Magnetic repulsion of solenoids

[Matt01]

I'm trying to understand the repulsive forces of solenoids. This is all paper as in I don't actually have them to test in real life. Take two solenoids that are equal to each other.

Specifications:
current is 10 amps
Number of turns is 300
Iron rod is 12 inches/ 304.8mm, with a Radius of 2 inches/50.8mm
Permeability of Iron (99.8% pure) µ (H/m) = 6.3×10−3
Relative permeability, μ/μ0 = 5000 (Source is wiki. I can give a link to the chart)

If I make it so that both the north polls are facing each other what will the repulsive forces be as I move them closer? For example- say the initial distance from one another is 5 inches/12.7 cm, then 2.5inches/6.35cm. I assume the force would increase as would a normal magnet.I'm unsure as to what formulas I need to apply, and depending on the complexity of said formula, how to reach an answer. Any help or advice would be very appreciated.

Thanks in advance,

Matt

 

[Charles Link]

To get you started on this calculation, the currents in the solenoids have the same geometry as the magnetic surface currents from a magnet of the same geometry with magnetization ## M ##, so that the forces should be the same as the corresponding magnet problem. (Substitute current per unit length ## n \, I=M/\mu_o ## where ## M ## is the magnetization,so that ## M=\mu_o n \, I ##. (## n ## is the number of turns per unit length) ). For the problem of the repulsive force between two magnets, you can consider one to be in the magnetic field of the other. ## \\ ## The energy ## U=-\int M \cdot H \, d^3 x ##. (Edit note: 9-5-17=made a correction to this last equation, changing the ## B ## to an ## H ## ). The force will be ## F=-\nabla U ## where ## U ## can be written as a function of the distance ## x ## between the magnets, because ## H ## is a function of this distance ## x ##. ## \\ ## The calculation for ## H ## can be simplified by the pole model=you don't need to compute it from the surface currents. The result is that ## H ## is just the field from the two poles of the magnet that has magnetization whose surface currents equal those of the solenoid. ## \\ ## To compute the magnetic poles, ## \sigma_m=M \cdot \hat{n}=M ##, and ## H(x)=\frac{1}{4 \pi \mu_o} \int \frac{\sigma_m(x') (x-x')}{|x-x'|^3} \, dA ## , (basically the inverse square law analogous to Coulombs law), and ## B(x)=\mu_o H(x) ##. If you can assume the poles are like points at their centers obeying the inverse square law, the computations simplify. i.e. ## Q_m=\sigma_m A ## where ##A ## is the area of the endface. Then ## H=\frac{Q_m}{4 \pi \mu_o s^2} ## where ## s ## is the distance from the pole, and again ## B=\mu_o H ##. ## \\ ## Finally, I think it is possible to show that the force ## F =-\nabla U ## is just that of the result of the sum of all the forces that would occur between the corresponding poles of both solenoids, using a magnetic force equation for the magnetic poles that is analogous to Coulombs law for electric charges: ## F=\frac{Q_1 Q_2}{4 \pi \mu_o r^2} ##. (I worked through the integral expression for ## U ## a couple of years ago for a simpler case of two very long solenoids spaced a distance ## x ## and that is the result that emerged.) ## \\ ## Editing: I did this for the case of two solenoids without any iron core in them. The calculations will be more complex in the case of an iron core. (I didn't see the iron core upon first reading it=I read it too quickly) ## \\ ## Additional note: To a good approximation, you should be able to assume the magnetization is not affected by the magnetic field from the other solenoid. In that case, the calculation proceeds very much like the above. ## \\ ## Additional note: A quick calculation with the numbers given, including the permeability show that ## M=60 ## T if I computed it correctly, so that the iron core would be saturated and ## M ## would simply be at its upper limit of about ## M=1.0 ## T . Meanwhile I compute ## A=\pi R^2=8 \, E-3 ## m^2 (approximately), so that ## Q_m ## =8 E-3. For ## x ## =4 inches=.1 m, I get force ## F=400 ## Newtons ## = ##100 lbs. (approximately) if my calculations are correct. ## \\ ## Inside of 4", the force will not continue to go up as an inverse square, and will probably not get too much greater than what it is at 4" , since the diameter of your endfaces is 4". In fact, the inverse square law will even start to level off somewhat at 4",(the force won't go to ## +\infty ## at ## x=0 ##, but will probably be just somewhat greater (perhaps by a factor of 2) than the value at ## x=4" ## ).

 

[Matt01]

Thank you Charles Link for your reply. That helped me out a lot.