91Ó°ÊÓ

1. Wire 1 carries a current of $i_1=0.750\text{A}$.
2. Wire 3 carries a current of$i_3=0.250\text{A}$
3. The x component per unit length of wire 2 is$\frac{F_{2x}}{L_2}=-0.627\text{μ±·/³¾}$as$x→∞$.
4. The horizontal axes is set by$x_s=12.0\text{cm}$

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 finding the value of distance $\mathit{d}$between wire 1 and 2 and substituting in Eq. 29-13, we can find the current through wire 2. Using information from the given graph, we can find the direction of the current in wire 2.

Formula:

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

$F_x=\frac{{\mathrm{μ}}_{0}L{i}_1{i}_2}{2\mathrm{π}d}$

The distance between wire 1 and 2 is
$d=\frac{x{i}_1}{i_3}$

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

The curve in Fi. 29-64(b) passes through zero; this implies that the currents in wire 1 and 3 exert forces in opposite direction on wire 2. Thus, current $i_1$points out of the page.

When wire 3 is at a great distance from wire 2, the only field that affects wire 2 is that caused by the current in wire 1 that is $i_1$; in this case the force is negative.

This means that wire 2 is attached to wire 1 which implies that wire 2’s current is in the same direction as wire 1’s current i.e. out of page.

With wire 3 initially far away, the force$F_{12}$per unit length is given. Thus, from Eq. 29-13,

$F_{12}=\frac{{\mathrm{μ}}_0{i}_1{i}_2}{2\mathrm{π}d}$

When wire 3 is at$x=0.04\text{m}$the curve passes through zero point, so the force between 2 and 3 must be equal.

From this, we can find the distance between wire 1 and 2

$\begin{array}{rcl}d& =& \frac{x{i}_1}{i_3}\\ & =& \frac{\left(0.04\text{m}\right)\left(0.750\text{A}\right)}{\left(0.250 \text{A}\right)}\\ & =& 0.12\text{m}\end{array}$

Therefore, from Eq. 29-13,

$\begin{array}{rcl}{F}_{12}& =& -\frac{F_{2x}}{L_2}\\ & =& \frac{{\mathrm{μ}}_0{i}_1{i}_2}{2\mathrm{Ï€}d}\\ 0.627\text{μ±·/³¾}& =& \frac{\left(4\mathrm{Ï€}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right)\left(0.750\text{A}\right)i_2}{2\mathrm{Ï€}\left(0.12\text{m}\right)}\end{array}$

Thus, the current through wire 2 is

$$\begin{gathered} i_2=\frac{2\mathrm{π}\left(6.27×{10}^{-7}\text{N/m}\right)\left(0.12\text{m}\right)}{\left(4\mathrm{π}×{10}^{-7}{\text{NA}}^{\text{- 2}}\right)\left(0.75\text{A}\right)} \\ =0.50\text{A} \end{gathered}$$

The direction of $i_2$is out of page as mentioned in part (a).