91Ó°ÊÓ

• Edge length is $a=8.50\text{cmor}0.085\text{m}$
• Each wire carries a current$i=15.0\text{A}$
• All currents are out of the page.

When two parallel current-carrying conductors are placed near each other, they exert a force on each other, either attractive or repulsive, depending on the direction of the current. The fields are expressed by Biot-Savart’s law. It is stated as,

$\mathit{d}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\mathbf{i}\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{×}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, $\mathit{d}\stackrel{\mathbf{→}}{\mathbf{B}}$ is the field produced by the current element $\mathit{i}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}$at a point located at $\mathit{r}$from the current element.

Using Biot-Savart’s law and Lorentz force equation, we can calculate the force on the current carrying conductor by another current carrying conductor.

By using Eq. 29-13 we can find the magnitude of forces $\left|{\stackrel{→}{F}}_{12}\right|$, $\left|{\stackrel{→}{F}}_{13}\right|$and$\left|{\stackrel{→}{F}}_{14}\right|$. By using these values, we can find the magnitude of net magnetic force per meter of wire length on wire 1. By calculating the position vector along the direction of ${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{1}}$, we can find the net magnetic force per meter of wire length in unit-vector notation.

Formulas:

From Eq. 29-13, the force per unit length of wire between two parallel currents is given by,

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

Here, $F_x$is x component of the force, ${\mathrm{μ}}_0$ is permeability constant, $d$ is the distance between the two parallel current conducting wires.

The net magnetic force per meter of wire length on wireis

$F_1=\left|{\stackrel{→}{F}}_{12}+{\stackrel{→}{F}}_{13}+{\stackrel{→}{F}}_{14}\right|$

Here, ${\stackrel{→}{F}}_{12}$is the force exerted by wire 2 on wire 1, ${\stackrel{→}{F}}_{13}$is the force exerted by wire 3 on wire 1, and ${\stackrel{→}{F}}_{14}$is the force exerted by wire 4 on wire 1.

From Eq. 29-13, the force per unit length of wire between two parallel currents is given by,

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

The force on wire 1 is along the diagonal and only the forces along the diagonal direction contribute. With$\mathrm{θ}=45°$, we can find thenet magnetic force per meter of wire length on wire 1.

$\begin{array}{rcl}{F}_1& =& \left|{\stackrel{→}{F}}_{12}+{\stackrel{→}{F}}_{13}+{\stackrel{→}{F}}_{14}\right|\\ & =& 2F_{12}\cos \mathrm{θ}+F_{13}\end{array}$

But,

$F_{12}=\frac{{\mathrm{μ}}_0{i}^2}{2\mathrm{π}a}$

and

$F_{13}=\frac{{\mathrm{μ}}_0{i}^2}{2\sqrt{2}\mathrm{π}a}$

On substituting the values, we get

$\begin{array}{rcl}{F}_1& =& 2\left(\frac{{\mathrm{μ}}_0{i}^2}{2\mathrm{π}a}\right)\cos 45°+\left(\frac{{\mathrm{μ}}_0{i}^2}{2\sqrt{2}\mathrm{π}a}\right)\\ & =& \frac{3}{2\sqrt{2}\mathrm{π}}\left(\frac{{\mathrm{μ}}_0{i}^2}{a}\right)\\ & =& \frac{3}{2\sqrt{2}\mathrm{π}}\left[\frac{\left(4\mathrm{π}{\displaystyle ×}{\displaystyle {{\displaystyle 10}}^{-7}}{\displaystyle }{\displaystyle \text{Tm/A}}\right){\displaystyle {{\displaystyle \left(15.0\text{A}\right)}}^2}}{{\displaystyle \left(0.085{\text{m}}^2\right)}}\right]\\ & =& 1.12×{10}^{-3}\text{N/m}\end{array}$

The direction of${\stackrel{→}{F}}_1$is along the position vector$\hat{r}=\frac{1}{\sqrt{2}}\left(\hat{i}-\hat{j}\right)$.

Therefore, in unit-vector notation, we have

$\begin{array}{rcl}{\stackrel{→}{F}}_1& =& \frac{\left(1.12×{10}^{-3}\text{N/m}\right)}{\sqrt{2}}\left(\hat{i}-\hat{j}\right)\\ & =& \left(7.94×{10}^{-4}\text{N/m}\right)\hat{i}-\left(7.94×{10}^{-4}\text{N/m}\right)\hat{j}\end{array}$

Therefore, the net magnetic force per meter of wire length on wire 1 is

${\stackrel{→}{F}}_1=\left(7.94×{10}^{-4}\text{N/m}\right)\hat{i}-\left(7.94×{10}^{-4}\text{N/m}\right)\hat{j}$