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 = 180 a, i2 = 190 a, i3 = 149 a, and i4 = 222 a. the distance shown is 0.0580 m. what is the magnitude of the force per unit length on wire 3? hint: remember that force is a vector.

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 3 due to wire 1

$F_{31}/L=\frac{\mu_0i_3i_1}{2\pi(2d)}$, with $i_1 = 180\ A$, $i_3=149\ A$ and $d = 0.0580\ m$.

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

$F_{32}/L=\frac{\mu_0i_3i_2}{2\pi d}$, with $i_2 = 190\ A$, $i_3 = 149\ A$ and $d = 0.0580\ m$.

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Step4: Calculate the force on wire 3 due to wire 4

$F_{34}/L=\frac{\mu_0i_3i_4}{2\pi(2d)}$, with $i_4 = 222\ A$, $i_3 = 149\ A$ and $d = 0.0580\ m$.

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Step5: Resolve the forces and find the net force per unit length

Wire 1 and wire 4 forces on wire 3 are along the y - direction and wire 2 force on wire 3 is along the x - direction.
The net force per unit length in the x - direction $F_x/L=F_{32}/L$ and in the y - direction $F_y/L=F_{31}/L - F_{34}/L$.

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The magnitude of the net force per unit length $F/L=\sqrt{(F_x/L)^2+(F_y/L)^2}$

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Answer:

$9.76\times 10^{-2}\ N/m$