# Torque on a Coil

1. Jun 30, 2009

### Maxwellkid

1. The problem statement, all variables and given/known data
a circular coil of wire has a diameter of 2 m and contains 10 loops. The current in each loop is 3 amps, and the coil is placed in a 2 tesla magnetic field. Determine the maximum torque exerted on the coil by the magnetic field.

2. Relevant equations
F = IL x B

torque = y x B

3. The attempt at a solution

I am trying to work this problem out from F = IL x B. I get stuck after I get to integrating the total torque on the circular coil. can you help???

2. Jun 30, 2009

### ideasrule

There's a general equation relating torque to the number of loops, current, coil area, and magnetic field strength. It's T=NIABsin theta; with this formula, your question should be trivial.

3. Jun 30, 2009

### Maxwellkid

i'm not in the habit of memorizing formulas and plugging in values. I'd rather derive starting from the basics.... please don't throw out no brainer formulas... it's a waste of time.

4. Jul 2, 2009

5. Jul 2, 2009

### queenofbabes

Don't get mired thinking in cartesians! You just need to integrate one equation going from 0 to 2pi. Start by thinking about a small element of the loop....

6. Jul 2, 2009

### Maxwellkid

yes, the 1 equation would be

$$\sum \vec F_{in} = \int d \vec F_{in} = \int -IB_0 R cos \theta d\theta \cdot \hat{k}$$
$$\sum \vec F_{out} = \int d \vec F_{out} = \int I B_0 R cos \theta d\theta \cdot \hat{k}$$

what do I do next????? and where do these forces apply so that I can properly apply the torque equation.

$$\vec {\tau} = \vec R \times \vec F_{perpendicular}$$

Last edited: Jul 2, 2009
7. Jul 2, 2009

### ideasrule

Since we're taking an integral, you should have dL=Rdθ rather than "L" inside the integrand. Since we're trying to calculate torque, calculating the total force wouldn't be particularly useful.

(1) The coil must rotate around an axis that passes through its center.
(2) What is the force on an infinitesimally small piece of coil? Express this in terms of θ (angular displacement from the axis), I, B, and dL=Rdθ.
(3) What's the moment arm for this small piece of coil? What's the torque?
(4) Integrate over the entire coil and you're done.

For an extra challenge, try deriving the formula I gave you, T=NIABsinθ, for an arbitrary shape. That's quite challenging!

8. Jul 2, 2009

### Maxwellkid

oooops! mistake! I've corrected it

9. Jul 2, 2009

### Maxwellkid

for each radius vector that extends out to the circle, there is a different magnitude of the Force component. How would you set it up???

10. Jul 2, 2009

### ideasrule

Consider just one tiny piece of coil of length dL=Rdθ. Since this piece is infinitesimally small, it's also infinitely straight, so the force on it would be F=IBdLsinθ. So, what's the moment arm and what's the torque on this small piece of coil?

11. Jul 3, 2009

### Maxwellkid

torque would be r x IBdLsin@. But dL sin@ is not the magneticforce on the coil. IBr sin@ is the magnetic force

12. Jul 5, 2009

### Maxwellkid

$$\vec F_{in} = -I L B_x cos\phi \cdot k$$
$$\vec F_{out} = I L B_x cos\theta \cdot k$$

$$dL = R d\phi$$
$$dL = R d\theta$$
$$d\vec F_{in} = -I R B_x cos\phi d\phi \cdot k$$
$$d\vec F_{out} = I R B_x cos\theta d\theta \cdot k$$

$$\sum \vec F_{in} = -I R B_x \int_{-\pi / 2}^{\pi / 2} cos\phi d\phi \cdot k$$
$$\sum \vec F_{out} = I R B_x \int_{-\pi / 2}^{\pi / 2} cos\theta d\theta \cdot k$$

$$\sum \vec F_{in} = -2I R B_x \cdot k$$
$$\sum \vec F_{out} = 2I R B_x \cdot k$$

$$\vec \tau_{1} = \vec R_{in} \times \sum \vec F_{in}$$

$$\vec \tau_{2} = \vec R_{out} \times \sum \vec F_{out}$$

THIS IS WHERE I AM STUCK!!!!! TORQUE 1 AND TORQUE 2 ADDED TOGETHER GIVES ME THIS!!!!! WHICH IS WRONG!!!

$$\vec R_{in} = <Rcos\phi , -Rsin\phi , 0> from -\pi / 2 \rightarrow 0$$
$$\sum \vec F_{in} = <0 , 0 , -2 I R B_x>$$

$$\vec R_{in} = <Rcos\phi , Rsin\phi , 0> from 0 \rightarrow \pi / 2$$
$$\sum \vec F_{in} = <0 , 0 , -2 I R B_x>$$

stuck!!!!!!

13. Jul 6, 2009

### Maxwellkid

help!!!

14. Jul 6, 2009

### ideasrule

I might as well give you the answer, since this question has been up for this long. I said before that the force on an infinitesimally small piece of coil would be dF=IBdLsinθ=IBRsinθdθ. The moment arm would be Rsinθ (if you can't see this, draw a diagram). Torque=force * moment arm=IBR2sin2(θ)dθ Integrate that from 0 to 2π and you get the answer: IBπR2. With N coils of wire, total torque would be T=NIBπR2

15. Jul 6, 2009

### Maxwellkid

there needs to be a pi R squared in the final equation. and what is the integral of sin squared of theta????

16. Jul 6, 2009

### ideasrule

But there is a pi R squared; look closely: T=NIBπR2. That's NIB times pi*R2.

As for the integral of sin squared theta, remember that cos 2θ=1-2sin2θ, so sin2θ=(1-cos 2θ)/2

17. Jul 6, 2009

### Maxwellkid

would you look through my steps and tell me what I'm doing wrong. I can get to the final equation if I don't take a step by step detailed approach....

Must I wait to take the integral of Cos until after I cross it with the radius to find the torque???

18. Jul 7, 2009

### Maxwellkid

I think I got it...but I ended up getting 2 pi radius squared in the final equation instead of Pi radius squared.

Darn!!!

19. Jul 7, 2009

### Maxwellkid

Also, I took the integral of $$cos^2 \theta$$ instead of $$sin^2 \theta$$

20. Jul 7, 2009

### ideasrule

How is the integral of (1-cos2θ)dθ/2 from 0 to 2pi 2piR^2? The integral of cos2θdθ from 0 to 2pi is 0 because we're integrating over two cycles. The integral of dθ/2 from 0 to 2pi is pi.