Calculating Torque on a Circular Coil in a Magnetic Field

In summary: 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 torque on the coil will require integrating both equations. The torque on the coil is F=ILxB, where F is the total torque, I is the current in the loop, L is the length of the loop, and B is the magnetic field strength. The maximum torque is found when the magnetic field is at its strongest and the current is flowing the most. The torque is equal to the product of the current and the magnetic field strength multiplied by the length of the loop.
  • #1
Maxwellkid
69
0

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?
 
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  • #2
Maxwellkid said:

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.
 
  • #3
ideasrule said:
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.
 
  • #4
Maxwellkid said:

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


[tex] \vec F = I\vec L \times \vec B_0[/tex]

dL = R [tex]d\theta[/tex]

[tex] \vec {\tau} = \vec {\mu} \times \vec B_0[/tex]

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.
[tex] \vec F = I \vec L_1 \times \vec B_0[/tex]
[tex] I\vec L_1 = <-IL_1 sin \theta , IL_1 cos \theta , 0> [/tex] from [tex] 0 \rightarrow \frac {\pi}{2}[/tex]
[tex] I\vec L_1 = <-IL_1 sin \theta , -IL_1 cos \theta , 0> [/tex] from [tex] \frac {\pi}{2} \rightarrow {\pi}[/tex]
[tex] I\vec L_1 = <IL_1 sin \theta , -IL_1 cos \theta , 0> [/tex] from [tex] \pi \rightarrow
\frac {-\pi}{2}[/tex]
and finally,
[tex] I\vec L_1 = <IL_1 sin \theta , IL_1 cos \theta , 0> [/tex] from [tex]\frac {-\pi}{2} \rightarrow 0[/tex]
[tex] \vec B_0 = <B_0, 0, 0> [/tex]

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?
 
  • #5
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
queenofbabes said:
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

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

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

[tex] \vec {\tau} = \vec R \times \vec F_{perpendicular}[/tex]
 
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  • #7
Maxwellkid said:
yes, the 1 equation would be

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

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

[tex] \vec {\tau} = \vec R \times \vec F_{perpendicular}[/tex]

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!
 
  • #8
ideasrule said:
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
 
  • #9
ideasrule said:
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?
 
  • #10
Maxwellkid said:
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?
 
  • #11
ideasrule said:
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
 
  • #12
[tex] \vec F_{in} = -I L B_x cos\phi \cdot k[/tex]
[tex] \vec F_{out} = I L B_x cos\theta \cdot k[/tex]


[tex] dL = R d\phi [/tex]
[tex] dL = R d\theta [/tex]
[tex] d\vec F_{in} = -I R B_x cos\phi d\phi \cdot k[/tex]
[tex] d\vec F_{out} = I R B_x cos\theta d\theta \cdot k[/tex]

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

[tex] \sum \vec F_{in} = -2I R B_x \cdot k[/tex]
[tex] \sum \vec F_{out} = 2I R B_x \cdot k[/tex]


[tex] \vec \tau_{1} = \vec R_{in} \times \sum \vec F_{in}[/tex]

[tex] \vec \tau_{2} = \vec R_{out} \times \sum \vec F_{out}[/tex]


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

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

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

stuck!
 
  • #13
help!
 
  • #14
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
ideasrule said:
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?
 
  • #16
Maxwellkid said:
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
 
  • #17
ideasrule said:
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?
 
  • #18
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
Maxwellkid said:
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 [tex] cos^2 \theta [/tex] instead of [tex] sin^2 \theta [/tex]
 
  • #20
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.
 
  • #21
Maxwellkid said:
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.
 
  • #22
I got it sir! THANK YOU VERY MUCH FOR YOUR GUIDANCE! I HOPE TO RETURN THE FAVOR SIR!
 

1. What is torque on a coil of wire?

Torque on a coil of wire refers to the rotational force that is exerted on a coil of wire when it is placed in a magnetic field and an electrical current is passed through it.

2. How is torque on a coil of wire calculated?

The torque on a coil of wire is calculated using the equation T = N * I * A * B * sin(theta), where T is the torque, N is the number of turns in the coil, I is the current flowing through the coil, A is the area of the coil, B is the strength of the magnetic field, and theta is the angle between the coil and the magnetic field.

3. What factors affect the torque on a coil of wire?

The torque on a coil of wire is affected by several factors, including the number of turns in the coil, the strength of the magnetic field, the current flowing through the coil, the size and shape of the coil, and the angle between the coil and the magnetic field.

4. How does the direction of the current affect the torque on a coil of wire?

The direction of the current flowing through the coil affects the direction of the torque on the coil. If the current is flowing in the same direction as the magnetic field, the torque will be in one direction, and if the current is flowing in the opposite direction, the torque will be in the opposite direction.

5. What are some real-world applications of torque on a coil of wire?

Torque on a coil of wire is used in a variety of real-world applications, including electric motors, generators, and loudspeakers. It is also used in devices such as galvanometers and electromechanical instruments to measure electric current.

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