Fig. 29-63 shows wire 1 in cross section; the wire is long and straight, carries a current of$\mathbf{4}\mathbf{.}\mathbf{00}\mathit{m}\mathit{A}$out of the page, and is at distance ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{40}\mathit{c}\mathit{m}$from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{c}\mathit{m}$from wire 1 and carries a current of$\mathbf{6}\mathbf{.}\mathbf{80}\mathit{m}\mathit{A}$into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?

Wire 1 carries a current$i_1=4.00\text{mA}=4.00\times {10}^{-3}\text{A}$

The distance of wire 1 from the surface is$d_1=2.40\text{cm}=0.0240\text{m}$

Wire 2 is parallel to wire 1

Wire 2 is at horizontal distance$d_2=5.00\text{cm}=0.0500\text{m}$

Wire 2 carries a current$i_2=6.80\text{mA}=6.80\times {10}^{-3}\text{A}$

We can find the $x$component of the magnetic force per unit length on wire2due to wire1.

The force between two parallel currents is

${\text{F}}_{21}=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi r}$

From Equation 29-13, the force between two parallel currents is

${\text{F}}_{21}=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi r}$

Since the distance between the wires is$r=\sqrt{d_1^2+d_2^2}$

Therefore, the x component of force is

$F_x={\text{F}}_{21}\cos \theta$, where$\cos \theta =d/\sqrt{d_1^2+d_2^2}$

Substituting the values, we get

$$\begin{gathered} F_x=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi \sqrt{d_1^2+d_2^2}}\frac{d}{\sqrt{d_1^2+d_2^2}} \\ {F}_x=\frac{{\mu }_{0}L{i}_1{i}_2{d}_2}{2\pi \left(d_1^2+d_2^2\right)} \end{gathered}$$

Therefore, the$x$component of the magnetic force per unit length on wire$2$due to wire$1$is

$$\begin{gathered} \frac{F_x}{L}=\frac{{\mu }_0{i}_1{i}_2{d}_2}{2\pi \left(d_1^2+d_2^2\right)} \\ \frac{F_x}{L}=\frac{\left(4\pi \times {10}^{-7}\right)\times \left(4.00\times {10}^{-3}\right)\times \left(6.80\times {10}^{-3}\right)\left(0.0500\right)}{2\pi \left({0.0240}^2+{0.0500}^2\right)} \\ \frac{F_x}{L}=8.84\times {10}^{-11}\text{N/m} \end{gathered}$$

Hence, $\frac{F_x}{L}=8.84\times {10}^{-11}\text{N/m}$