- #1
flyingpig
- 2,579
- 1
Homework Statement
[PLAIN]http://img814.imageshack.us/img814/4456/84684200.png
The Attempt at a Solution
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So from the symmetry I can conclude these
[tex]Tcos\phi= mg[/tex]
[tex]Tsin\phi = F[/tex]
F is the force (reaction force) exerted by the other wire. I am only looking at one of the wires
So I can solve for it in terms of force per mass and I get
[tex]gtan\phi = \frac{F}{m}[/tex]
Now the reaction force is also
[tex]\vec{F} = I\vec{d} \times \vec{B}[/tex]
Now I define [tex]\lambda = \frac{m}{d}[/tex] and then [tex]d = \frac{m}{\lambda} [/tex]
So now
[tex]\vec{F} = I\vec{d} \times \vec{B}[/tex]
[tex]\frac{F}{d} = IB[/tex]
[tex]\frac{F}{d} = I\frac{\mu_0 I}{2\pi x}[/tex]
Where x is the distance between the two wires and I had to use the law of cosine to get it
[tex] x = l\sqrt{2 - 2cos\theta}[/tex]
[tex]\frac{F}{\frac{m}{\lambda} } = I\frac{\mu_0 I}{2\pi x}[/tex]
[tex] \lambda gtan\phi = \frac{\mu_0 I^2}{2\pi x}[/tex]
Solving for I, I get
[tex] \sqrt{\frac{2\pi \lambda x gtan\phi}{\mu_0}}= I[/tex]
The book gives me 67.8A, which I don't understand why
I also tried
[tex] \sqrt{\frac{2\pi x gtan\phi}{\mu_0 \lambda }}= I[/tex]
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