Magnetic Fields and Forces
This is material you don't need for the standard physics course. Electromagnetism is such a big field that they cannot cover everything. How much force can a magnet exert on a chunk of iron? You probably won’t find that in your physics book, but if you have had electrostatics, you will be able to follow this stuff.
The electric field of an electric dipole (point charges + and -q separated
by distance L) is worked out in standard physics texts and the derivations
won't be repeated here. E = 2kp/x3, for example, is the field on a
line through the charges at a large distance x from the center of a dipole
having dipole moment p = qL. The Coulombs law
constant, k, is also known as
1/(4p e o), where eo is called the permittivity of space. Note: With some browsers, the pi symbol shows up as p.
k= 1/(4pi times eo)
A bar magnet is a magnetic dipole, and its field varies in the space around the magnet in the same way as the electric field varies around an electric dipole. So we can use the results of the E calculations to find the magnetic field B and forces between magnets.
A word of caution: "Pole strength" of a magnet is sort of like electric charge in an electric dipole, but it is not a real charge-like thing. You might think that you could cut off a pole and have something like a single pole. No, you would have two magnets, each with two poles. If you cut a bar magnet into 10 identical pieces, you would have 10 identical bar magnets. Electric charge is real; a magnetic pole is just a convenient crutch. Find a magnetic particle corresponding to a + or - charge (the "magnetic monopole") and you will be rich and famous and a probable Nobel Prize winner. It has not been found, so we think it does not exist.
I will call the pole strength of the bar magnet qm, the fictitious magnetic equivalent to the charge q. Then the dipole moment m of the magnet is qmL, where L is its length. If you have not studied the dipole moment yet, it is related to the torque on a dipole in an external field. For a coil of N turns, each loop of area A, m = NiA, where i is the current in the coil. We can use this to find qm = NiA/L. The magnetic field at the end a coil is Bo = m oNi/[2(L2 + R2)0.5], (see this stuff and use cosa = L/(L2 + R2)0.5 and cosb = cos90o = 0) so
qm = 2BoA(L2+R2)0.5 /(m oL) (1)
This is valid for a cylindrical bar magnet, also, and it should be a good approximation for one with a rectangular cross-section. Measure the field Bo at a pole. (m and mo are two completely different things; mo is a universal constant called the permeability of space, 4p x 10-7 tesla∙meter/ampere, that’s 4 times pi x 10-7 )
physics books have problems on electric charges and the forces on them and the
electric field in there vicinity. Now you can use those methods for the
corresponding magnetic problem using equation 1. For example, finding the electric field at
the center of the square formed by 4 charges (below) is done by adding 4
vectors, each of which is found by kq/r2. The corresponding magnetic problem is done by
calculating qm by equation 1, then using
the constant mo/(4p) in place of k and qm
in place of q, do the same calculation. (Note: kàmu-sub-zero/(4 times pi))
+ - N S
+ - N S
E problem corresponding B problem
For a very short coil,
Bo = moNi/(2R), so a refrigerator magnet has
qm = 2BoAR/(moL) = NiA/L = m/L (2)
Now recalling the electric field formula, E = 2kp/x3 at large x
along the axis, we write the corresponding magnetic field:
B=2(m o/4p )qmL/x3 = BoA(L2+R2)0.5/(p x3), along a line through the poles at a distance x from the center, with x large compared to the length of the magnet. A = cross-sectional area. (Again, it should pi in two places above)
B = BoA(L2+R2)0.5/(p x3) (large x) (3)
What to do for small x? Again with a cylindrical bar magnet, B on a line through the poles is Bo(L2+R2)1/2/L times the difference between two cosines: of the angle between the center axis and a line to the edge of one face of the magnet and the angle between the center axis and the line drawn to the edge of the other face. In other words, if x is the distance to the closer pole,
B=[Bo(L2+R2)1/2/L][(x+L)(R2+(x+L)2)-1/2- x(R2+x2)-1/2]. (4)
It can be shown that this approaches equation 3 for large x. Another approach to these things: try this.
Along a perpendicular bisector of the bar magnet, it is easy to show in a similar way that if you measure r not to the center but the diagonal distance to an end, then
B = BoAL/(2p r3). (5)
This is valid close to the magnet, as well. Of course for large r the diagonal distance is approximately the same as the distance to the center, so this is half of equation 3 at large r.
We model it after the electric force: Consider a capacitor. The force on one plate due to the other is q(E/2), the charge of the one times the field due to the other (half of the total field). But recall that the total field is charge per area divided by e o, so q is equal to e oAE. Hence F = e o AE2/2. In magnetism, the counterpart to e o is 1/m o, so the force between two attracting surfaces which are very close together (two magnets or magnet and a chunk of iron) is given by
F = AB2/(2m o), (6)
where B is the magnetic field between them in Tesla. (1 T = 104 Gauss.) You could also derive this using an energy approach if you know that the energy/volume in a magnetic field is B2/2m o. Think about the (unstated) assumptions necessary for all this to be valid.
look at the force between two identical electric dipoles, collinear with each
other as shown.
+ - x + -
The dipole on the left has a force to the right = two attractions minus two repulsions. If the closest distance is x and the distance between charges on each dipole is L, then
F = kq2[1/x2 + 1/(x+2L)2 – 2/(x+L)2] (7)
For a pair of identical magnets, replace k with mo/4p and replace q with qm from equation 1 for the general case:
F = [Bo2A2 (L2+R2) /(pm o L2)] [1/x2 + 1/(x+2L)2 – 2/(x+L)2] (8)
I fed the 2nd bracket into Excel using L=0.1, and fit it to a
power law, and it was
≈ 0.4x -2.2 for x very small to 0.1, ~x -3.76 from 0.1 to 0.5, and x-3.97 from 0.5 to 5. This is L dependent, so do your own for a particular problem. For very large x, use
(a+b)n = an + nan-1b + n(n-1)an-2b2/2 +… on the 2nd bracket, and you will find that it goes like x -4. As a practical matter, very large x is not very useful for forces, because the force is so small. Am. J. Phys. 74 (6) June 06, p. 510, equation 5, is equivalent to my equation 8 for the case of large x. In that article, the authors discuss methods of finding these forces experimentally.
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