Magnetic field at a pt created from two wires.

[unhip_crayon]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

B2 = (μo/2π) 14

Thats it, I am 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...