Constant angular speed loop of wire

AI Thread Summary
The discussion focuses on calculating the torque required for a loop of wire rotating at a constant angular velocity in a magnetic field. The relevant equations include torque, induced electromotive force (EMF), and current, with participants exploring the relationships between these variables. There is uncertainty regarding the direction of forces acting on the loop as it rotates, and participants emphasize the importance of visualizing the scenario through sketches. Ultimately, a consensus emerges on using the torque equation and substituting in the current derived from Faraday's law. The conversation concludes with a sense of confidence in the approach to solving the problem.
darksyesider
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Homework Statement



You have a loop of wire (dimensions c and d) which is oriented vertically on the y-axis (the y-axis splits the rectangle in two equal pieces), and rotates about the y-axis at constant angular ##\omega##. The magnetic field is in the +i direction.

What would be the torqque needed to have the loop rotate at a constant w?


Homework Equations

##\tau = NIAB\sin\theta##

The Attempt at a Solution



##\mathcal{E} = B(cd)\omega \sin( \omega t )## from Faraday's law.

##I = \dfrac{B(cd)\omega\sin (\omega t)}{R}##

I am not sure how to proceed from here...what do I set the torque, ##\tau = NIAB\sin\theta## equal to?
 
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You have a current flow in a magnetic field. Do you see any forces from this?
 
I see ##F_B = ILB\sin\theta##, however, I can't tell which direction they are going in, and doesn't it change as the loop rotates?
 
however, I can't tell which direction they are going in
Did you draw a sketch?
In principle, you don't need that, if you use the cross-product.[/size].

and doesn't it change as the loop rotates?
Sure it does.
 
Yeah, I drew a sketch.
How about ##F_B = ILB \sin\omega t##?
From my sketch, it seems that all the forces cancel though... hint please :smile:
 
darksyesider said:
Yeah, I drew a sketch.
How about ##F_B = ILB \sin\omega t##?
Looks good.

From my sketch, it seems that all the forces cancel though... hint please :smile:
Then your sketch might be wrong. As I don't have the sketch, I can't say which part.
 
OH wait, would it be:

##\tau = IA\times B = IAB\sin (\omega t)##
## I(cd)B\sin\omega t##

Then plug in I from the equation in the original post ? i am fairly confident on this now.
 
Looks good.
 
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