Calculate the interaction force between two portions of straight filamentary conductors, parallel and of the same length l, carrying the currents i_{1} si i_{2}. The conductors are placed in air, at the distance a between each other.
Solution
The magnetic (Laplace) force exerted on a straight conductor has the expression:
F = I (l x B). 
According the the results of problem 1, the magnetic flux density produced by a straight segment of the first conductor has (with the notations of the present problem) the expression:
where x is the current coordinate along the second conductor.
Since the magnetic flux density B_{1} is normal on the current's direction, the force exerted on the second conductor can be obtained by integration, with the formula:
Indicate in which conditions for the length l the result thus obtained has physical meaning.
> restart:
> B[1]:= mu[0]/4*i[1]/Pi/a*(x/sqrt(x^2+a^2)(xl)/sqrt((xl)^2+a^2));
B_{1} : =
1
4
m_{0} i_{1} (
x
Ö{x^{2} + a^{2}}

x  l
Ö{x^{2}  2 x l + l^{2} + a^{2}}
)
p a
> F[21]:=i[2]*Int(B[1],x=0..l);
F_{21} : = i_{2}
ó
õ
l
0
1
4
m_{0} i_{1} (
x
Ö{x^{2} + a^{2}}

x  l
Ö{x^{2}  2 x l + l^{2} + a^{2}}
)
p a
dx
> f:=value(F[21])/l; # Perunitlength force
f : =
1
2
i_{2} m_{0} i_{1} (
Ö
l^{2} + a^{2}

Ö
a^{2}
)
p a l
> fA:=limit(f,l=infinity); # Perunitlength force between infinitely long conductors (Ampere's force)
fA : =
1
2
i_{2} m_{0} i_{1}
p a
Numerical application (the SI definition of the Ampere)
> i[1]:=1; i[2]:=1; a:=1: mu[0]:=4*Pi*10^(7):
i_{1} : = 1
i_{2} : = 1
> evalf(fA,1);
.2 10^{6}
>
When the conductors are carrying currents of 1 A and are placed at a distance of 1m, the perunitlength force has therefore the value
2 10^{(  7)} N/m = 0.2 mN/m.