1. Apr 29, 2013

### 215

How do i calculate the magnet field for an Solenoid with an ironcore with x windings and y amps..

The same for an horseshoe magnet, where there is an air gap about 2mm..

For some reason does my calculation not make sense..

Last edited: Apr 29, 2013
2. Apr 29, 2013

### Staff: Mentor

It would be easier to look for a mistake if you added your calculations here.

3. Apr 29, 2013

### 215

I have choosen to use 1A as constant.

For the Horseshoe magnet

B = $\frac{i*n*µ0*µr}{lj+2vµr+2xµr}$

lj = length of the ironcore = 0.034+0.011+0.034
µ0 = 4pi *10^-7
x= length of one of the airgaps
v= length of the other airgap
µr = Relative permability = µ0(1+XM)=5000
I = 1
For the solonoid

B = µrH

H=$\frac{N*I}{l}$
l = 0.065
I = 1

I know B = 0.4
I've been trying to solve for N, windings, and for the solonoid do i get 22, and for the horseshoe do i get 1688..

4. Apr 29, 2013

### Staff: Mentor

And where is the issue with those solutions?

I don't understand the horseshoe geometry with multiple (4?) air gaps.

5. Apr 29, 2013

### physwizard

looks like a homework problem. move to other thread??

6. Apr 29, 2013

### 215

Homework... Well, is more like a project.. I am trying to build an Electromagnet which capable off pulling a certain amount of newton..
I done conversion from Newton to Tesla...
Now do i have define the tesla value..

http://snag.gy/oL7D5.jpg

7. Apr 30, 2013

### physwizard

horseshoe magnet will have only one air gap. how come you have 4? please draw the detailed diagram of the magnet.

8. Apr 30, 2013

### 215

Here is a picture of the one I've been working with
http://snag.gy/A7zJd.jpg

There is an airgap between each leg, each leg has to pull a piece of metal with 3N, in the distance of 2 mm, i am quite unsure about the next one, but would the airgap between the the metal pieces have something to say?

Last edited: Apr 30, 2013