ0 NI

The field outside the solenoid has to be

Boutside = μ0 NI - 0.5 μ0 NI (cos β2 – cos β1)

Bob S

3. Nov 5, 2009

### mo09890

Thanks for the very speedy reply Bob! I have one question though - is that the equation for the field outside the solenoid but still on the axis or will this work for any point in a 3D space? (Am i asking too much for it to be as simple as this for 3D? :P)

4. Nov 5, 2009

### Bob S

Hi mo09890-
The result quantitativly applies to any line integral outside the coil. To get the actual B field anywhere outside the coil, you will have to choose the appropriate path of the line integral, meaning that although the line integral is correct, the value of B anywhere is not determined, except very close to the coil.
Bob S

5. Nov 24, 2009

### mary8789

Magnetic Field from a Finite Solenoid

Hello
I need to calculate the magnetic fied from a coil. The solenoid has a finite side, and its length and number of turns are known. However, I can't find out how to do the calculation. I know that the B-field in the middle is equal to u0*n*I but I don't understand how to find the result, using Ampere's law.

6. Nov 24, 2009

### Bob S

Re: Magnetic Field from a Finite Solenoid

The Biot Savart Law is exact, and can be used on and off the axis of the solenoid. For an infinite solenoid, the Ampere's Law agrees with the Biot-Savart solution. The on-axis solution for a finite solenoid is also exact (see my previous post). Ampere's Law does not give the correct field for a finite length solenoid, even on axis. For fields outside a finite solenoid, I would probably have to resort to integrating the vector potential of a solenoid based on a circular loop. See Smythe (ibid) page 290-291. There must be an easier way.

[added] you might try some inexpensive magnetic field mapping software. Look at
http://www.vizimag.com/
Bob S

Last edited: Nov 24, 2009
7. Nov 25, 2009

### Ikoro

Simple use faradays law since you are running a current through it:
e = (integral)E DL = dI (b) / dt
dI (b)= the changing magnetic field.
Now you can use this equation an integrate over the distance ....one way is to plot it in a graph and take points away from the center and integrate it just like finding the E filed of a wire segment. GOOD luck