What current is needed for an equilateral triangle formation with three wires?

[dongeto]

Homework Statement

The figure is a cross section through three long wires with linear mass density 52.0g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. (please see attachment) Just in case attachment is not working (top wire is going out of page and above lower wires, which are going into the page.) What current will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

View attachment 3 wires.doc

Homework Equations

F=uLI^2/(2pid)

F=mg

(please see attached for better format for equations)

ATTACH]13181[/ATTACH]

The Attempt at a Solution

since current in lower wires are in opposite direction from top wires, therefore they repel each other. The y-components from each lower two wires must equal force of gravity of top wire:

mg = 2 x uLI^2/(2pid)sin30
solve for I, assume L=1 m and where d=0.04m, mg=0.5096N
I=319A...but why is this not the right answer?

 

[Doc Al]

The y-components from each lower two wires must equal force of gravity of top wire:

mg = 2 x uLI^2/(2pid)

Looks like you forgot to take the y-component of the magnetic forces.

 

[dongeto]

oh sorry...I forgot to put sin 30 in the equation but i still got 319A

I= sqrt [(0.5096*pi*0.04)/((4pi*10^-7)*sin 30)]=319A

 

[Doc Al]

I get the same answer. Who says it's wrong?

 

[dongeto]

i entered that answer in mastering physics and its incorrect

 

[dongeto]

can someone please help...this is due tmr...thanks

 

[Doc Al]

The angle of each force with the horizontal is 60 degrees, not 30; so you should use sin(60). (D'oh! I made the same error.)

 

[dongeto]

k...i get it...thank you very much Doc Al!