Answer
$B_{net} = 2.56\times 10^{-7}~T$
Work Step by Step
The distance from each current to the point $P$ is $5.00~m$
If the angle above the horizontal from $P$ to the upper current is $\theta$, then $cos~\theta = 0.800$
By the right hand rule and symmetry, the horizontal components of the magnetic fields due to the currents cancel out, and the net magnetic field at $P$ is the sum of the vertical components.
We can find the vertical component of the magnetic field at $P$ due to the upper current:
$B = \frac{\mu_0~i}{2~\pi~R}~cos~\theta$
$B = \frac{(4\pi\times 10^{-7}~H/m)(4.00~A)}{(2~\pi)~(5.00~m)}~(0.800)$
$B = 1.28\times 10^{-7}~T$
By symmetry, the vertical component of the magnetic field at $P$ due to the lower current is also this value.
We can find the net magnetic field at $P$:
$B_{net} = (2)(1.28\times 10^{-7}~T)$
$B_{net} = 2.56\times 10^{-7}~T$
