Magnetic field at a pt created from two wires.

[unhip_crayon]

1. The problem statement, all variables and given/known data

The two wires carrying the currents are infinite in length. i1 = 5.00 A out
of the paper and i2 = 7.00 A into the paper. a = 40.0 cm and b = 30.0 cm.
Calculate the direction of the magnetic field at P, measured counterclockwise
from the +x axis.

Answer: 126.4°

http://answerboard.cramster.com/answer-board/image/5769206ead06ab440f5adef680da9eba.jpg

2. Relevant equations

http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-20088221440246335501282431425007016.gif

3. The attempt at a solution

B1 = (μo /2π) (5 A) / (0.30m) = (μo/2π) 16.6667

B2 = (μo/2π) 14

Thats it, Im stuck. I tried breaking this into components and finding the angle but it did not work.

Please Help

[dynamicsolo]

All right, you recognize that you need components, which is good. Using the right-hand rule for B1 at point P, what are the components? (This one isn't bad.)

For dealing with B2 then, we know the components will be

B_{2x} = B_2 \cdot cos \theta and B_{2y} = B_2 \cdot sin \theta.

So what is the angle? It must point perpendicular to the hypotenuse of the right triangle formed by the two wire cross-sections and P; as you also recognized, this is a 3-4-5 right triangle. So a vector pointing along the hypotenuse from wire 2 to P would have components proportional to
< -4/5 , 3/5 >. B2 points perpendicular to this vector into the "first quadrant", so the components of B2 are proportional to < 3/5, 4/5 >. Those are the cosine and sine values for our components of B2, so we can get the components themselves from that.

We're not quite done, since we still have to add the x- and y-components of the two magnetic fields at P to get the components for the total field... then there's getting the direction in which the total field vector points. But I imagine you can take it from there...