How Does Earth's Magnetic Field Combine with a Nearby Current-Carrying Wire?

[veroniknstudying]

A long horizontal wire carries 22.0A of current due north. What is the net magnetic field 20.0 cm due west of the wire if the Earth's field there points downward, 40 degrees below the horizontal, and has magnitude 5.0 E-5 T?

My approach:
1. Drew north to the right and the wire lying flat, with current pointing toward right.
2. Used thumb rule to determine that the B field in the front of the wire goes down and the portion behind the wire goes up.
3. Made west above the wire and drew the B field at this point to be pointing directly toward me.
4. <Bk, Bj> where Bk equals ((μI/2pid)+5.0E-5cos(40)) and Bj equals -5.0E-5sin(40).
5. Used pythagorean theorem to find the magnitude of the B field and arctan to find the angle.

I saw other problems online for this, and it seems like my sin and cos may be reversed, but I don't understand why it would be the other way around. If someone can explain that to me, that would be great. The solution is 4.0 E-5 T, 15 degrees below horizontal. This isn't for homework answers. I'm studying for a test, so please feel free to go in depth with explanations and links. Thank you!

 

[fresh_42]

Mentor's note: moved form a technical forum.

 

[gneill]

Hi veroniknstudying,

Welcome to Physics Forums.

Sometimes it can help to try different perspectives for your drawing to make sure that you're using the right angle and trig function. Concentrate on the Earth field alone and draw it in profile. That is, let left to right be S to N as before, but let bottom to top be upwards $\hat{k}$ direction.