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 = 171 a, i2 = 146 a, i3 = 233 a, and i4 = 194 a. the distance shown is d = 0.0720 m. what is the magnitude of the force per unit length on wire 1? hint: remember that force is a vector. save for later submit answer 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

$F_{21}/L=\frac{\mu_0i_1i_2}{2\pi d}$, substituting $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$, $i_1 = 171\ A$, $i_2 = 146\ A$ and $d = 0.0720\ m$, we get $F_{21}/L=\frac{4\pi\times10^{-7}\times171\times146}{2\pi\times0.0720}=6.99\times10^{-2}\ N/m$. The direction is along the negative x - axis according to the right - hand rule for parallel current - carrying wires (opposite currents repel).

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

$F_{23}/L=\frac{\mu_0i_2i_3}{2\pi\sqrt{2}d}$, substituting $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$, $i_2 = 146\ A$, $i_3 = 233\ A$ and $d = 0.0720\ m$, we get $F_{23}/L=\frac{4\pi\times10^{-7}\times146\times233}{2\pi\times\sqrt{2}\times0.0720}=3.47\times10^{-2}\ N/m$. The x - component of $F_{23}/L$ is $F_{23x}/L=(F_{23}/L)\cos45^{\circ}$ and the y - component is $F_{23y}/L=(F_{23}/L)\sin45^{\circ}$.

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

$F_{24}/L=\frac{\mu_0i_2i_4}{2\pi d}$, substituting $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$, $i_2 = 146\ A$, $i_4 = 194\ A$ and $d = 0.0720\ m$, we get $F_{24}/L=\frac{4\pi\times10^{-7}\times146\times194}{2\pi\times0.0720}=7.79\times10^{-2}\ N/m$. The direction is along the positive x - axis according to the right - hand rule for parallel current - carrying wires (opposite currents repel).

Step5: Sum up the x - components of the forces

$F_x/L=F_{24}/L - F_{21}/L+F_{23x}/L$.
$F_{23x}/L=(F_{23}/L)\cos45^{\circ}=\frac{4\pi\times10^{-7}\times146\times233}{2\pi\times\sqrt{2}\times0.0720}\times\frac{\sqrt{2}}{2}=2.45\times10^{-2}\ N/m$.
$F_x/L = 7.79\times10^{-2}-6.99\times10^{-2}+2.45\times10^{-2}=3.25\times10^{-2}\ N/m$.
$F_y/L = F_{23y}/L=(F_{23}/L)\sin45^{\circ}=2.45\times10^{-2}\ N/m$.

Step6: Calculate the magnitude of the net force per unit length

$F/L=\sqrt{F_x^2 + F_y^2}=\sqrt{(3.25\times10^{-2})^2+(2.45\times10^{-2})^2}=4.07\times10^{-2}\ N/m$.

Answer:

$4.07\times10^{-2}\ N/m$