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futurepocket
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Homework Statement
A square loop of wire with side 0.60 m carries a current of 9.0 A as shown in the figure (http://img847.imageshack.us/i/springquestion.jpg/). When there is no applied magnetic field, the plane of the loop is horizontal and the nonconducting, nonmagnetic spring (k = 550 N/m) is unstretched. A horizontal magnetic field of magnitude 1.3 T is now applied. At what angle theta is the wire loop's new equilibrium position? Assume the spring remains vertical because theta is small. (HINT: Set the sum of the torques from the spring and the magnetic field equal to 0)
Homework Equations
tau = NIABsin(theta)
tau = rFsin(theta)
The Attempt at a Solution
Well, the hint says it all, except I can't seem to figure out the torque from the spring. I assume that since F = -kx, we put that in for the force for the torque, the 'r' is the distance from the axis of rotation to the spring (which I am assuming is 0.6 / 2 since it says its a square but the image doesn't look like a square). Unfortunately, I don't get the right answer when I do that because the sin(theta) can just cancel out on both sides.
Therefore, I considered that we assume the spring stays vertical, so sin(theta) for the torque from the spring doesn't exist.
So we get:
-rkx + NIABsin(theta) = 0
but we don't know 'x', the stretch of the spring. So I used Newton's laws, and summed up the forces and figured out that the force due to spring must be equal to the magnetic force on the spring... so I did that, and still didn't get the right answer. I am totally lost, any help?