# Magnetic fields problem

1. Oct 29, 2013

### rbighouse

1. The problem statement, all variables and given/known data

The unit of magnetic flux is named for Wilhelm Weber. A practical-size unit of magnetic field is named for Johann Karl Friedrich Gauss. Along with their individual accomplishments, Weber and Gauss built a telegraph in 1833 that consisted of a battery and switch, at one end of a transmission line 3 km long, operating an electromagnet at the other end. Suppose their transmission line was as diagrammed in the figure below. Two long, parallel wires, each having a mass per unit length of 48.5 g/m, are supported in a horizontal plane by strings scripted l = 5.95 cm long. When both wires carry the same current I, the wires repel each other so that the angle between the supporting strings is θ = 15.8°.

2. Relevant equations

Not sure how to input them here...

FB/L = $\mu$I1I2/2pia

F = ma

3. The attempt at a solution

This is my first time posting here. Bare with my formatting please!

Draw free body diagram of one half of the problem. We are in equilibrium therefore Fnet is 0. That means T cos 7.9 = mg/L and T sin 7.9 = FB/L , where L is the 3km lenght.

so this can be reduced to tan 7.9 = FB/mg , but for my attempt I kept the /L on both sides since we are given a formula for FB/L and a mass per lenght.

I end up getting: tan 7.9 = $\mu$I^2 / (2$\pi$a mg/L), which I think is right.

My issue comes with the mathematics involved at this point. I end up with:

I = $\sqrt{2pia mg/L tan 7.9/ mu}$ , but the right answer has the tan 7.9 outside of the square root. How can that be?

Last edited: Oct 29, 2013
2. Oct 29, 2013

### Simon Bridge

Welcome to PF;

We should get a look at your working.
I also don't see the diagram. I think I know what it shows - the two wires are approximately straight so when they repel each other the whole length swings out the same amount and the cross-section looks like repelling pendulums. That about right?

Meantime - here's a hand with your formatting (use the "quote" button at the bottom of this post to see how it's done) $$\frac{F}{\Delta L} = \frac{\mu_0 I_1 I_2}{2\pi r}$$
$$I=\sqrt{ \frac{ 2\pi amg }{ L \tan(7.9/\mu) } }$$
... that what you meant to write?

How did you incorporate the distance between the wires?

Last edited: Oct 30, 2013