Answer
a. $B_{net}=(2 \times10^{-5} T /A)(I-25A)$
b. $B_{net}=(2 \times10^{-5} T /A)(I+25A)$
Work Step by Step
a. The currents are in the same direction. The 2 magnetic fields, halfway between the currents, will point in opposite directions. Their magnitudes are subtracted.
$$B_{net}=\frac{\mu_oI_1}{2\pi r}-\frac{\mu_oI_2}{2\pi r}$$
$$B_{net}=\frac{4\pi \times10^{-7} T\cdot m/A}{2\pi (0.010m)}(I-25A)$$
$$B_{net}=(2 \times10^{-5} T /A)(I-25A)$$
b. Now the currents run in opposite directions. The 2 magnetic fields, halfway between the currents, point in the same direction. Their magnitudes are added.
$$B_{net}=\frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}$$
$$B_{net}=\frac{4\pi \times10^{-7} T\cdot m/A}{2\pi (0.010m)}(I+25A)$$
$$B_{net}=(2 \times10^{-5} T /A)(I+25A)$$