91Ó°ÊÓ

First, by using Eq. 29-13, we find the components of ${\mathit{F}}_{\mathbf{4}\mathbf{x}}$and ${\mathit{F}}_{\mathbf{4}\mathbf{y}}$per meter of wire length. By using these components, we can find the separation force ${\mathit{F}}_{\mathbf{4}}$per meter of wire length. Now by finding angle $\mathit{ϕ}$ that ${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{4}}$makes with the positive x axis, we can find the net magnetic force per meter of wire length on wire 4 in the unit vector notation.

Formulas:

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

${\text{F}}_4={\left({\text{F}}_{4\text{x}}^2+{\text{F}}_{4\text{y}}^2\right)}^{1/2}$

$\mathrm{Ï•}=ta{n}^{-1}\left(\frac{F_{4y}}{F_{4x}}\right)$

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

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

For $i_1=i_2=i$and for force per meter of wire length, we take $L=1\text{m}$.

Therefore,

$F_x=\frac{{\mathrm{μ}}_0{i}^2}{2\mathrm{π}d}$

By superposition of forces, the magnetic forceper meter of wire lengthon wire 4 is

${\stackrel{→}{F}}_4={\stackrel{→}{F}}_{14}+{\stackrel{→}{F}}_{24}+{\stackrel{→}{F}}_{34}$

With, $\mathrm{θ}=45°$, the situation is shown in the figure below

The x component is

$F_{4x}=-F_{43}-F_{42}cos\mathrm{θ}$

Using Eq. 29-13, we have

$F_{4x}=-\frac{{\mathrm{μ}}_0{i}^2}{2\mathrm{π}a}-\frac{{\mathrm{μ}}_0{i}^2}{2\sqrt{2}\mathrm{π}a}cos45°$

$F_{4x}=-\frac{3{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}$

And y component is

$F_{4y}=F_{41}-F_{42}sin\mathrm{θ}$

Using Eq. 29-13, we have

$F_{4y}=\frac{{\mathrm{μ}}_0{i}^2}{2\mathrm{π}a}-\frac{{\mathrm{μ}}_0{i}^2}{2\sqrt{2}\mathrm{π}a}sin45°$

$F_{4y}=\frac{{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}$

Thus, the separation force$F_4$per meter of wire length isgiven by

$F_4={\left(F_{4x}^2+F_{4y}^2\right)}^{1/2}$

$F_4={\left[{\left(-\frac{3{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}\right)}^2+{\left(\frac{{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}\right)}^2\right]}^{1/2}$

$\begin{array}{rcl}{F}_4& =& {\left[{\left(-\frac{3\left(4\mathrm{π}×{10}^{-7}\right){\left(7.50\right)}^2}{4\mathrm{π}\left(0.135\right)}\right)}^2+{\left(\frac{\left(4\mathrm{π}×{10}^{-7}\right){\left(7.50\right)}^2}{4\mathrm{π}\left(0.135\right)}\right)}^2\right]}^{1/2}\\ & =& 1.32×{10}^{-4}\text{N/m}\end{array}$

The force${\stackrel{→}{F}}_4$makes an angle$\mathrm{ϕ}$with the positive x axis where

$\begin{array}{rcl}\mathrm{ϕ}& =& ta{n}^{-1}\left(\frac{F_{4y}}{F_{4x}}\right)\\ & =& ta{n}^{-1}\left[\frac{\left(\frac{{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}\right)}{\left(-\frac{3{\mathrm{μ}}_0{i}^2}{4\mathrm{π}a}\right)}\right]\\ & =& ta{n}^{-1}\left(-\frac{1}{3}\right)\\ \mathrm{ϕ}& =& 162°\end{array}$

In unit vector notation, we have

$$\begin{gathered} {\stackrel{→}{F}}_4=F_4\left[cos\mathrm{Ï•}\hat{i}+sin\mathrm{Ï•}\hat{j}\right] \\ =\left(1.32×{10}^{-4}\text{N/m}\right)\left[cos162°\hat{i}+sin162°\hat{j}\right] \\ =\left(-125\text{μ±·/³¾}\right)\hat{i}+\left(41.7\text{μ±·/³¾}\right)\hat{j} \end{gathered}$$

Final Statement:

By using the equation for the force between two parallel current carrying conductors, we can find the net magnetic force per meter of wire length on thewire.