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Magnetic Forces/Fields: help!

  • #1
Hey, thanks in advance for checking my work. :smile:

Ok here's my question:

The drawing (attachment) shows two wires that carry the same current of I=85.0 A are are oriented perpendicular to the plane of the paper. The current in one wire is directed out of the paper, while the current in the other is directed into the paper. Find the magnitude and direction of thenet magnetic field at point P.

For those who can't get the attacment, the drawing is simple:

X       0
A equilateral triangle with sides of 0.150 m, X being current into and 0 being out of.

Ok it seems that I just have to use: B=uI/2pieR

Using the right hand rules I draw two circlular fields reaching point P sort of like venn diagrams. Ok the radius is really just the distance from the currents to the P, so R=0.150m.

Plug in numbers:

B=4pie x 10^-7 (85.0A)/ 2pie (0.150m)
B=1.133x10^-4 T.

Ok, I know B2=B1? so All I have to do is find the net

which is B3 (going downards)=SQRT(B1^2+B2^2)
and B3 turns out to be SQRT(2.55x10^-8)
which is 1.60x10^-4 T?

Anyone want to say correct? or wrong... :uhh:

Answers and Replies

  • #3
I few questions,
What is B in relationship to B_1 and B_2? You show a calculation for B, what is this?
Where is the third magnetic field coming from?
The two fields are actually opposite in polarity since the current are flowing in opposite directions, did you take this into consideration?
  • #4
B is just the magnitude of the field for one current, or ciruclar field. Since they are both the same just in diff directions, and since the directins converge toward P or towards the center, I figured I could use pythagoras to find the resulting magnitude of B3...Am I wrong lol?
  • #5
Well, both fields eminate in a circular fashion, so at point P, the magnetic field from EACH wire would have magnitude B = muI/2pi(.15). Since the triangle between the three is equilateral, you can add up the components of the magnetic field to produce a field that I believe would eminate upwards. I'm not really sure though, you definitely want a third opinion on this. I don't remember E&M very well but I'm just giving you questions that I would ask myself.
  • #6
NotaPhysicsMan said:
Ok, I know B2=B1? so All I have to do is find the net

which is B3 (going downards)=SQRT(B1^2+B2^2)
The magnetic field is a vector. Your equation would work only if B1 and B2 were perpendicular to each other. Here, they're not. Hint: how do you add vectors, in general?
  • #7
Break it down into it's x and y's and then sum? Off to bed :) Thanks for that hint, I'll try it tomorrow.
  • #8
Ok, so basically I have two right angle triangles split in half, with the x-directions cancelling each other and the y-direction adding together going downwards. Ok so each of the angles will give 60/2=30. So 1.133x10^-4 T(cos30)=9.812x10^-5T for the y and for x=5.665x10^-5 T. X's will cancel. And for Y's or the net magnetic field is:

(2 x 9.812x10^-5T)=1.96x10^-4 T downwards?

Correct now? :bugeye:
  • #9
Anyone want to say "correct or wrong" ? lol. *bump*
  • #10
Does it at least appear correct ? :cry:
  • #11
*BUMP* !! Come on, maybe I'll just re-post it then lol.
  • #12
Add only the y component of B_1 and B_2 since there x components cancel.
  • #13
Yes, that's what I proposed and did. Maybe an error?
  • #14
Ok, a freind of mine got 1.133x10^-4 T down??? That doesn't make any sense, that means all he did was plug in the numbers in the B formula...too simple?

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