Magnetic force between two parallel hanging wires

[c4795]

Magnetic force between two parallel hanging wires (solved)

Homework Statement

Two long, parallel wires hang by 4.00-cm-long cords from a common axis. The wires have a mass per unit length of 1.30×10⁻² kg/m and carry the same current in opposite directions. What is the current in each wire if the cords hang at an angle 6° with the vertical?

Here is the diagram: http://i.imgur.com/alycY.jpg

Homework Equations

Magnetic force between two parallel lines of current

$$F_{B} = L \frac{\mu_{0}I_{1}I_{2}}{2\pi r}$$

The Attempt at a Solution

Attempt 1

Well first I tried setting the gravitational potential energy equal to the work done by the magnetic force.

$$L \times g \times d \times (1.3 \times 10^{-2}) = \int_a^b{F_{B} \mathrm{d}r}$$

Where L is the length of the wires.

I solved for the distance the wires would move against gravity, d, using trig, and since the currents are the same I₁I₂ becomes I².

$$L \times g \times (0.04 - 0.04\mathrm{cos}(6^{\circ})(1.3 \times 10^{-2}) = \int_a^b{F_{B} \mathrm{d}r}$$

The integral simplifies to a constant times the integral of 1/r, and the L's cancel out.

The limits of the integral can be solved by trigonometry, what I thought they'd be is 0 to 2*0.04*sin(6°). However, 1/r does not converge at 0 so this blows up.

Attempt 2

I thought I might be over complicating it so I attempted setting the force of gravity equal to the magnetic force:

$$L \times g \times (1.3 \times 10^{-2}) = L \frac{\mu_{0}I^{2}}{2\pi r}$$

Again the L's cancel out, and if I use r = 2 * (0.04 - 0.04*cos(6°)) I get the answer I = 16.7 A, which doesn't work.

Attempt 3

I tried drawing a FBD, where there's tension T along the cords and it's equal to <-F_B, -F_g>. I'm not sure I can go anywhere from here.

Any help will be greatly appreciated.

 

[Inna]

What do you have to find?

 

[c4795]

What do you have to find?

Oh so sorry, the question is to find the current I in each wire, if the cords hang at 6 degrees with the vertical (as in the picture). I'll edit my post.

 

[Inna]

Free body diagram for one of the wires is a great idea. You are right that veritical component of tension should be equal to the force of gravity and horizontal component is equal to magnetic force between the wires. This gives you two equations with two unknowns.
Good luck!

 

[c4795]

Free body diagram for one of the wires is a great idea. You are right that veritical component of tension should be equal to the force of gravity and horizontal component is equal to magnetic force between the wires. This gives you two equations with two unknowns.
Good luck!

I am having trouble understanding how the two equations relate so that I can solve them.

 

[Inna]

Can you try to write these equations?

 

[c4795]

Yes, they are:

$$T_{x} = -F_{B} = -L \frac{\mu_{0}I^{2}}{4\pi(0.04-0.04sin(\theta))}$$

with r = 2 * (0.04 - 0.04sin(t)), and

$$T_{y} = -F_{g} = L (1.3\ast 10^{-2}) g$$

I am not sure where to go from here. Does it perhaps involve torque?

 

[Inna]

T_x = Tsin(6), T_y = Tcos(6)

I am not sure if your expression for the distance between the wires is correct if theta is 6 degrees. Would not it be r = 2*(.04)*sin(6)?

 

[c4795]

T_x = Tsin(6), T_y = Tcos(6)

I am not sure if your expression for the distance between the wires is correct if theta is 6 degrees. Would not it be r = 2*(.04)*sin(6)?

Thank you very much, I was able to solve it. I can't believe I overlooked that . And yes you're correct it's r = 2*0.4*sin(6). Thanks again, I'm going to make the TeX a bit nicer and mark the thread resolved .