# Torque and magnetism

## Homework Statement

The 10-turn loop of wire shown in the figure lies in a horizontal plane, parallel to a uniform horizontal magnetic field, and carries a 2.0 current. The loop is free to rotate about a nonmagnetic axle through the center. A 50 mass hangs from one edge of the loop.

What magnetic field strength will prevent the loop from rotating about the axle?

F = ILxB
torque = FR

## The Attempt at a Solution

First I attempt to sort out the torques to see what cancels what. Since the wire loops are suppose to stay parallel to the magnetic field, the sum of the torques should be zero. Another thing to note is that the magnetic field only acts on the wire where they go perpendicular to the field. Since there are 10 loops here, that means that we have 10 times the force on each side of the wire loop (F = 10ILxB). So

mgr-10ILBr-10ILBr=0
mgr-20ILBR=0
mg=20ILB
(mg)/(20IL) = B

(.05*9.8)/(20*2*.1) = .1225 Tesla

Does this seem right?

Thanks!

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