**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/x^{3}, 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 e_{o}
is called the permittivity of space. Note: With some browsers, the pi symbol
shows up as p.

k= 1/(4pi times e_{o})

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 q_{m},
the fictitious magnetic equivalent to the charge q. Then the dipole moment m of
the magnet is q_{m}L, 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 q_{m} = NiA/L. The magnetic field at the end a coil is B_{o}
= m _{o}Ni/[2(L^{2} + R^{2})^{0.5}], (see this stuff and use cosa = L/(L^{2} + R^{2})^{0.5}
and cosb
= cos90^{o} = 0) so

q_{m}_{ }= 2B_{o}A(L^{2}+R^{2})^{0.5}
/(m _{o}L)
(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 B_{o}
at a pole. (m
and m_{o}
are two completely different things; m_{o} is a
universal constant called the permeability of space, 4p x 10^{-7} tesla∙meter/ampere, that’s 4 times pi x 10^{-7 )}

Standard
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/r^{2}. The corresponding magnetic problem is done by
calculating q_{m} by equation 1, then using
the constant m_{o}/(4p) in place of k and q_{m}
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,

B_{o} = m_{o}Ni/(2R), so a
refrigerator magnet has

q_{m} = 2B_{o}AR/(m_{o}L) = NiA/L = m/L (2)

Now recalling the electric field formula, E = 2kp/x^{3} at large x
along the axis, we write the corresponding magnetic field:

B=2(m _{o}/4p )q_{m}L/x^{3} = B_{o}A(L^{2}+R^{2})^{0.5}/(p x^{3}), 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 = B_{o}A(L^{2}+R^{2})^{0.5}/(p x^{3}) (large x) (3)

What to do for small x? Again with a cylindrical bar magnet, B on a line
through the poles is B_{o}(L^{2}+R^{2})^{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=[B_{o}(L^{2}+R^{2})^{1/2}/L][(x+L)(R^{2}+(x+L)^{2})^{-1/2}-
x(R^{2}+x^{2})^{-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 = B_{o}AL/(2p r^{3}). (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.

**Forces**

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 _{o}AE. Hence F = e
_{o }AE^{2}/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 = AB^{2}/(2m _{o}), (6)

where B is the magnetic field between them in Tesla.
(1 T = 10^{4} Gauss.) You could also derive this using an energy
approach if you know that the energy/volume in a magnetic field is B^{2}/2m _{o}. Think about the (unstated)
assumptions necessary for all this to be valid.

Let’s
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 = kq^{2}[1/x^{2}
+ 1/(x+2L)^{2} – 2/(x+L)^{2}] (7)

For
a pair of identical magnets, replace k with m_{o}/4p and
replace q with q_{m} from equation 1 for the
general case:

x

F = [B_{o}^{2}A^{2}
(L^{2}+R^{2}) /(pm _{o }L^{2})]
[1/x^{2} + 1/(x+2L)^{2} – 2/(x+L)^{2}] (8)

I fed the 2^{nd} 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} = a^{n} + na^{n-1}b
+ n(n-1)a^{n-2}b^{2}/2 +… on the 2^{nd} 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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