Two long parallel wires carrying currents 8A and 15A in opposite...

Here, the magnetic fields produced by the two wires will be perpendicular to one another.

Magnetic field due to wire carrying current $8A$ is ${B}_{1} = \dfrac{{\mu}_{0}{i}_{1}}{2\pid}$

and magnetic field due to wire carrying current $15A$ is ${B}_{2} = \dfrac{{\mu}_{0}{i}_{2}}{2\pid}$

The net magnetic field is

${B}_{net} = \sqrt{{B}_{1}^{2} + {B}_{2}^{2}} = \dfrac{{\mu}_{0}}{2\pid}\sqrt{{i}_{1}^{2} + {i}_{2}^{2}}$

$= \dfrac{4\pi \times {10}^{- 7}}{2\pi \times \left(\dfrac{7}{\sqrt{2}}\right) \times {10}^{- 2}} \times \sqrt{{8}^{2} + {15}^{2}}\left(d = \dfrac{7}{\sqrt{2}}cm\right)$

$= 68 \times {10}^{- 6}T$.