Question

Two long, straight wires are separated by 0.12 m. The wires carry currents of 4.0 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field at the point A.

Answer 1

If we have I= 7.5 A:

I think my solution might just help you answer the problem on your own:

You have the formulas correct, watch your signs and BRACKETS. 

B = μ0/(2π) (Current) / (Perpendicular distance) 
Since μ0=4π E -7 Tm/A, we have: 
B1 = (4πE-7 Tm/A)(7.5 A)/[2π (0.030 m)] = 5E-5 T 
B2 = (4πE-7 Tm/A)(-7.5 A)/[2π (0.150 m)] = -1E-1 T 
So BA = B1 + B2 = ? 
(It looks like you just left out the square brackets, hence multiplying Pi and 0.03 and 0.15 instead of dividing them.) 

For the point B, the two distances are -0.060 m and +0.060 m. Be careful with the signs. Unlike point A, the two components will have the same sign.

Answer 2

Answer: Hello mate!

a wire that is transporting an I current generates a magnetic field B at an r distance by the next equation:

$B = \frac{uI}{2*pi*r}$ where u is a constant, and pi = 3.141592..

The generated field is always perpendicular to the direction of the flow of the electricity, you can see the direction of the field doing the next thing:

put your thumb in the direction where the electricity is flowing, now if your palm is facing up, then the magnetic field is pointing up if your palm is facing down, then the field is pointing down.

them if both currents have currents in opposite directions, the fields generated in the middle point between them also have opposite directions. And knowing that the distance between each wire and the middle point is the same and that each wire has a 4.0 A current, it is easy to see that the magnetic field is zero in that point, but let's compute this.

I₁ = 4A and I₂ = -4A,  and if we define r= 0 at the first wire, then the distances to the middle point can be calculated as:

r₁ = 0.06m and r₂ = 0.12m - 0.6m = 0.6m  

then $B = \frac{u4A}{2pi*0.06m}  + (- \frac{u4A}{2pi*0.06m} ) = 0$