91Ó°ÊÓ

1. Wire separation is$d_1=0.75\text{cm}=7.5×{10}^{-3}\text{m}$
2. Wire 1 carries current$6.5\text{A}$
3. $d_2=1.50\text{cm}=1.50×{10}^{-2}\text{m}$
   

Use the formula for magnetic field at a certain point away from the semi-infinite wire as follows:

$\mathit{B}\mathbf{=}\frac{\mathit{μ}\mathit{I}}{\mathbf{4}\mathit{π}\mathit{R}}$

Given the net magnetic field is zero which means that the magnetic field by wire 1 must cancel with magnetic field by wire 2.

$\mathit{B}\mathbf{=}\frac{{\mathit{μ}}_{\mathbf{0}}\mathbf{×}\mathit{I}}{\mathbf{4}\mathit{π}\mathit{R}}$

Magnetic field at a certain point from the wire is as follows

$B=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}R}$

So, for wire 1 magnetic field at point p is as follows

$B_1=\frac{{\mathrm{μ}}_0{I}_1}{4\mathrm{π}\left(d_1+d_2\right)}$

So, for wire 2 magnetic field at point p is as follows

$B_2=\frac{{\mathrm{μ}}_0{I}_2}{4\mathrm{π}\left(d_2\right)}$

The net magnetic field is zero means

$$\begin{gathered} B_1=B_2 \\ \frac{{\mathrm{μ}}_0{I}_1}{4\mathrm{π}\left(d_1+d_2\right)}=\frac{{\mathrm{μ}}_0{I}_2}{4\mathrm{π}{d}_2} \end{gathered}$$

Substitute the values and solve:

$$\begin{gathered} \frac{{\mathrm{μ}}_0×6.5}{4\mathrm{π}\left(7.5×{10}^{-3}+1.50×{10}^{-2}\right)}=\frac{{\mathrm{μ}}_0×I_2}{4\mathrm{π}×1.50×{10}^{-2}} \\ {I}_2=4.3\text{A} \end{gathered}$$

hence the current is, $4.3\text{A}$.

Direction of current $I_2$is out of the page or opposite $I_1$ to create an opposite magnetic field. That is out of the page.