Calculating Torque on a Circular Coil in a Magnetic Field

[Maxwellkid]

Homework Statement

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.

Homework Equations

F = IL x B

torque = y x B

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?

 

[ideasrule]

Homework Statement

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.

Homework Equations

F = IL x B

torque = y x B

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?

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.

 

[Maxwellkid]

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.

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.

 

[Maxwellkid]

Homework Statement

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.

Homework Equations

\vec F = I\vec L \times \vec B_0

dL = R d\theta

\vec {\tau} = \vec {\mu} \times \vec B_0

The Attempt at a Solution

I am trying to work this problem out from F = IL x B. FIRST, let's set up the loop of coil on the XY plane. The magnetic field will be in the positive x direction. The current will be flowing counter clockwise.
\vec F = I \vec L_1 \times \vec B_0
I\vec L_1 = <-IL_1 sin \theta , IL_1 cos \theta , 0> from 0 \rightarrow \frac {\pi}{2}
I\vec L_1 = <-IL_1 sin \theta , -IL_1 cos \theta , 0> from \frac {\pi}{2} \rightarrow {\pi}
I\vec L_1 = &lt;IL_1 sin \theta , -IL_1 cos \theta , 0&gt; from \pi \rightarrow <br /> \frac {-\pi}{2}
and finally,
I\vec L_1 = &lt;IL_1 sin \theta , IL_1 cos \theta , 0&gt; from \frac {-\pi}{2} \rightarrow 0
\vec B_0 = &lt;B_0, 0, 0&gt;

if you cross the Length vector with the 4 intervals with the magnetic field vector, you get the total force exerted by the magnetic field on the wire inward and outward. This is where I am stuck.

where would I place the total force inward and total force outward on this loop? Am I able to choose the two opposite ends of the loop and say all the inward force is at the right end of the loop and the outward force is on the left end of the loop?

 

[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...

 

[Maxwellkid]

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...

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}

 

[ideasrule]

yes, the 1 equation would be

\sum \vec F_{in} = \int d \vec F_{in} = \int -IL_1 B_0 R cos \theta \cdot \hat{k}
\sum \vec F_{out} = \int d \vec F_{out} = \int IL_1 B_0 R cos \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}

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.

Think about it this way:

(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!

 

[Maxwellkid]

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.

Think about it this way:

(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!

oooops! mistake! I've corrected it

 

[Maxwellkid]

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.

Think about it this way:

(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!

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?

 

[ideasrule]

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?

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?

 

[Maxwellkid]

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?

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

 

[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} = &lt;Rcos\phi , -Rsin\phi , 0&gt; from -\pi / 2 \rightarrow 0
\sum \vec F_{in} = &lt;0 , 0 , -2 I R B_x&gt;

\vec R_{in} = &lt;Rcos\phi , Rsin\phi , 0&gt; from 0 \rightarrow \pi / 2
\sum \vec F_{in} = &lt;0 , 0 , -2 I R B_x&gt;

stuck!

 

[Maxwellkid]

help!

 

[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

 

[Maxwellkid]

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

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

 

[ideasrule]

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

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

 

[Maxwellkid]

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

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?

 

[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!

 

[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?

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

 

[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.

 

[ideasrule]

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?

In your previous attempt, you found the total force, not the total torque. You have to find the torque on an infinitesimally small piece of wire and integrate over the coil to find the total torque.

 

[Maxwellkid]

I got it sir! THANK YOU VERY MUCH FOR YOUR GUIDANCE! I HOPE TO RETURN THE FAVOR SIR!