Question

In Figure, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length $\mathit{a}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{cm}}$. Each wire carries$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{A}}$, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3. In unit vector notation, what is the net magnetic force per meter of wirelengthon wire 4?

Short answer

The net magnetic force per meter of wire length on wire 4 is

${\overrightarrow{F}}_4=\left(-125\text{μN/m}\right)\hat{i}+\left(41.7\text{μN/m}\right)\hat{j}$

Step-by-step solution

Given

1. The edge length of the square formed by four wires is $a=13.5\text{cm}$
2. Each wire carries current $i=7.50\text{A}$.

Understanding the concept

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 $\mathit{\varphi }$ that ${\overrightarrow{\mathbf{F}}}_{\mathbf{4}}$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:

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

$F_x=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi d}$

2. The separation force $\overrightarrow{F_4}$ per meter of wire length is given by

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

3. An angle $\varphi$ is

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

The net magnetic force per meter of wire length on wire  4

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

$F_x=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi d}$

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

Therefore,

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

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

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

With, $\theta =45°$, the situation is shown in the figure below

The x component is

$F_{4x}=-F_{43}-F_{42}cos\theta$

Using Eq. 29-13, we have

$F_{4x}=-\frac{{\mu }_0{i}^2}{2\pi a}-\frac{{\mu }_0{i}^2}{2\sqrt{2}\pi a}cos45°$

$F_{4x}=-\frac{3{\mu }_0{i}^2}{4\pi a}$

And y component is

$F_{4y}=F_{41}-F_{42}sin\theta$

Using Eq. 29-13, we have

$F_{4y}=\frac{{\mu }_0{i}^2}{2\pi a}-\frac{{\mu }_0{i}^2}{2\sqrt{2}\pi a}sin45°$

$F_{4y}=\frac{{\mu }_0{i}^2}{4\pi a}$

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

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

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

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

The force${\overrightarrow{F}}_4$makes an angle$\varphi$with the positive x axis where

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

In unit vector notation, we have

$$\begin{gathered} {\overrightarrow{F}}_4=F_4\left[cos\varphi \hat{i}+sin\varphi \hat{j}\right] \\ =\left(1.32\times {10}^{-4}\text{N/m}\right)\left[cos162°\hat{i}+sin162°\hat{j}\right] \\ =\left(-125\text{μN/m}\right)\hat{i}+\left(41.7\text{μN/m}\right)\hat{j} \end{gathered}$$

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.