Question

the figure below shows a cross - sectional view of 5 wires separated by a distance 0.0291 m. the currents in the wires are into the page for wires 1, 3, and 5 and out of the page for wires 2 and 4. the amount of current through each of the wires is as follows: i1 = 204 a, i2 = 235 a, i3 = 162 a, i4 = 250 a, and i5 = 129 a. what is the magnitude of the force per unit length on wire 2 due to the other wires? n/m

Explanation:

Step1: Recall force - per - unit - length formula

The force per unit length between two parallel current - carrying wires is given by $F/L=\frac{\mu_0i_1i_2}{2\pi r}$, where $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$, $i_1$ and $i_2$ are the currents in the two wires, and $r$ is the distance between them.

Step2: Calculate the force on wire 2 due to wire 1

$F_{21}/L=\frac{\mu_0i_1i_2}{2\pi d}$, with $i_1 = 204\ A$, $i_2 = 235\ A$, and $d = 0.0291\ m$.
$F_{21}/L=\frac{4\pi\times10^{-7}\times204\times235}{2\pi\times0.0291}=\frac{2\times10^{-7}\times204\times235}{0.0291}\ N/m$.

Step3: Calculate the force on wire 2 due to wire 3

$F_{23}/L=\frac{\mu_0i_2i_3}{2\pi d}$, with $i_3 = 162\ A$, $i_2 = 235\ A$, and $d = 0.0291\ m$.
$F_{23}/L=\frac{4\pi\times10^{-7}\times235\times162}{2\pi\times0.0291}=\frac{2\times10^{-7}\times235\times162}{0.0291}\ N/m$.

Step4: Calculate the force on wire 2 due to wire 4

$F_{24}/L=\frac{\mu_0i_2i_4}{2\pi(2d)}$, with $i_4 = 250\ A$, $i_2 = 235\ A$, and $2d=0.0582\ m$.
$F_{24}/L=\frac{4\pi\times10^{-7}\times235\times250}{2\pi\times0.0582}=\frac{2\times10^{-7}\times235\times250}{0.0582}\ N/m$.

Step5: Calculate the force on wire 2 due to wire 5

$F_{25}/L=\frac{\mu_0i_2i_5}{2\pi(3d)}$, with $i_5 = 129\ A$, $i_2 = 235\ A$, and $3d = 0.0873\ m$.
$F_{25}/L=\frac{4\pi\times10^{-7}\times235\times129}{2\pi\times0.0873}=\frac{2\times10^{-7}\times235\times129}{0.0873}\ N/m$.

Step6: Sum up the forces

The net force per - unit length on wire 2 is $F/L=F_{21}/L - F_{23}/L+F_{24}/L - F_{25}/L$.
$F_{21}/L=\frac{2\times10^{-7}\times204\times235}{0.0291}=\frac{2\times10^{-7}\times47940}{0.0291}\approx3.30\times10^{-1}\ N/m$.
$F_{23}/L=\frac{2\times10^{-7}\times235\times162}{0.0291}=\frac{2\times10^{-7}\times38070}{0.0291}\approx2.62\times10^{-1}\ N/m$.
$F_{24}/L=\frac{2\times10^{-7}\times235\times250}{0.0582}=\frac{2\times10^{-7}\times58750}{0.0582}\approx2.02\times10^{-1}\ N/m$.
$F_{25}/L=\frac{2\times10^{-7}\times235\times129}{0.0873}=\frac{2\times10^{-7}\times30315}{0.0873}\approx0.69\times10^{-1}\ N/m$.
$F/L=(3.30 - 2.62+2.02 - 0.69)\times10^{-1}\ N/m=2.01\times10^{-1}\ N/m$.

Answer:

$0.201\ N/m$