# Torque and magnetism

#### hover

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
F = ILxB
torque = FR

3. 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!

#### Attachments

• 12.3 KB Views: 798
Related Introductory Physics Homework News on Phys.org

#### hover

Never freaking mind. An hour later and nothing posted. I got it right anyway.

#### skr1p7k1dd

Then post what you did to get it right.

EDIT: Oh whoops. Nevermind. Your original post is 100% correct. I just screwed up my calculation.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving