Physics
posted by Jim .
Two straight and parallel wires of length 1.0 m carry a current  the first carries a currecnt of 9.0 A while the other wire carries a current of 4.0 A. What distance must separate the two straight and parallel copper wires if the force between them is to be 6.0 x 106 N?
My Answer:
Magnetic field created by the second wire:
Fm = IlBsinθ
6*106= (4)(1)Bsin(90)
6*106= (4)B
B =6*106/4
B=1.5*106 T
This means the first wire will be in a magnetic field of 1.5*106 T.
First wire:
B=μ I / 2π r
r= μ I / 2πB
r=(4π*107)(9)/ 2π(1.5*106)
r= (1.130973355*105)/(9.424777961*106)
r=1.9999979
The wires must be separated a distance of 1.2m.
Is this correct? thanks for your help.

I believe your formula is incorrect.
F/L = μ I1*I2/(2 pi r)
= 2*10^7 * 36/(2*pi*r) = 6*10^6 N
r = [1/(2 pi)]*6*10^1
That is about 0.095 meters
Reference:
http://theory.uwinnipeg.ca/physics/mag/node10.html 
Thanks a lot for your help, drwls. I just have one more problem. Why did you substitute (2*107) for μ. I thought μ= (4pi*107). If I substitute(4pi*107), my answer come out correct. Did you mean to cancel the 2pi in the numerator and denominator. Could you help me understand? Thanks again.
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