# Homework Help: Need help with torque on a rectangular coil

1. Dec 19, 2004

### Sanosuke Sagara

i have think very long on this question but still can't figure out what is meant by the question.I have my question and my doubt on the attachment that followed.

#### Attached Files:

• ###### rectangular.doc
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Last edited: Dec 19, 2004
2. Dec 19, 2004

### Sanosuke Sagara

I really need somebody to help me figure out what the question want and I really need someone to give me the explaination to the question that I asked.Thanks for anybody that spend sometime on this question.

3. Dec 19, 2004

### Staff: Mentor

"When the coil is in equilibrium" means that the net force (and torque) on the coil is zero. (What are the forces acting on the coil?) The arrow on the coil shows the direction of the current.

4. Dec 19, 2004

### Sanosuke Sagara

Thanks for mentor help.I really appreaciate it.

5. Dec 20, 2004

### Sanosuke Sagara

I want to ask whether the angle tetha is made between 'the magnetic field B and the plane coil' or between 'the magnetic field B and the normal of the plane' ?The reason I ask this phrase 'the magnetic field B and the normal of the plane' is because the dotted line in the diagram made me confuse whether the dotted line means the normal of the plane or there is another meaning on the dotted line.

The attachment that followed is still the same but I forgot to put on a word 'B' on the arrow that are perpendicular to the axis XY.

I hope that somebody will help me figure out my doubt and give explaination to my question.Thanks for anybody that spend some time on this question.

#### Attached Files:

• ###### rectangular.doc
File size:
20 KB
Views:
76
6. Dec 20, 2004

### HallsofIvy

You said "rectangular coil". If that is true then theta, in the picture, must be 90 degrees, a right angle. In any case, it is irrelevant to the problem. The only thing that matters to the force is the angle between the plane of the coil and the magnetic field which is 90 degrees.

7. Dec 20, 2004

### Sanosuke Sagara

My solution the question but still can't find the answer.

I have my solution to the question in the attachment that followed.Unfortunately ,I can't find the answer that match the answer behind the textbook.

I hope that someone will help me figure out where have I done wrong and explain to me.Thanks for anybody that spend some time on this question.

#### Attached Files:

• ###### Doc1.doc
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91
8. Dec 20, 2004

### Staff: Mentor

As I interpret the diagram, the angle theta is the angle that the plane of the coil makes with the vertical axis (not the normal to the plane).

9. Dec 20, 2004

### Staff: Mentor

For some reason, you set a torque equal to a force, which is incorrect. (The units won't even match!)

In any case, the torque on the coil (measured about the axis of rotation) due to the weight of the coil (2 sides of length a and one of length b contribute to this torque) must equal the opposing torque due to the magnetic force on the current.

10. Dec 20, 2004

### Sanosuke Sagara

I'm sorry mentor that I have used wrong solution to solve this question and have misunderstand with the used of formula.

Does mentor get this answer tan tetha = 2IBab / mg ?

11. Dec 20, 2004

### Staff: Mentor

Almost. The answer I get is: $tan\theta = 2IBb/mg$.

12. Dec 20, 2004

### learningphysics

I also get tan theta=2IBb/mg.

At equilibrium: total torque about any axis = 0.

Calculate the total torque about the XY axis, and set it equal to zero.

Remember that
$$\vec \tau = \vec r X \vec F$$
$$|\vec \tau |= |\vec r| X |\vec F| sin\alpha$$

13. Dec 21, 2004

### Sanosuke Sagara

Need someone to confirm my solution to the question is right or wrong

I have my solution in the attachment that followed.Thanks for anyone who help me figure out with my solution.

#### Attached Files:

• ###### Doc2.doc
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28.5 KB
Views:
111
14. Dec 21, 2004

### Gamma

Torqe produced by the weight is not Fa Sin (theta). Rather, it is F a/2 Sin (theta)
Also, Torque is defined as follows: F X L. Not 2 * F X L. Making these corrections, you will get,
F a/2 Sin (theta) = I b B Cos (theta)

This gives, tan(theta) = 2 IbB / mg

gamma.

15. Dec 21, 2004

### Staff: Mentor

Here's how I would analyze this problem: First realize that the weight exerts a "downward" torque while the magnetic force exerts an "upward" torque. Let's find the two torques and set them equal.

(1) Downward torque due to the weight of the coil: The weight of the coil is mg acting at its center of mass, so the downward torque equals:
$$\tau_{down} = mg (a/2) sin\theta$$

(2) Upward torque due to the magnetic force: Only the force on the "b" side will contribute to the torque. The magnetic force is $IbB$ (pointing outward); thus the upward torque is:
$$\tau_{up} = IbBa cos\theta$$

Now set $\tau_{down} = \tau_{up}$:
$$mg (a/2) sin\theta = IbBa cos\theta$$
Thus:
$$tan\theta = 2IbB/mg$$

(Edit: Original solution posted was needlessly complicated. )

Last edited: Dec 21, 2004
16. Dec 21, 2004

### Sanosuke Sagara

Finally ,I have understand with the question and why my solution is not accepted.Lastly , I want to sany thank you to everybody that try to figure out my problem including Learning physics ,Gamma , Doc Al and HallsofIvy mentor. I have finally understand with this question.