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The Magnetic Field-HELP

  • #1
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The Magnetic Field--HELP

I was hoping someone could help me on the following problem.

http://www.quantumninja.com/hw/question.jpg

I tried doing it with torques but nothing canceled out. The answer is supose to be 1.63A

If someone could walk me through it, that would be great... or at least help me get started.

(note: a B kinda got cut off in the diagram)
 

Answers and Replies

  • #2
Doc Al
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The net torque on the cylinder must be zero. What torque-producing forces act on it? (Show what you did that didn't work.)
 
  • #3
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[tex] N i r L B sin (90-\theta) = MR^2 [/tex]

[tex] i= \frac{r}{sin(90-\theta} [/tex]
 
  • #4
Doc Al
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Tom McCurdy said:
[tex] N i r L B sin (90-\theta) = MR^2 [/tex]
I assume this is your torque equation? Is the left hand side supposed to be the torque produced by the magnetic field on the current loop? What's the right hand side? (What is r? R?)

Again, what torque-producing forces act on the cylinder? We know one: the magnetic force. (That torque should be [itex]NILB(2R)\sin\theta[/itex].) What's the other?

Note: Don't forget that there must be translational equilibrium as well--consider forces parallel to the incline.
 
  • #5
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torque due to gravity?
 
  • #6
Doc Al
Mentor
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Tom McCurdy said:
torque due to gravity?
The weight of the cylinder acts through its center, so it exerts no torque about that point.

What other forces act on the cylinder?
 

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