91Ó°ÊÓ

1. The five parallel wires are separated by distance$d=8.00\text{cm}  \text{or}  8.00×{10}^{-2}\text{m}$
2. The length of the wires is$L=10.0\text{m}$
3. The current through the wires is$i=3.00\text{A}$

A force will operate between two parallel wires carrying a current because of the interaction between the magnetic fields of the two wires. Each wire is being pulled in the same direction but with an equal amount of force.

By using Eq. 29-13 and the given summary, we can find the net magnetic force on each wire.

Formula:

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

${\stackrel{→}{F}}_x=\frac{{\mathrm{μ}}_0{i}_1{i}_{2}L}{2\mathrm{π}d}$

Where, ${\mathrm{μ}}_0$is the magnetic permeability of free space $(4\mathrm{π}×{10}^{-7}{\text{NA}}^{\text{- 2}})$

$i_1$ is the current in wire 1

$i_2$ is the current in wire 2

$d$ is the distance separating the conductors

$L$ is the length of conductor

Let us label these wires 1 through 5 from left to right.

Since the current through each wire is the same, and each wire experiences a magnetic force due to the currents in the other wires.

By using Eq. 29-13, the magnetic force on wire 1 is

${\stackrel{→}{F}}_1=\frac{{\mathrm{μ}}_0{i}^{2}L}{2\mathrm{π}}\left(\frac{1}{d}+\frac{1}{2d}+\frac{1}{3d}+\frac{1}{4d}\right)\hat{j}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{F_1}& =& \frac{\left(4\mathrm{Ï€}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right){\left(3 \text{A}\right)}^2\left(10 \text{m}\right)}{2\mathrm{Ï€}}\left(\frac{1}{\left(8×{10}^{-2} \text{m}\right)}+\frac{1}{2\left(8×{10}^{-2} \text{m}\right)}+\frac{1}{3\left(8×{10}^{-2} \text{m}\right)}+\frac{1}{4\left(8×{10}^{-2} \text{m}\right)}\right)\hat{\text{j}}\\ & =& \frac{\left(4\mathrm{Ï€}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right){\left(3 \text{A}\right)}^2\left(10\right)}{2\mathrm{Ï€}}\left(12.5+6.25+4.166+3.125\right)\hat{\text{j}}\\ & =& \left(4.69×{10}^{-4}N\right)\hat{\text{j}}\end{array}$

Similarly, for wire 2,the force between parallel currents is

${\stackrel{→}{F}}_2=\frac{{\mathrm{μ}}_0{i}^{2}L}{2\mathrm{π}}\left(\frac{1}{2d}+\frac{1}{3d}\right)\hat{j}$

$\begin{array}{rcl}{\stackrel{→}{F}}_2& =& \frac{\left(4\mathrm{π}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right){\left(3 \text{A}\right)}^2\left(10 \text{m}\right)}{2\mathrm{π}}\left(\frac{1}{2\left(8×{10}^{-2}\text{m}\right)}+\frac{1}{3\left(8×{10}^{-2}\text{m}\right)}\right)\hat{\text{j}}\\ & =& \frac{\left(4\mathrm{π}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right){\left(3 \text{A}\right)}^2\left(10\right)}{2\mathrm{π}}\left(6.25+4.166\right)\hat{\text{j}}\\ & =& \left(1.88×{10}^{-4}N\right)\hat{\text{j}}\end{array}$

From Fig. 29-64 and from the summary for wire 3, we have

${\stackrel{→}{F}}_3=0$

For wire 4 we have

${\stackrel{→}{F}}_4=-{\stackrel{→}{F}}_2$

${\stackrel{→}{F}}_4=\left(-1.88×{10}^{-4}\text{N}\right)\hat{\text{j}}$

For wire 5, we have

${\stackrel{→}{F}}_5=-{\stackrel{→}{F}}_1$

${\stackrel{→}{F}}_5=\left(-4.69×{10}^{-4}\text{N}\right)\hat{\text{j}}$