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Magnetic forces exerted by current-carrying wires

  • Thread starter SA32
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  • #1
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Homework Statement


The figure (http://img.photobucket.com/albums/v80/northerndancer/thefigure.jpg?t=1173558797 [Broken]) is a cross section through three long wires with linear mass density 50 g/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. What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

Homework Equations


F=I1LB2 (magnetic force between two parallel wires)
F=mg (force due to an object's weight)



The Attempt at a Solution


The two bottom wires exert attractive magnetic forces on each other because they are parallel and both of their currents move in the same direction. In order for the third wire to "float", it must be repelled by the two lower wires by a magnetic force equal and opposite to its weight. But:

-The upper wire is not parallel to the lower wires, so I'm not sure how to calculate the force each lower wire exerts on the upper wire.
-How do I calculate the mass of the upper wire if I don't know its length?

Thanks for any hints.
 
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Answers and Replies

  • #2
Doc Al
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In order for the third wire to "float", it must be repelled by the two lower wires by a magnetic force equal and opposite to its weight.
Good!

But:

-The upper wire is not parallel to the lower wires, so I'm not sure how to calculate the force each lower wire exerts on the upper wire.
Sure they are parallel. They are not in the same vertical line, but all three wires are parallel. What direction is the force from each lower wire on the top wire? Hint: Consider the vertical and horizontal components of those forces.
-How do I calculate the mass of the upper wire if I don't know its length?
Think in terms of force per unit length and mass per unit length.
 
  • #3
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Okay I think I see, the top wire is parallel to each of the bottom wires, just the force from each has both an x-component and y-component. But the x-components cancel out and so the net magnetic force on the top wire is just straight up, isn't it?

And looking again at the equation for the force between parallel wires, I think I get what you mean about the mass. The equation for magnetic force involves L, and then if I equate it to mg and make "m" = (50 g/m)*L the lengths should cancel out?
 
  • #4
Doc Al
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Exactly right.
 

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