Magnetic field calculations for solenoid with iron core and horseshoe magnet

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Confused about electromagnet??

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..
 
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215 said:
For some reason does my calculation not make sense..
It would be easier to look for a mistake if you added your calculations here.
 
ohh sorry my bad..

I have choosen to use 1A as constant.

For the Horseshoe magnet

B = [itex]\frac{i*n*µ0*µr}{lj+2vµr+2xµr}[/itex]


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=[itex]\frac{N*I}{l}[/itex]
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..
 
215 said:
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..
And where is the issue with those solutions?

I don't understand the horseshoe geometry with multiple (4?) air gaps.
 
looks like a homework problem. move to other thread??
 
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
 
horseshoe magnet will have only one air gap. how come you have 4? please draw the detailed diagram of the magnet.
 
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?
 
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