Question 5.21: A very long transmission line consists of two parallel wires......
A very long transmission line consists of two parallel wires separated by a distance d[m] in free space. A steady current of I [A] flows in the opposite directions as shown in Fig. 5.36. Find the magnetic force per unit length of the line.
From Ampere’s circuital law, B on wire 1 due to current I in wire 2 is
\pmb{B}_{1-2} = – \frac{\mu _{o} I}{2\pi d}\pmb{a}_{z} (5-139)
Differential length vector on wire 1 is
d \pmb{l}_{1} = dx \pmb{a}_{x}From Eq. (5-137), the magnetic force per unit length of wire 1 is
\pmb{F}_{1} = I \oint_{C_{1}}{d \pmb{l}_{1} \times \pmb{B}_{1-2}} (5-137)
\pmb{F}_{1} = I \oint_{C_{1}}{d \pmb{l}_{1} \times \pmb{B}_{1-2}} = I \int_{x=0}^{x=1}{ dx \pmb{a}_{x} \times \left\lgroup- \frac{\mu _{o} I}{2\pi d}\pmb{a}_{z}\right\rgroup }\\\quad\quad\quad\quad\quad\quad\quad\quad \quad \quad \quad = \frac{\mu _{o} I^{2}}{2\pi d}\pmb{a}_{y}In view of the opposite currents in the wires, the magnetic forces acting on the wires are repulsive. Thus the magnetic force on wire 2 is
\pmb{F}_{2} = – \pmb{F}_{1} =- \frac{\mu _{o} I^{2}}{2\pi d}\pmb{a}_{y}