Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58(b), using the rules of vector addition to sum the contributions from each wire.

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

Current through wire A is,${\mathrm{I}}_{\mathrm{A}}=10.0\mathrm{A}$

Current through the wire B is,${\mathrm{I}}_B=10.0\mathrm{A}$

Current through the wire C is, ${\mathrm{I}}_C=5.0\mathrm{A}$

Current through the wire D is,${\mathrm{I}}_D=5.0\mathrm{A}$

The magnetic field created by a long straight wire can be calculated using the following formula,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}.........\left(1\right)$

Where ${\mathrm{\mu }}_0$ is the magnetic permeability of free space.

Consider the given diagram

Point O, the middle of the square created by four wires, will be the equidistant point.

The distance of point O from each wire is,

$$\begin{gathered} \mathrm{r}=\frac{1}{2}\sqrt{0.{20}^2+0.{20}^2} \\ \mathrm{r}=14.14\times {10}^{-2}\mathrm{m} \end{gathered}$$

Substituting the values from the figure for wire A in equation (1), we get

$$\begin{gathered} {\mathrm{B}}_{\mathrm{A}}=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·m/A\right)\left(10.0\mathrm{A}\right)}{2\mathrm{\pi }(14.14\times {10}^{-2}\mathrm{m})} \\ =1.414\times {10}^{-5}\mathrm{T} \end{gathered}$$

Substituting the values from the figure for wire B in equation (1), we get

$$\begin{gathered} {\mathrm{B}}_B=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·m/A\right)\left(10.0\mathrm{A}\right)}{2\mathrm{\pi }(14.14\times {10}^{-2}\mathrm{m})} \\ =1.414\times {10}^{-5}\mathrm{T} \end{gathered}$$

Substituting the values from the figure for wire C in equation (1), we get

$$\begin{gathered} {\mathrm{B}}_C=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·m/A\right)\left(5.0\mathrm{A}\right)}{2\mathrm{\pi }(14.14\times {10}^{-2}\mathrm{m})} \\ =0.707\times {10}^{-5}\mathrm{T} \end{gathered}$$

Substituting the values from the figure for wire D in equation (1), we get

$$\begin{gathered} {\mathrm{B}}_D=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·m/A\right)\left(5.0\mathrm{A}\right)}{2\mathrm{\pi }(14.14\times {10}^{-2}\mathrm{m})} \\ =0.707\times {10}^{-5}\mathrm{T} \end{gathered}$$

The magnetic field vector diagram from the right-hand thumb rule is shown at point 0

The component of the net magnetic field will be zero, as shown in the vector diagram. The net magnetic field is made up of only the x components of all four magnetic fields. As a result, the net magnetic field has positive x-direction.

The magnitude of the net magnetic field can be calculated using the following formula:

$$\begin{gathered} {\mathrm{B}}_{\mathrm{net}}=\left({\mathrm{B}}_{\mathrm{A}}+{\mathrm{B}}_{\mathrm{B}}+{\mathrm{B}}_{\mathrm{C}}+{\mathrm{B}}_{\mathrm{D}}\right)\cos {45}^{\circ } \\ =\left(1.414+1.414+0.707+0.707\right)\times {10}^{-5}\times \cos {45}^{\circ } \\ =3.0\times {10}^{-5}\mathrm{T} \end{gathered}$$

Therefore, the magnetic field at the point equidistant from the wires is $3.0\times {10}^{-5}\mathrm{T}$in the positive x-direction.