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Torque on Circular Current 
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#1
Feb1609, 02:56 PM

P: 16

1. The problem statement, all variables and given/known data
Note getting the right answer. Question: a). What is the magnitude of the torque on the circular current loop in the figure? b). What is the loop's equilibrium position. 2. Relevant equations Torque = [(I*A)*B*(sin(theta))] L= 2.0cm a= 2.0mm Iwire = 2.0A Iloop = 0.20A 3. The attempt at a solution First, I tried to find theta using tan^1(x/y). Second, I tried to find B, using B= [(1.257E6T*(m/A)*(I)] /(2*pi*r). Finally, I tried Torque = [(I*A)*B*(sin(theta))]. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Feb1609, 03:10 PM

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#3
Feb1609, 03:17 PM

P: 16

So my angle is 90deg, and I didn't need to solve for it?



#4
Feb1609, 03:28 PM

P: 183

Torque on Circular Current
What did you use for r? It's not L, you have to use the pythagorean theorem with L and a as your legs.



#5
Feb1609, 03:38 PM

P: 16

For r, I did the square root of [(L)^2+(a/2)^2].



#6
Feb1609, 09:19 PM

P: 183

OK good, you've got that. Maybe:
1) Did you account for both torques, one for each wire in the loop? 2) Did you use the correct area for the loop? 3) Are the directions correct? I think the force on the bottom piece would be up and to the left, and the force on the top would be up and to the right.. If not that I can't see what else might be wrong. 


#7
Feb1609, 10:13 PM

P: 16

1). Aren't both torques the same, so I would double it?
2). Isn't the area A= [(pi)*(R)^2]? 3). I don't think direction is important, because it just wants the magnitude of torque. Let me know I made any bad assumptions on these 3. **Also, can someone confirm that the angle in my calculation for Torque = [(I*A)*B*(sin(theta))] is 90deg because of the figure being perpendicular. Thanks so far 


#8
Feb1709, 09:26 PM

P: 16

Still could use a reply, to my last post (especially about the 90deg).
Thanks 


#9
Feb1809, 06:21 AM

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P: 41,316




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