Two long parallel wires carrying currents $8$A and $15$A in opposite directions are placed at a distance of $7$ cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnetic field at P is ___ $\times 10^{-6}$ T. (Given: $\sqrt{2} = 1.4$, $\mu_0 = 4\pi \times 10^{-7}$ SI unit)

Let the distance of point P from each wire be $r$. Since the lines joining P to the wires are perpendicular, they form a right-angled isosceles triangle with the distance between the wires as the hypotenuse.

$r^2 + r^2 = d^2 \Rightarrow 2r^2 = (7 \times 10^{-2})^2 \Rightarrow r = \dfrac{7}{\sqrt{2}} \times 10^{-2}$ m

The magnetic fields due to the two wires at point P are:

$B_1 = \dfrac{\mu_0 I_1}{2\pi r}$ and $B_2 = \dfrac{\mu_0 I_2}{2\pi r}$

Since the position vectors from the wires to point P are perpendicular to each other, the magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ are also perpendicular to each other.

The net magnetic field at P is:

$B = \sqrt{B_1^2 + B_2^2} = \dfrac{\mu_0}{2\pi r} \sqrt{I_1^2 + I_2^2}$

Substituting the given values:

$B = \dfrac{2 \times 10^{-7}}{\dfrac{7}{\sqrt{2}} \times 10^{-2}} \sqrt{8^2 + 15^2}$

$B = \dfrac{2 \sqrt{2} \times 10^{-5}}{7} \times 17$

$B = \dfrac{34 \times 1.4}{7} \times 10^{-5}$

$B = 34 \times 0.2 \times 10^{-5} = 6.8 \times 10^{-5} \text{ T} = 68 \times 10^{-6} \text{ T}$

Answer: $68$