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In this chapter, we evaluate André Ampère's magnetic-force equation, which possesses cylindrical symmetry and contains the permeability constant µ0, which, like ε0, does not exist in nature either. Notice the similarity of the development of this chapter with respect to the previous one.
Ampère's magnetic-force equation defines the value of the SI unit of electrical current, the ampere, A, which is a coulomb per second, C·s-1, as follows:
Fem s-1 = µ0 i2 (2 π d)-1 (43
In this equation, we let the electrical current, i, be the number of coulombs of electrons in one second that passes by a point on either of two, parallel, conductive wires of indefinite length that are spaced one meter, d, apart such that a magnetic force, Fem, of (2 × 10-7 N) occurs between interfacing, one-meter, s, lengths of the two wires. This current is defined to be one ampere, A.
Because the magnitudes of the lengths, d and s, are the same, we can remove them from the equation that defines the ampere, as follows:
Fem = µ0 i2 (2 π)-1 (44
Inserting the values of the factors into the equation gives:
(2 × 10-7 N) = µ0 (1 A)2 [2 π]-1 (45
Solving for the SI value for the permeability constant, µ0, gives:
µ0 = (2 × 10-7 N) (2 π) (1 A)-2 =
(4 π × 10-7 kg·m·t-2) (1 C·t-1)-2 =
4 π × 10-7 kg·m·C-2 (46
The permeability constant in SE units of measure is derived, as follows:
µ0 = 4 π × 10-7 kg·m·C-2 (47
µ0 = (4 π × 10-7)(1.7 × 10-10 me)(4.1 × 1011 λe) ×
(6.2 × 1018 qe)-2 (48
µ0 = (68.5)-1 me·λe·qe-2 = 2 β-1 me·λe·qe-2 (49
Rearranging the factors in preparation for removing them from the permeability constant (making it disappear by converting its value to dimensionless unity):
µ0 = 2 (β-1 me·λe·te-2) (qe-2·te2) (50
To see these numbers in context, let us put them in place of the historical value of the permeability constant, µ0, in the SE version of Ampère's force equation as follows:
Fee = [2 (β-1 me·λe·te-2) (qe-2·te2)] [i2 (2 π)-1] (51
where the first set of square brackets encloses the SE value of the permeability constant, µ0, and the second set, the rest of the right side of the equation. Of course, the variable, i, uses SE units of measure. Separating this variable's SE value from its SE units of measure gives:
Fee = [2 (β-1 me·λe·te-2) (qe-2·te2)] ×
[|i|e2 (2 π)-1 (qe2·te-2)] (52
In Equation 52, the SE-based permeability constant, µ0, consists of three groups of factors: (2), (β-1 me·λe·te-2), and (qe-2·λe2). We eliminate each group in turn in the same manner as we did with the permittivity constant, ε0 , in the previous chapter.
The number, 2, belongs with the i2 factor. In sum, the current in one conductor contributes i2 to the force, and the current in the second conductor does the same for a total of (2 i2). Placing the number, 2, with the two currents gives:
Fee = [(β-1 me·λe·te-2) (qe-2·te2)] ×
[2 |i|e2 (2 π)-1 (qe2·te-2)] (53
Each of the two remaining groups of factors in the permeability constant, µ0, represents the SE unit of magnetic force, but is expressed in terms of the action that created the force reaction and not the unit of force, itself, as it should be. The SE unit of inertial force, fie, replaces (me·λe·te-2), and the SE unit of magnetic force, fee(t), replaces (qe2·te-2) to give:
Fee = [(β-1 fie) (fee(t)-1)] [2 |i|e2 (2 π)-1 (fee(t))] (54
which shows more-easily that [(β-1 fie) (fee(t)-1)] exists to convert the unit of force that is used in the equation from magnetic to inertial; therefore:
(β-1 fie) = (fee(t)) (55
and
[(β-1 fie) (fee(t)-1)] = 1 (56
and the permeability constant, µ0, no longer exists as such.
Ampère's SE magnetic-force equation is now:
Fee = 2 |i|e2 (2 π)-1 fee(t) (57
or, by converting the reaction, fee(t), into the action that caused it, qe2·te-2, and reinserting these units back into |i|e2, the final form becomes,
Fee = 2 i2 (2 π)-1 (58
We compare the SI- and SE-based magnetic-force equations using the SI newton, N, as the comparison unit of force, where i equals qe·te-1. The four steps of this procedure are as follows:
1) Determine the force Fem (in newtons), using the historic form of Ampère's force equation:
Fem = µ0 (qe2·te-2)(2 π)-1 = (1.3 × 10-6 kg·C-2·m) ×
[(1.6 × 10-19 C)2 (8.1 × 10-21 s)-2] [2 π]-1 =
7.8 × 10-5 N (59
2) Determine the force Fee, (in fee(t)), using the new SE-based form of Ampère's force equation (without a constant of proportionality):
Fee = 2 (1 qe2) (1 te2)-1 (2 π)-1 = 0.32 qe2·te-2 (60
or, converting from the action to reaction form,
Fee = 0.32 fee(t) (61
3) Convert the SE value of fee(t) to the SI value:
fee(t) = (2 π) 2-1 Fem = [π (7.8 × 10-5 N)] =
2.5 × 10-4 N (62
Notice that the magnitude of fee(t) is the same as that of fee(λ), which we calculated in the previous chapter, although, supposedly, two different phenomena created these forces.
4) Convert the SE value of Fee to the SI value:
Fem = fee(t) Fee = [(2.5 × 10-4 N) (fee(t)-1)] (0.32 fee(t)) =
7.8 × 10-5 N (63
and Equations 59 and 63 equal each other.
Thanks to the work of James Maxwell, electronic and magnetic forces are shown to be paired aspects of the same phenomenon--electromagnetic force.
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