Question 5.18: Two very long conducting wires of an equal radius a are para......
Two very long conducting wires of an equal radius a are parallel to each other in free space, with the axis-to-axis distance of b(b>>a). They carry uniform currents flowing in the opposite directions as shown in Fig. 5.33. Compute the external and internal inductances per unit length of the pair of wires.
In the region a < y < (b – a) in the yz–plane, B due to the left conductor is obtained by Ampere’s circuital law as
\pmb{B}_{1} = -\frac{\mu _{o}I}{2 \pi y }\pmb{a}_{x }In the same region, B due to the right conductor is
\pmb{B}_{2} = -\frac{\mu _{o}I}{2 \pi(b – y) }\pmb{a}_{x }Under the condition b >> a, the magnetic flux linkage with the pair of wires, per unit length, is given by the surface integral of \pmb{B}_{1} + \pmb{B}_{2} over the surface S shown in Fig. 5.33, that is,
\Lambda _{ex} = \int_{S}{ (\pmb{B}_{1} + \pmb{B}_{2}) \pmb{\cdot } d \pmb{s}} = \frac{\mu _{o}I}{2 \pi }\int_{y=a}^{y = b-a}{\left[\frac{1}{y} + \frac{1}{b – y} \right] dy } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{\mu _{o}I}{ \pi }\ln \left\lgroup\frac{b – a}{a}\right\rgroup \cong \frac{\mu _{o}I}{ \pi }\ln \frac{b}{a}The external inductance per unit length of the wires is therefore
L_{ex} =\frac{ \Lambda _{ex}}{I} = \frac{\mu _{o}}{ \pi }\ln \frac{b}{a} (5-111)
In the interior of the left conductor, for instance, there exist two B’s originating from two different sources: one is due to the internal current I and the other is due to the current flowing in the other conductor. Under the condition b >> a, B due to I flowing in the other conductor is ignored compared with that due to the internal current. Thus the internal inductance of a wire, per unit length, is equal to \mu _{o}/8\pi as given by Eq. (5-110). The internal inductance of the pair of wires is therefore
L = \frac{\Lambda_{1} +\Lambda_{2}}{I} =\frac{\mu _{o}}{2\pi}\ln \frac{b}{a} + \frac{\mu _{o}}{8\pi }\\ \quad \quad \quad \quad \quad \equiv L_{ex}+L_{in} [H/m] (5-110)
L_{in} = \frac{\mu _{o}}{4 \pi } [H/m] (5-112)
The total inductance per unit length of a pair of wires is therefore
L = L_{ex} + L_{in} = \frac{\mu _{o}}{\pi }\left[\ln \frac{b}{a} + \frac{1}{4} \right] [H/m] (5-113)