Question

In Figure, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length $\mathit{a}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{50}\mathbf{cm}$. Each wire carries$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{A}}$, and all the currents are out of the page. In unit-vector notation, what is the net magnetic force per meter of wire lengthon wire 1?

Short answer

The net magnetic force per meter of wire length on wire 1 is
${\overrightarrow{F}}_1=\left(7.94\times {10}^{-4}\text{N/m}\right)\hat{i}-\left(7.94\times {10}^{-4}\text{N/m}\right)\hat{j}$

Step-by-step solution

Identification of given data

• 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.

Understanding the concept

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,

$\mathit{d}\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{i}\mathbf{d}\overrightarrow{\mathbf{s}}\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, $\mathit{d}\overrightarrow{\mathbf{B}}$ is the field produced by the current element $\mathit{i}\mathit{d}\overrightarrow{\mathbf{s}}$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 $\left|{\overrightarrow{F}}_{12}\right|$, $\left|{\overrightarrow{F}}_{13}\right|$and$\left|{\overrightarrow{F}}_{14}\right|$. 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 ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$, 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,

$F_x=\frac{{\mu }_0{i}^2}{2\pi d}$

Here, $F_x$is x component of the force, ${\mu }_0$ 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

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

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

Determining the net magnetic force per meter of wire length on wire  .

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

$F_x=\frac{{\mu }_0{i}^2}{2\pi d}$

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

$\begin{array}{rcl}{F}_1& =& \left|{\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{14}\right|\\ & =& 2F_{12}\cos \theta +F_{13}\end{array}$

But,

$F_{12}=\frac{{\mu }_0{i}^2}{2\pi a}$

and

$F_{13}=\frac{{\mu }_0{i}^2}{2\sqrt{2}\pi a}$

On substituting the values, we get

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

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

Therefore, in unit-vector notation, we have

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

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

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