
#1
Apr2913, 01:51 PM

P: 25

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.. 



#2
Apr2913, 02:23 PM

Mentor
P: 10,864





#3
Apr2913, 02:33 PM

P: 25

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.. 



#4
Apr2913, 02:53 PM

Mentor
P: 10,864

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



#5
Apr2913, 03:03 PM

P: 153

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




#6
Apr2913, 03:24 PM

P: 25

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
Apr3013, 05:34 AM

P: 153

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




#8
Apr3013, 12:50 PM

P: 25

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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