Calculating the direction and magnitude of the magnetic field in a 3D plane

[mike115]

Homework Statement

Two long, straight wires cross each other at right angles, as shown in Figure P19.49. (a) Find the direction and magnitude of the magnetic field at point P, which is in the same plane as the two wires. (b) Find the magnetic field at a point 30.0 cm above the point of intersection (30.0 cm out of the page, toward you).

Homework Equations

B = uo*I/(2*pi*r)
Second Right Hand Rule - thumb in direction of current, fingers curl in direction of magnetic field

The Attempt at a Solution

I got part (a) right.

For part (b), I can determine the magnitude of the two magnetic fields.

B_left = uo*3/(2*pi*sqrt(.3^2 + .3^2))

B_right = uo*5/(2*pi*sqrt(.3^2 + .4^2))

B_left makes an angle of 45 degrees above the plane. B_right makes an angle of 36.9 degrees above the plane.

Now, I am thinking that the magnetic field for each one would have to be perpendicular to these angles. That would give a component of the magnetic field along the x-y plane and a component perpendicular to the x-y plane (along the z axis). I'm not really sure what to do next or if I am even approaching the problem correctly. Can anybody please help?

 

[haruspex]

above the point of intersection

… not above point P.

 

[Delta2]

Yes, @haruspex is right. There are two components of the B-field above the point of intersection:
$$B_{down}=\frac{\mu_0}{2\pi}\frac{5}{0.3}$$
$$B_{right}=\frac{\mu_0}{2\pi}\frac{3}{0.3}$$

The angle between them is 90 degrees so we can use pythagorean theorem to find the total B as $$B=\sqrt{B_{down}^2+B_{right}^2}$$ and the angle it makes with the horizontal has $$\tan\theta=\frac{B_{down}}{B_{right}}$$