Magnetic Field required to hold a loop in static equilibrium

[mmest]

Homework Statement

There is a square loop with mass of 4.0 kg and side length 4.5 m that is tilted upwards from the ground at a 25 degree angle with one side resting on the surface. The loop is a conductor that has 25 A of current going through it. The question asks to find the strength a uniform horizontal magnetic field to keep the loop in static equilibrium in that position.

Homework Equations

F = IL x B
F = qv X B

The Attempt at a Solution

F[/B] = IL x B = ILBsin(theta)
B = F/(25A*4.5m*sin(25 degrees)
I tried solving for B by getting force as equal to mg = 4 kg * 9.8 m/s^2, but the resulting B is incorrect. Where did I mess up?

 

[BvU]

with one side resting on the surface

Hello mmest, !

 

[Delta2]

Seems @BvU net/power went down while he was writing the message...

My advice would be to forget temporarily about the magnetic field, and view this as a pure mechanics problem. What hypothetical force (in magnitude and direction) and at which side of the square ,must be applied in order to keep the square loop at static equilibrium?

Once you solve the mechanics problem and find this hypothetical force, i believe you ll find easily the magnetic field that is required to have this force applied by the magnetic field.

On second thought you can't forget the magnetic field completely, given that it is horizontal (and if the general configuration is as I think it is, you should provide a schematic to be 100% sure)that should give you a hint on which side and at what direction the hypothetical force should be.

 

[rude man]

Remember that you have to account for the weight of all three sides. Does the fourth (ground) side matter at all?
Does the B field exert any force on the two upward-pointing sides? Does gravity?
You need to draw a body diagram of the loop and the forces applied to it, and the ensuing torques.

 

[kuruman]

It's probably more direct to balance the gravitational torque against the magnetic torque $\vec \mu \times\vec B$ where $\vec \mu~$is the loop's magnetic moment.

 

[rude man]

It's probably more direct to balance the gravitational torque against the magnetic torque $\vec \mu \times\vec B$ where $\vec \mu~$is the loop's magnetic moment.

Yeah, but magnetic moment is somewhat of an advanced concept, at least compared to F = mg and F = iL x B.