# Torque on Current Loops

## Homework Statement

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

Torque = u * B
Torque = (I*A) B

## The Attempt at a Solution

Ive tried all sorts of things I konw that I is 2.0 A and A is .1 * .05 so that gives .01 * B = Torque and Ive set torque equal to just the F of the weight times the radius and diameter and Ive also tried doubling the F because the force down on both sides. No matter what Ive tried it doesn't seem to work. Answers that I have gotten are 1.225. 2.45, 4.9, and .6125 so if you try this problem and get those they are wrong. Thanks for the help.

#### Attachments

• knight_Figure_32_80.jpg
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If there are 10 turns in the loop, wouldn't you use the equation that incorporates the # of turns?

As far as I know, the 2nd equation you stated holds true for rectangular loops (I can't see your attachment so I don't really know what the problem set up is).

Additionally, if the loops is parallel to the magnetic field...what would sin(theta) be?

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I can't find any equations in my book for turns in the loop so if you could help me out with that Id appriciated it and I was wondering that myself as for the picture it is set up as a rectangular loop so I wasn't sure that the turns ment anything and wouldn't it make the sin of theta 90 degrees?

For any other loop that is not a rectangle, torque=NIABsin(theta), where N=# of turns.

ok got it thanks for the help.

ok got it thanks for the help.[/QUO
What is the answer for this question for the existing numbers? Can you help?

Did anybody come up with an answer for this question?