Page 181 - Electromagnetics

Thus

ˆ
2
φ µ 0 Iρ/2πa ,ρ ≤ a,
B(r) = (3.169)
ˆ
φ µ 0 I/2πρ, ρ ≥ a.
The force density within the wire,
2
µ 0 I ρ
dF = J × B =−ˆρ ,
2 4
2π a
is directed inward and tends to compress the wire. Integration over the wire volume gives
F = 0 because
2π
ˆ ρ dφ = 0;
0
however, a section of the wire may experience a net force. For instance, we can compute
the force on one half of the wire split down its axis by using ˆρ = ˆ x cos φ + ˆ y sin φ to
obtain F x = 0 and
µ 0 I 2 a 2 π µ 0 I 2
F y =− 2 4 dz ρ dρ sin φ dφ =− 2 dz.
2π a 0 0 3π a
The force per unit length
F µ 0 I 2
=−ˆ y (3.170)
2
l 3π a
is directed toward the other half as expected.
If the wire takes the form of a loop carrying current I, then (3.167)becomes

dF(r) = Idl(r) × B(r) (3.171)
and the total force acting is

F = I dl(r) × B(r).

We can write the force on J in terms of the current producing B. Assuming this latter
current J occupies region V , the Biot–Savart law (3.164)yields

µ r − r

F = J(r) × J(r ) × dV dV. (3.172)
3
4π V V |r − r |
This can be specialized to describe the force between line currents. Assume current 1,
following a path 1 along the direction dl, carries current I 1 , while current 2, following
path 2 along the direction dl , carries current I 2 . Then the force on current 1 is

µ r − r

F 1 = I 1 I 2 dl × dl × .
3
4π |r − r |
1 2
This equation, known as Ampere’s force law, can be written in a better form for compu-
tational purposes. We use (B.7)and ∇(1/R) from (3.57):
µ 1 µ r − r

F 1 = I 1 I 2 dl dl ·∇ − I 1 I 2 (dl · dl ) 3 . (3.173)
4π |r − r | 4π |r − r |

2 1 1 2
The first term involves an integral of a perfect differential about a closed path, producing
a null result. Thus
µ r − r

F 1 =−I 1 I 2 (dl · dl ) 3 . (3.174)
4π |r − r |
1 2
© 2001 by CRC Press LLC