Answer:
Magnetic field shall be zero at exactly in between the wires.
Step-by-step explanation:
We can find the magnetic field by biot Savart law as follows
![\overrightarrow{dB}=(\mu _(0)I)/(4\pi )\int \frac{\overrightarrow{dl}* \widehat{r}}{r^(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/ppaegldwn8l9cawqez89wik7yg7txjz9tz.png)
For current carrying wire in positive y direction we have
![\overrightarrow{dB_(1)}=(\mu _(0)Idl)/(4\pi )\int \frac{\widehat{j}* \widehat{r_(1)}}{r_(1)^(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/npqlddpxcvqkhzb8gc73lepm2ff91drkjo.png)
Similarly for wire carrying current in -y direction we have
![\overrightarrow{dB_(2)}=(-\mu _(0)Idl)/(4\pi )\int \frac{\widehat{j}* \widehat{r_(2)}}{r_(2)^(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/b3ne30j6v5b45u0hm5srvmnapupyd8nhtm.png)
Thus the net magnetic field at any point in space is given by
![\overrightarrow{dB_(1)}+\overrightarrow{dB_(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/c211vk3j3wwhgcf4ssnqi7ad3xetgqiu3m.png)
![(\mu _(0)Idl)/(4\pi )\int \frac{\widehat{j}* \widehat{r_(1)}}{r_(1)^(2)}+(-\mu _(0)Idl)/(4\pi )\int \frac{\widehat{j}* \widehat{r_(2)}}{r_(2)^(2)}=0\\\\\Rightarrow \overrightarrow{r_(1)}=\overrightarrow{r_(2)}](https://img.qammunity.org/2020/formulas/physics/high-school/5j4mlsd1hwl4d8n6kyats8w2dh6o85utfv.png)
For points with same position vectors from the 2 wires we have a net zero magnetic field. These points are exactly midway between the 2 wires