Figure 29-82 shows, in cross section, two long parallel wires spaced by distance d=10.0cm; each carries 100A, out of the page in wire 1. Point Pis on a perpendicular bisector of the line connecting the wires. In unit-vector notation, what is the net magnetic field at Pif the current in wire 2 is (a) out of the page and (b) into the page?

1. Distance between the parallel lines, d=10.0cm
2. Current carried by the each wire,i=100A
3. Current through wire 1 is out of the page.

The net magnetic field at any point due to the current flowing along the current wires contributes to the net field. As the current flows through the straight wires, the current direction will give the magnetic field direction as per Fleming's left-hand rule. The net vertical fields cancel out when the currents are in the same direction, while, the horizontal fields cancel out when the currents are in the opposite direction.

Formula:

The magnetic field for a current carrying wire,

$B=\frac{{\mu }_{0}i}{2\pi r}\cos \theta$ …(i)

The distance of each wire from point P is given as follows:

$$\begin{gathered} r=\frac{d}{\sqrt{2}} \\ =\frac{0.10 \text{m}}{\sqrt{2}} \\ =0.071 \text{m} \end{gathered}$$

With the currents being parallel, application of the right-hand rule reveals that the vertical components cancel and the horizontal components add to yield the net magnetic field at point P by both the wires that are given using equation (i) as follows:

(As the point P is at an angle $\theta ={45}^0$from each wire)

$$\begin{gathered} B=\frac{2\left(4\pi \times {10}^{-7} {\text{N/A}}^2\right)\left(200 \text{A}\right)}{2\pi \left(0.071 \text{m}\right)}\cos \left({45}^0\right)\left(\because {\mu }_0=4\pi \times {10}^{-7} {\text{N/A}}^2\right) \\ =3.98\times {10}^{-4} \text{T} \\ \approx \text{4}\times {10}^{-4} \text{T} \end{gathered}$$

The direction of the magnetic field is in -x direction.

Hence, the net magnetic field is $(\text{4}\times {10}^{-4}\text{\hspace{0.17em}T})\stackrel{⌢}{i}$.

Now, with the currents anti-parallel, application of the right-hand rule shows that the horizontal components cancel and the vertical components add. Thus, the net magnetic field at point P by both the wires that are given using equation (i) as follows:

$$\begin{gathered} B=\frac{2\left(4\pi \times {10}^{-7} {\text{N/A}}^2\right)\left(200 \text{A}\right)}{2\pi \left(0.071 \text{m}\right)}\cos \left({45}^0\right)\left(\because {\mu }_0=4\pi \times {10}^{-7}{\text{N/A}}^2\right) \\ =3.98\times {10}^{-4} \text{T} \\ \approx \text{4}\times {10}^{-4} \text{T} \end{gathered}$$

The direction of the magnetic field is in +y direction.

Hence, the net magnetic field is $(\text{4}\times {10}^{-4}\text{\hspace{0.17em}T})\stackrel{⌢}{j}$.