91Ó°ÊÓ

1. Permeability of free space, ${\mathrm{μ}}_0=4\mathrm{π}×{10}^{-7}\frac{\text{Tm}}{\text{A}}$.
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:

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}d}$

Consider the net magnetic field is given as follows:

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

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

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

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

Here, $B_3=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}d}$and $B_1^{\text{'}}=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}d\text{'}}$

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

There is one common term localid="1663230110052" $\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}}$; 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:

$d\text{'}=0.464×d$

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

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.