In Fig.29-64, five long parallel wires in an xy plane are separated by distance $\mathbf{}\mathbf{}\mathit{d}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{cm}}$. The currents into the page are ${\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{A}}\mathbf{,}\mathbf{}{\mathit{i}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{}\mathbf{\text{A}}\mathbf{,}\mathbf{}\mathbf{}{\mathit{i}}_{\mathbf{4}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{A}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{}{\mathit{i}}_{\mathbf{5}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{A}}$; the current out of the page is ${\mathit{i}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{A}}$. What is the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires?

• Five long parallel wires are separated by distance$d=50.0\text{cm}$
• The current in wire 1 is$i_1=2.00\text{A}$.
• The current in wire 2 is$i_2=4.00\text{A}$.
• The current in wire 3 is$i_3=0.250\text{A}$
• The current in wire 4 is$i_4=4.00\text{A}$
• The current in wire 5 is$i_5=2.00\text{A}$
• The current into the page are$i_1$, $i_3$,$i_4$, and$i_5$.
• The current out of the page is $i_2$.

Using Eq. 29-13 and considering the given summary, we can find the magnitude of net force per unit length acting on wire 3 due to current in the other wires.

Formula:

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

$F_x=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi d}$

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

$F_x=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi d}$

Since currents into the page are$i_1$, $i_3$,$i_4$, and$i_5$, wire 3 is, therefore, attached to all but wire 2.

Therefore, from Eq. 29-13, themagnitude of net force per unit length acting on wire 3 due to the current in the other wires is

$\frac{\left|\overrightarrow{F}\right|}{L}=\frac{{\mu }_0{i}_3}{2\pi }\left(-\frac{i_1}{2d}+\frac{i_2}{d}+\frac{i_4}{d}+\frac{i_5}{2d}\right)$

Substituting given values, we get

$\begin{array}{rcl}\frac{\left|\overrightarrow{F}\right|}{L}& =& \frac{\left(4\pi \times {10}^{-7}\right)\left(0.250\right)}{2\pi }\left[-\frac{\left(2.00\right)}{2\left(0.500\right)}+\frac{\left(4.00\right)}{\left(0.500\right)}+\frac{\left(4.00\right)}{\left(0.500\right)}+\frac{\left(2.00\right)}{2\left(0.500\right)}\right]\\ \frac{\left|\overrightarrow{F}\right|}{L}& =& \frac{\left(4\pi \times {10}^{-7}\right)\left(0.250\right)}{2\pi }\left(16\right)\\ \frac{\left|\overrightarrow{F}\right|}{L}& =& 8.00\times {10}^{-7}\text{N/m}\end{array}$

Hence, the magnitude of the net force per unit length acting on wire 3 due to current in the other wires is $\frac{\left|\overrightarrow{F}\right|}{L}=8.00\times {10}^{-7}\text{N/m}$