91Ó°ÊÓ

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

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$i_2=6.80\text{mA}=6.80×{10}^{-3}\text{A}$

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

The force between two parallel currents is

${\text{F}}_{21}=\frac{{\mathrm{μ}}_{0}L{i}_1{i}_2}{2\mathrm{π}r}$

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

${\text{F}}_{21}=\frac{{\mathrm{μ}}_{0}L{i}_1{i}_2}{2\mathrm{π}r}$

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

Therefore, the x component of force is

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

Substituting the values, we get

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

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

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

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