Page 202 - Engineering Electromagnetics, 8th Edition

184 ENGINEERING ELECTROMAGNETICS

Figure 7.3 An infinitely long straight filament
carrying a direct current I. The field at point 2 is
H = (I/2πρ)a φ .

We illustrate the application of the Biot-Savart law by considering an infinitely
long straight filament. We apply (2) first and then integrate. This, of course, is the
same as using the integral form (3) in the first place. 2
Referring to Figure 7.3, we should recognize the symmetry of this field. No
variation with z or with φ can exist. Point 2, at which we will determine the field,
is therefore chosen in the z = 0 plane. The field point r is therefore r = ρa ρ . The
source point r is given by r = z a z , and therefore

R 12 = r − r = ρa ρ − z a z
so that

ρa ρ − z a z
ρ + z 2
2
a R12 =
We take dL = dz a z and (2) becomes

Idz a z × (ρa ρ − z a z )

dH 2 =
4π(ρ + z )
2 3/2
2
Because the current is directed toward increasing values of z , the limits are −∞ and

∞ on the integral, and we have

∞ Idz a z × (ρa ρ − z a z )
H 2 = 4π(ρ + z )
2 3/2
2
−∞
I ∞ ρdz a φ

=
2
2 3/2
4π −∞ (ρ + z )
2 The closed path for the current may be considered to include a return filament parallel to the first
filament and infinitely far removed. An outer coaxial conductor of infinite radius is another theoretical
possibility. Practically, the problem is an impossible one, but we should realize that our answer will be
quite accurate near a very long, straight wire having a distant return path for the current.