- #1

veroniknstudying

- 1

- 0

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!