91Ó°ÊÓ

For two long, parallel transmission lines in which the currents are in the same direction the magnetic field is given as${\mathit{B}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{1}}}{\mathbf{2}\mathbf{π}\mathbf{(}\mathbf{d}\mathbf{−}\mathbf{x}\mathbf{)}}\mathbf{\text{and}}{\mathit{B}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{2}}}{\mathbf{2}\mathbf{π}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$and

For two long, parallel transmission lines in which the currents are in the opposite direction the magnetic field is given as ${\mathit{B}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{1}}}{\mathbf{2}\mathbf{π}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{\text{and}}{\mathit{B}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{I}}_{\mathbf{2}}}{\mathbf{2}\mathbf{π}\mathbf{(}\mathbf{d}\mathbf{+}\mathbf{x}\mathbf{)}}$were, I is the current passing through the wire, ${\mathit{μ}}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{π}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is permeability of free space.

According to the right-hand rule, the only place where the magnetic fields of the two wires are in

opposite directions are between the wires. Assume that point where the magnetic field is zero P. Let x be the distance from the lower wire to this point, so the distance between the upper wire and this point is d — x. When the magnetic field is zero$B_1=B_2$ . Substitute the values we get,

$\begin{array}{r}\frac{{\mathrm{μ}}_0{l}_1}{2\mathrm{π}(d−x)}=\frac{{\mathrm{μ}}_0{l}_2}{2\mathrm{π}x}\\ {\mathrm{I}}_2(d−x)=I_{1}x\\ x=\frac{I_{2}d}{l_1+I_2}\\ =\frac{75.0×0.400}{25.0+75.0}=0.300\mathrm{m}\end{array}$

Therefore, in the same direction it is 0.300m

According to the right-hand rule, the only place where the magnetic fields of the two wires are in opposite directions is between the wires. As shown in the following figure, the magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires, but only at points above both wires the magnetic field is zero, since 12 > Il. Let x be the distance between the upper wire and the point P When the magnetic field is zero$B_1=B_2$ . Substitute the values we get,

$\begin{array}{r}\frac{{\mathrm{μ}}_0{I}_1}{2\mathrm{π}\left(x\right)}=\frac{{\mathrm{μ}}_0{I}_2}{2\mathrm{π}(d+x)}\\ (d+x)I_1=x{I}_2\\ x=\frac{I_{1}d}{I_2−I_1}\\ =\frac{25.0×0.400}{75.0−25.0}\\ =0.200\mathrm{m}\end{array}$

Therefore, in the opposite direction it is 0.200m