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Magnetic force between two parallel hanging wires (solved)
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
Magnetic force between two parallel lines of current
F_{B} = L \frac{\mu_{0}I_{1}I_{2}}{2\pi r}
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 <-FB, -Fg>. I'm not sure I can go anywhere from here.
Any help will be greatly appreciated.
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 <-FB, -Fg>. I'm not sure I can go anywhere from here.
Any help will be greatly appreciated.
Last edited:
. 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 
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