Posted by **Jim** on Wednesday, December 8, 2010 at 12:19am.

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 10-6 N?

My Answer:

Magnetic field created by the second wire:

Fm = IlBsinθ

6*10-6= (4)(1)Bsin(90)

6*10-6= (4)B

B =6*10-6/4

B=1.5*10-6 T

This means the first wire will be in a magnetic field of 1.5*10-6 T.

First wire:

B=μ I / 2π r

r= μ I / 2πB

r=(4π*10-7)(9)/ 2π(1.5*10-6)

r= (1.130973355*10-5)/(9.424777961*10-6)

r=1.9999979

The wires must be separated a distance of 1.2m.

Is this correct? thanks for your help.

- Physics -
**drwls**, Wednesday, December 8, 2010 at 12:38am
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

- Physics -
**Jim**, Wednesday, December 8, 2010 at 12:00pm
Thanks a lot for your help, drwls. I just have one more problem. Why did you substitute (2*10-7) for μ. I thought μ= (4pi*10-7). If I substitute(4pi*10-7), 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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