# I Magnet Interaction

1. Mar 25, 2016

### NTL01

Lets us say that two magnets of unequal strength have one pole each lined up facing the other magnets pole. For convention magnet 1 is an electromagnet with an air core and a 200 gauss field strength. Magnet 2 is a neodymium permanet magnet rated at 20,000 gauss. Assume the "north pole of magnet 1 is opposed to the south pole of magnet 2

The magnets will attract each other when the electromagnet is energized , the question is how to solve for the force of attraction

Wiki tells me that if the two magnets are both permanent magnets a reasonable approximation can be given by the formula

where F is the force qm1 the gauss of the first magnet , qm2 that of the second magnet and u0 the permeability of space, and r the distance the two magnets are apart.

So my answer might be if the magnets are 1 inch apart

F= 1/4pie * (20,000*200)/1 which would roughly be
318471.3

My question is I cannot find a formula for interaction of a electromagnet with a permanent magnet and in this case the electromagnet has an air core.

Some people think in this case the strength of the electromagnet sets a limit on the force equation

Seeking comments and a formula if available

2. Mar 25, 2016

I found what I think is a more reliable formula, because it is more clear what the units will be. $F=B1*B2*A/(2 \mu_o)$ where $\mu_o=4 \pi E-7$. The magnetic fields must be expressed in Wb/m^2. 1.0 E+4 gauss=1.0 Wb/m^2. The A is the area of the surfaces, (assumed to be equal and the gap must be small. The force you get will be in Newtons. (I need to research this further. Should be able to see what the units in your original formula need to be. )Just an estimate, if it is an electromagnet with an iron core, the magnetic field and the force will be about 100x as strong.

Last edited: Mar 25, 2016
3. Mar 25, 2016

I have a formula using the magnetic fields and the surface area of the endface (pole) of each of the two magnets that is similar to your original formula, but is in c.g.s. units and the force will be in dynes. The distance r is in centimeters (don't forget to use r^2). F=(B1)(B2)(A1)*(A2)/r^2. The magnetic field strength B is in gauss. (The B*A is essentially the pole strength Q of your original formula. The c.g.s. units are simpler than M.K.S. for this problem.) The areas A1 and A2 must be expressed in cm^2. Just to put in some numbers: Let A1=A2=5 cm^2 r^2=6.25 cm^2 B1=2E+4 B2=2E+2 ==>> F=1.6 E+7 dynes=160 Nt =160*(2.2/9.8) lbs. =36 lbs. if I did my conversions correctly. The formula is not exact, but a good approximation, and it looks like the numbers obtained are quite reasonable.

Last edited: Mar 25, 2016
4. Mar 25, 2016

### NTL01

Charles

A few things if you have a minute or two more:

1. What does the E term represent. It appears to relate to the strength of the magnets, but could you amplify a little.
2. Do you have any sense of what the field strength expressed in gauss would be for the field strength terms of your equation so I can get a sense of how they related to real world magnets.
3. I assume that your point of view is that whether one magnet is an electromagnet and the other a permanent magnet doesn't matter, the formula holds
4. If I assume correct in 3 , then you would agree the electromagnet does not set a force limit. Any field strength more than 1 ( whether its dynes or teslas or gauss) produces a field strength that would at least be as strong as that of the permanent magnet acting on iron at the same distance

A word on air cores versus iron cores in the electromagnets since you asked.

In the application I am designing , the goal is to create a movement in the magnets toward and then away from each other by reversing the direction of current flow in the electromagnet , changing its pole facing the permanent magnet back and forth from north to south. If an iron core was part of the elctromagnet it would be as you say 100 times stronger, but I am thinking the much stronger permanent magnet would "captivate" the iron core and the magnets would not separate upon polarity reversal. I think I will just try this and see , but wondering if you have any thoughts on that

best regards
Richard

5. Mar 25, 2016

Comment on the above formula (post #3). I derived it in a bit of hurry-more accurately it should read $F=(B1)(B2)(A1)(A2)/((2 \pi)^2r^2)$ (This comes from $Q=M*A$ and $M=B_o/2 \pi$ where $B_o$ is the magnetic field at the surface of the endface, Q is the strength of the magnetic pole, and M is the magnetization per unit volume. It all follows from an equation $B=H+4 \pi M$, along with $F=(Q1)(Q2)/r^2$, but that's a lot of extra detail.) You normally don't see the force between two magnetic poles in a standard physics text because it is not an exact equation and it comes with some approximations. For this formula, the magnets are assumed to be two long cylindrical magnets and the computed force is the force between two poles that can be somewhat close to each other, but not right next to each other. The electromagnet with or without air core has a similar magnetic field behavior, so it can be treated as a cylindrical pole magnet. The equation assumes the magnetic field B for each magnet is at the surface of the "pole" or at the mouth of the electromagnet. (If the electromagnet interior magnetic field is given, that value needs to be divided by 2 to get the value at the mouth.) The notation e.g. E+7 means 10^7 or 10,000,000. (and E-7 =10^-7=1/(10^7) etc.) If the magnets are brought in very close proximity, the calculations become more complex and also less precise. Meanwhile, an electromagnet is likely to have a practical upper limit which is approximately the saturation level of its iron core which will typically be in the range of a high quality permanent magnet-somewhere in the neighborhood of 20,000 gauss. The $(2 \pi)^2$ factor in the denominator of the equation will change the result of my sample calculation in post #3 to give force F=1.0 lbs. (approximately) instead of 36 lbs. Suggestion for you that I would be very much interested in seeing the results: Take a couple of long cylindrical magnets whose B values are known and connect them with a spring balance at a distance of an inch or two apart and see what kind of accuracy the above equations give...

Last edited: Mar 25, 2016
6. Mar 27, 2016

### NTL01

Charles
Thank you for the above.Very helpful

I will be setting up test with the two magnets and the spring balance, so I will let you know the results and will email you directly ( assuming I can do that , I have not checked yet). Will probably be about a week until I have the answer.

As you say , the formula is more complex when the magnets are very close together. I am primarily interested in the force when they are about 1/8 inch apart.I dont immediately see why your equation would not work at this close distance. If you have some comments on that I would be grateful for them.

Thanks
Richard

7. Mar 27, 2016

### NTL01

Hi Charles

Trying to work the equation and replicate your numbers. I am getting about 5 pounds of force using your inputs( I think). If you have a moment , could you look this over and see where I have gone astray. I did assume you were expressing the magnet strength in gauss. I also do not understand the 2.2/9.8 raction that reduces the newtons to pounds

B force of PM Gauss 20000
B force of EM Gauss 200
B force PM webers
B force EM webers
A= pier *r>2 5

value of 2 pier squared 39.4384
R>2 ( distance between magnets 6.25

Force between magnets
F=(B1B2)( A1A2)/(( 2*3.14)>2)+r>2 2535606.071 dynes
Force in newtons =dynes/ 100000 25.35606071
forc e in pounds N* (2.2/9.8) 5.692176895
B1b2 in gauss 4,000,000.00
aia2 in centimeters 25.00
b1b2a1a2+ 100,000,000.00

8. Mar 27, 2016