1. Geometry of the point $P$

Let the two wires be at points $A$ and $B$, with separation $AB=7\text{ cm}.$

Point $P$ is equidistant from both wires and the lines joining $P$ to the wires are perpendicular. So in triangle $APB$:

• $AP=BP=r$
• $\angle APB = 90^\circ$

Thus, $AB$ is the hypotenuse of an isosceles right triangle: $AB = r\sqrt{2}$ Hence, $r = \frac{AB}{\sqrt{2}} = \frac{7}{1.4} = 5\text{ cm} = 0.05\text{ m}.$

2. Magnetic field due to each wire at $P$

Magnetic field due to a long straight wire: $B = \frac{\mu_0 I}{2\pi r}$ with $\frac{\mu_0}{2\pi} = 2\times 10^{-7}\,\text{T m/A}.$

So, $B_1 = 2\times 10^{-7}\cdot \frac{8}{0.05} = 32\times 10^{-6}\,\text{T}$

and $B_2 = 2\times 10^{-7}\cdot \frac{15}{0.05} = 60\times 10^{-6}\,\text{T}.$

3. Angle between the two magnetic fields

The currents are in opposite directions. At point $P$, the radii $PA$ and $PB$ are perpendicular. Since magnetic field around each wire is tangential to the circle centered on that wire, each magnetic field is perpendicular to its corresponding radius.

Because the currents are opposite, the two field directions at $P$ turn out to be perpendicular to each other.

Therefore, resultant field is $B = \sqrt{B_1^2 + B_2^2}.$

4. Resultant magnetic field

$B = \sqrt{(32)^2 + (60)^2}\times 10^{-6}$ $= \sqrt{1024 + 3600}\times 10^{-6}$ $= \sqrt{4624}\times 10^{-6}$ $= 68\times 10^{-6}\,\text{T}.$

5. Final answer

The required value is $\boxed{68}.$

6. Comparison with stored answer

Stored correct answer = $68$.

So my derived answer agrees with the stored answer.