Solenoid Field
Most recent answer: 04/02/2010
Q:
hello, I am building a solenoid and would like to know the
magnetic field strength it will produce at its center. on each end of my solenoid are 4 inch diameter flanges 1/4
inch thick. The core that i am wraping on is 1 1/2 inch in diameter with a 1 inch hole in its center leaving a 1/4 inch wall.My total wraping length equals 4.080 inches.
I am using 10 gauge coated wire ( magnet wire ) to wrap this form.10 turns per inch. There will be 480 turns on this form. It will be powered by 10 amps initially. Then i will try car batteries in parallel. The solenoid is made of soft iron
which was obtained from a machine shop since i have no way of knowing soft iron from any other iron. Now i have searched the internet for a solution but the problem seems to have numerous solutions each different. The average answer seems to be that the magnet field strength depends on the current applied and the number of turns of wire per unit length but when i see examples of the above and work them i get completely different answers. I would appreciate any help. Also in their equations they are using a symbol ( uo ) i know that this has something to do with permability.But what ? There are also numerous permabilities stated for of soft iron
- ron harris (age 56)
cordova,south carolina,29039
- ron harris (age 56)
cordova,south carolina,29039
A:
First a safety tip: With too much current you could overheat the coils and start a fire, You should start with a smaller current and work your way up, making sure that nothing is getting too hot.
Wikipedia actually has a nice clear explanation of the field calculation with the relevant formulas:
The different permeabilities you're finding for iron reflect the large sensitivity of its permeability to metallurgical details. By the way, μo = 4π * 10-7 Henries per meter is the permeability of free space.
It looks to me, based on your numbers (and using the same formula that appears in the wikipedia link) that your bare field (without the iron) would be about 5*10-2 (whoops- older version was wrong, see below) Tesla if you manage to safely get to 10 A current. Your iron would be driven beyond the range where its response to the field is linear. It would be near saturation- as magnetized as it can get. That would mean a field of around 1.5 Tesla internally for the most magnetized part. Once you get near saturation, there's little point in increasing the current- another reason to start with a small current and work your way up.
Without a picture, it's a little hard for me to tell the geometry well enough to figure out what the fields will be at the ends of your flanges. Generally, to get a large field you want the solenoid core to be completely filled with the iron, and you don't want it to get thicker at the ends, since that spreads out the field lines.
Mike W.
Wikipedia actually has a nice clear explanation of the field calculation with the relevant formulas:
The different permeabilities you're finding for iron reflect the large sensitivity of its permeability to metallurgical details. By the way, μo = 4π * 10-7 Henries per meter is the permeability of free space.
It looks to me, based on your numbers (and using the same formula that appears in the wikipedia link) that your bare field (without the iron) would be about 5*10-2 (whoops- older version was wrong, see below) Tesla if you manage to safely get to 10 A current. Your iron would be driven beyond the range where its response to the field is linear. It would be near saturation- as magnetized as it can get. That would mean a field of around 1.5 Tesla internally for the most magnetized part. Once you get near saturation, there's little point in increasing the current- another reason to start with a small current and work your way up.
Without a picture, it's a little hard for me to tell the geometry well enough to figure out what the fields will be at the ends of your flanges. Generally, to get a large field you want the solenoid core to be completely filled with the iron, and you don't want it to get thicker at the ends, since that spreads out the field lines.
Mike W.
(published on 04/02/2010)
Follow-Up #1: mistake
Q:
What was used to arrive at the estimate of 5*10-3 Tesla for the bare field?
I have tried to reproduce it using the referenced formulas, but was unable to.
B = u_o N I / L.
u_o = 4*Pi * 10^(-7)
N = 480 Turns
I = 10 Amp
L = 0.104 or 0.114 meters depending on if you count the flanges.
This works out to around 0.0528 Telsa, which is off by a factor of 10.
Could you kindly point out my mistake?
- Jason (age 29)
Mi
- Jason (age 29)
Mi
A:
Jason- You've kindly pointed out mine. It's fixed now.
Many thanks!
Mike W.
Many thanks!
Mike W.
(published on 05/22/2010)