Answer
(a) Point B
(b) Point A ,$2.1\mu T$; Point B, $13\mu T$
Work Step by Step
(a) We know that the magnitudes of the fields due to each wire are the same at each point but their directions are opposite at A and the magnitude of the net magnetic field is greater at point B.
(b) We know that
$B_A=\frac{\mu_{\circ}}{2\pi r}(I_1-I_2)$
$B_A=\frac{4\pi\times 10^{-7}}{2\pi(0.16m)}(6.2A-4.5A)$
$B_A=2.1\mu T$
Now the magnitude of the net magnetic field at point B is given as
$B_B=\frac{\mu_{\circ}}{2\pi r}(I_1+I_2)$
We plug in the known values to obtain:
$B_B=\frac{4\pi \times 10^{-7}}{2\pi(0.16m)}(6.2A+4.5A)$
$\implies B_B=13\mu T$
