Question

the figure below shows, in cross - section, four wires that are parallel, straight and very long. the currents in the wires are as follows: i1 = 235 a, i2 = 273 a, i3 = 142 a, and i4 = 227 a. the distance shown is 0.0550 m. what is the magnitude of the force per unit length on wire 2? 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 each of the other wires

The force on wire 2 due to wire 1: $F_{21}/L=\frac{\mu_0i_1i_2}{2\pi d}$, with $i_1 = 235\ A$, $i_2 = 273\ A$, and $d = 0.0550\ m$.
$F_{21}/L=\frac{4\pi\times10^{-7}\times235\times273}{2\pi\times0.0550}=\frac{2\times10^{-7}\times235\times273}{0.0550}\ N/m$.
The force on wire 2 due to wire 3: $F_{23}/L=\frac{\mu_0i_2i_3}{2\pi\sqrt{2}d}$, with $i_3 = 142\ A$. Since the distance between wire 2 and wire 3 is $r=\sqrt{d^{2}+d^{2}}=\sqrt{2}d$.
$F_{23}/L=\frac{4\pi\times10^{-7}\times273\times142}{2\pi\times\sqrt{2}\times0.0550}=\frac{2\times10^{-7}\times273\times142}{\sqrt{2}\times0.0550}\ N/m$.
The force on wire 2 due to wire 4: $F_{24}/L=\frac{\mu_0i_2i_4}{2\pi d}$, with $i_4 = 227\ A$.
$F_{24}/L=\frac{4\pi\times10^{-7}\times273\times227}{2\pi\times0.0550}=\frac{2\times10^{-7}\times273\times227}{0.0550}\ N/m$.

Step3: Resolve the forces and find the net force

The force $F_{21}/L$ and $F_{24}/L$ act along the x - axis, and $F_{23}/L$ has components along the x and y axes.
The x - component of the net force per unit length on wire 2:
$F_{x}/L=F_{21}/L + F_{24}/L+F_{23}/L\cos45^{\circ}$
$F_{21}/L=\frac{2\times10^{-7}\times235\times273}{0.0550}\approx2.35\times10^{-1}\ N/m$
$F_{24}/L=\frac{2\times10^{-7}\times273\times227}{0.0550}\approx2.27\times10^{-1}\ N/m$
$F_{23}/L=\frac{2\times10^{-7}\times273\times142}{\sqrt{2}\times0.0550}\approx0.63\times10^{-1}\ N/m$
$F_{23,x}/L = F_{23}/L\cos45^{\circ}=\frac{F_{23}/L}{\sqrt{2}}\approx0.44\times10^{-1}\ N/m$
$F_{x}/L=(2.35 + 2.27+0.44)\times10^{-1}\ N/m=5.06\times10^{-1}\ N/m$
The y - component of the net force per unit length on wire 2: $F_{y}/L=-F_{23}/L\sin45^{\circ}\approx - 0.44\times10^{-1}\ N/m$
The magnitude of the net force per unit length on wire 2:
$F/L=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(5.06\times10^{-1})^{2}+(- 0.44\times10^{-1})^{2}}\approx5.08\times10^{-1}\ N/m$

Answer:

$5.08\times10^{-1}\ N/m$