Figure 29-52 shows, in cross section, four thin wires that are parallel, straight, and very long. They carry identical currents in the directions indicated. Initially all four wires are at$\mathbf{distance}\mathbf{}\mathbf{d}\mathbf{=}\mathbf{}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$from the origin of the coordinate system, where they create a net magnetic field .(a) To what value of xmust you move wire 1 along the xaxis in order to rotate counter clockwise by $\mathbf{30}\mathbf{°}$? (b) With wire 1 in that new position, to what value of xmust you move wire 3 along the xaxis to rotate by$\mathbf{30}\mathbf{°}$back to its initial orientation?

1. Permeability of free space, ${\mu }_0=4\pi \times {10}^{-7}\frac{\text{Tm}}{\text{A}}$.
2. Distance of each wire from origin,$d=15.0\text{cm}$.
3. Wire shown by cross sign is inward direction. Wire 2.
4. Current in each wire is the same.

Formula:

Magnetic field due to long wire carrying current is:

$B=\frac{{\mu }_{0}i}{2\pi d}$

Consider the net magnetic field is given as follows:

$B_{net,y}=0$and localid="1663230136338" $B_{net,x}=B_2+B_4=2\left(\frac{{\mu }_{0}i}{2\pi d}\right)$

To obtain the condition of localid="1663230123483" $30°$, we must write

$B_{net,y}=B_{net,x}\text{tan}\left({30}^0\right)$

$B_1\text{'}-B_3=2\left(\frac{{\mu }_{0}i}{2\pi d}\right)tan\left(30\right)$

Here, $B_3=\frac{{\mu }_{0}i}{2\pi d}$and $B_1^{\text{'}}=\frac{{\mu }_{0}i}{2\pi d\text{'}}$

$\frac{{\mu }_{0}i}{2\pi d\text{'}}-\frac{{\mu }_{0}i}{2\pi d}=\frac{2\left(\frac{{\mu }_{0}i}{2\pi d}\right)1}{\sqrt{3}}$

There is one common term localid="1663230110052" $\frac{{\mu }_{0}i}{2\pi }$; after canceling out, resolve as follows:

$\frac{1}{d\text{'}}-\frac{1}{d}=\frac{2}{d\sqrt{3}}$

$\frac{1}{d\text{'}}=\frac{2}{d\sqrt{3}}+\frac{1}{d}$

$\frac{1}{d\text{'}}=\frac{2+\sqrt{3}}{d\sqrt{3}}$

$d\text{'}=\frac{d\sqrt{3}}{2+\sqrt{3}}$

Solve further as:

$d\text{'}=0.464\times d$

Given localid="1663228564741" $d=-15.0\mathrm{cm}$for wire 1. Using this in the above equation, solve as:$$\begin{gathered} d\text{'}=0.464\times -15.0 \\ d\text{'}=-7.0\text{cm} \end{gathered}$$

Move the wire 1 to localid="1663228536959">x=-7.0cm.

To restore the initial setting, we need to move wire 3 to$x=+7.0\text{cm}$because the two wires have the same current and they must be at the same distance from the point so that the initial symmetry is restored.