Answer:
The magnetic field will be [tex]\large{\dfrac{1.4 \times 10^{-4}}{d}} T[/tex], '2d' being the distance the wires.
Explanation:
From Biot-Savart's law, the magnetic field ([tex]\large{\overrightarrow{B}}[/tex]) at a distance '[tex]r[/tex]' due to a current carrying conductor carrying current '[tex]I[/tex]' is given by
[tex]\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}[/tex]
where '[tex]\overrightarrow{dl}[/tex]' is an elemental length along the direction of the current flow through the conductor.
Using this law, the magnetic field due to straight current carrying conductor having current '[tex]I[/tex]', at a distance '[tex]d[/tex]' is given by
[tex]\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}[/tex]
According to the figure if '[tex]I_{t}[/tex]' be the current carried by the top wire, '[tex]I_{b}[/tex]' be the current carried by the bottom wire and '[tex]2d[/tex]' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the [tex]\bigotimes[/tex] symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by [tex]\bigodot[/tex] symbol.
Given [tex]\large{I_{t} = 19.5 A}[/tex] and [tex]\large{I_{B} = 12.5 A}[/tex]
Therefore, the magnetic field ([tex]\large{B_{t}}[/tex]) at 'P' due to the top wire
[tex]B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}[/tex]
and the magnetic field ([tex]\large{B_{b}}[/tex]) at 'P' due to the bottom wire
[tex]B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}[/tex]
Therefore taking the value of [tex]\mu_{0} = 4\pi \times 10^{-7}[/tex] the net magnetic field ([tex]\large{B_{M}}[/tex]) at the midway between the wires will be
[tex]\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T[/tex]
