I did everything right, but answer Key is wrong right?

[flyingpig]

Homework Statement

[PLAIN]http://img814.imageshack.us/img814/4456/84684200.png

The Attempt at a Solution

http://img845.imageshack.us/img845/9151/11477498.th.png

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So from the symmetry I can conclude these

$$Tcos\phi= mg$$

$$Tsin\phi = F$$

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

$$gtan\phi = \frac{F}{m}$$

Now the reaction force is also

$$\vec{F} = I\vec{d} \times \vec{B}$$

Now I define $$\lambda = \frac{m}{d}$$ and then $$d = \frac{m}{\lambda}$$

So now

$$\vec{F} = I\vec{d} \times \vec{B}$$

$$\frac{F}{d} = IB$$

$$\frac{F}{d} = I\frac{\mu_0 I}{2\pi x}$$

Where x is the distance between the two wires and I had to use the law of cosine to get it

$$x = l\sqrt{2 - 2cos\theta}$$

$$\frac{F}{\frac{m}{\lambda} } = I\frac{\mu_0 I}{2\pi x}$$

$$\lambda gtan\phi = \frac{\mu_0 I^2}{2\pi x}$$

Solving for I, I get

$$\sqrt{\frac{2\pi \lambda x gtan\phi}{\mu_0}}= I$$

The book gives me 67.8A, which I don't understand why

I also tried

$$\sqrt{\frac{2\pi x gtan\phi}{\mu_0 \lambda }}= I$$

 

[SammyS]

The angle between the vertical and the string is θ/2, so your first two equations are in error.

 

[flyingpig]

What do you mean θ/2? WHy did you cut it in half?

 

[flyingpig]

Oh wait, never mind I gotcha.

 

[flyingpig]

ahahaha i got the answer now!