physics question please help mee?

Question

Two long, straight wires cross each other at right angles, as shown in Figure P19.53.

Figure P19.53
http://www.webassign.net/sf5/p19_49.gif

(a) Find the direction and magnitude of the magnetic field at point P, which is in the same plane as the two wires.
Magnitude
T
Direction
into the page
59.0° counterclockwise from the -x direction
59.0° counterclockwise from the +x direction
out of the page
59.0° clockwise from the +x direction
31.0° clockwise from the +x direction


(b) Find the magnetic field at a point 29.5 cm above the point of intersection (29.5 cm out of the page, toward you).
Magnitude
T
Direction
into the page
59.0° counterclockwise from the +x direction
31.0° clockwise from the +x direction
59.0° clockwise from the +x direction
59.0° counterclockwise from the -x direction
out of the page

Answer 1

ok, now I have the idea. First, let's calculate the magnetic field at point P, from the wire with 5 A.

Uo = 4pi*10^-7

B1 = Uo*5 / 2pi*(40 / 100)

B1 = 20pi*10^-7 / 80pi / 100 = 10^-5 / 4 (Tesla)

Using the right hand rule :

B1 is on the "-y" direction B1 = 10^-5 / 4 (-j) T

Calculating magnetic field for the wire with 3 A

B2 = Uo*3 / 2pi*(30 / 100)

B2 = 4pi*10^-7*3 / 2pi*(20 / 100) = 3*10^-6 T

Using the right hand rule, I found that the direction of the magnetic field is directly into the page, you can consider it like "-z" direction

B2 = 3*10^-6 (-k) T

Ok, the final result of the magnetic field in P, is another vector, and we have to use pitagoras for it.

Bp = (3*10^-6)^2 +( 10^-5 / 4 )^2

Bp = 9/100*10^-10 + 10^-10 / 16

Bp = 10^-5(0.39) = 3.9*10^-6 (Tesla)

Using trigonometry :

3.9*10^-6sin(alfa) = 10^-5 / 4

alfa = 39º

So the magnetic field is : 10^-5 / 4 (Tesla), into the page, with 39º with the x axiss.