In Figure, a long straight wire carries a current $\mathbf{}{\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathit{A}$and a rectangular loop carries current$\mathbf{}{\mathit{i}}_{\mathbf{2}}$=20.0 A. Take $\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{L}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$. In unit-vector notation, what is the net force on the loop due to ${\mathit{i}}_{\mathbf{1}}$ ?

• The current through wire 1 is $i_1=30.0A$.
• The current through wire 2 is $i_2=20.0A$.
• The length of wire is $L=30.0cm=0.300m$
• The dimension$a=1.00cm=0.0100m$
• The dimension$b=8.00cm=0.0800m$

The force between the two parallel current carrying conductors is proportional to the length of the conductor, currents through the conductors, and inversely proportional to the separation between the conductors. Using this relation,we can findthe force on each side of the loop.But the force acting on two perpendicular sides of the loop of length b will be equal and opposite, and hence, they will cancel out. Thus, the net force is now the only force on each side of the loop lying parallel to the long wire. Therefore, usingEq.29-13, we can find the net force on the loop due to current $\mathit{i}$.

Formula:

The force between two parallel current-carrying conductors is given by,

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

From equation (i), the force between two parallel currents is given by

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

We can find the forces on the sides of the rectangle that are parallel to the long straight wire with current$i_1$. But the force on each side lying perpendicular to it would be calculated by integrating as follows:

role="math" localid="1663226281267" $F_{\perp ,sides}={\int }_a^{a+b}\frac{{\mu }_0{i}_1{i}_2}{2\pi y}dy$

This force acting on two perpendicular sides of length$d$cancel out. For the remaining two parallel sides of length$L$, we obtain,

$F=\frac{{\mu }_{0}L{i}_1{i}_2}{2\pi }\left(\frac{1}{a}-\frac{1}{a+b}\right)$

Here, $\frac{1}{d}=\left(\frac{1}{a}-\frac{1}{a+b}\right)$

Thus,

$F=\frac{{\mu }_{0}L{i}_1{i}_{2}b}{2\pi a\left(a+b\right)}$

Substituting the given values, we get

$$\begin{gathered} F=\frac{\left(4\pi \times {10}^{-7}\right)\left(0.300\text{m}\right)\left(30.0\text{A}\right)\left(20.0\text{A}\right)\left(0.0800m\right)}{2\pi \left(0.0100\text{m}\right)\left(0.0100\text{m}+0.0800\text{m}\right)} \\ =3.20\times {10}^{-3}\text{N} \end{gathered}$$

And $\overrightarrow{F}$ points toward the wire, i.e., along $+\hat{j}$.

Therefore, the force is,$\overrightarrow{F}=\left(3.20\times {10}^{-3}N\right)\hat{j}$

Thus, the net force on the loop due to $i_1$is $\overrightarrow{F}=\left(3.20\times {10}^{-3}N\right)\hat{j}$.