How Can a Simple Electric Motor Be Created Using a Coil in a Magnetic Field?

[CAF123]

Homework Statement

A coil is formed by wrapping 80 cm of thin wire round a non-conducting square of size 2 × 2 cm which is placed in a horizontal magnetic field of strength 20mT.

A)Sketch the orientation of the coil that will result in the maximum torque acting, and the value of this torque when the coil is carrying a current of 200mA.
B) Explain how the above system can be modified to produce a simple electric motor where the coil rotates in the magnetic field.

The Attempt at a Solution

A)We have derived in class that the torque on a coil suspended in a B field is $\tau = AIB\sin \theta$. So this will be maximal when $\theta = \pi/2$ which means $\tau = (4 \times 10^{-4} m^2 \cdot (200 \times 10^{-3} A) \cdot (20 \times 10^{-3})T \approx 1.6 \times 10^{-6}Nm$

Is this correct? I haven't used the length of the wire here but I guess I assumed it covered the whole sqaure. If it is 80cm, then each full turn will give 4cm of the wire used which means to wrap the whole wire around the sqaure I need 20 turns. However, I don't know the thickness of the wire so I am unable to tell how much of the 2x2 square those 20 turns take up. Is that 80cm a distractor?

B) I get the general idea about having to switch the direction of current every 180 degrees so as to reverse the magnetic force acting on the coil. But I have a question. Say the B field is horizontal to the left. Put the square parallel to the field such that a current runs clockwise. Then the magnetic force tends to tilt the sqaure upwards until it is perpendicular to the B field. The torque is then zero and the square will fall back down. How then is it possible to get the square to exceed 90 + degrees? I thought it could simply be inertia which makes it just exceed the 90 mark, but then if this is the case then since the current in the square is now flipped, this tends to push the sqaure back towards the 90 mark. So how is it even possible to get the sqaure to 180 degrees? (I hope this makes sense, if not I will attach a sketch).

Many thanks.

 

[rude man]

Homework Statement

A coil is formed by wrapping 80 cm of thin wire round a non-conducting square of size 2 × 2 cm which is placed in a horizontal magnetic field of strength 20mT.

A)Sketch the orientation of the coil that will result in the maximum torque acting, and the value of this torque when the coil is carrying a current of 200mA.
B) Explain how the above system can be modified to produce a simple electric motor where the coil rotates in the magnetic field.

The Attempt at a Solution

A)We have derived in class that the torque on a coil suspended in a B field is $\tau = AIB\sin \theta$. So this will be maximal when $\theta = \pi/2$ which means $\tau = (4 \times 10^{-4} m^2 \cdot (200 \times 10^{-3} A) \cdot (20 \times 10^{-3})T \approx 1.6 \times 10^{-6}Nm$

Is this correct? I haven't used the length of the wire here but I guess I assumed it covered the whole sqaure. If it is 80cm, then each full turn will give 4cm of the wire used which means to wrap the whole wire around the sqaure I need 20 turns. However, I don't know the thickness of the wire so I am unable to tell how much of the 2x2 square those 20 turns take up. Is that 80cm a distractor?

No. You need to take into consideration the number of turns in the coil, which you have correctly computed to be 20.

B) I get the general idea about having to switch the direction of current every 180 degrees so as to reverse the magnetic force acting on the coil. But I have a question. Say the B field is horizontal to the left. Put the square parallel to the field such that a current runs clockwise. Then the magnetic force tends to tilt the sqaure upwards until it is perpendicular to the B field. The torque is then zero and the square will fall back down. How then is it possible to get the square to exceed 90 + degrees? I thought it could simply be inertia which makes it just exceed the 90 mark, but then if this is the case then since the current in the square is now flipped, this tends to push the sqaure back towards the 90 mark. So how is it even possible to get the sqaure to 180 degrees? (I hope this makes sense, if not I will attach a sketch).

Many thanks.

Better attach a sketch ...

 

[CAF123]

Hi rude man,

No. You need to take into consideration the number of turns in the coil, which you have correctly computed to be 20.

I thought as much, but I don't know the thickness of the wire. So assuming I start winding the wire round right at the top of the square then to get the whole wire around the square I require 20 turns. But I believe in order to find the area of the coil I need to find $A = 2 \times x$, $0 < x \leq 2$.

Better attach a sketch ...

Ok, I will do.

Edit: See attached sketch

 

[rude man]

Don't worry about the thickness of the wire. Pretend it's zero.

 

[CAF123]

I presume we are also meant to assume that the winding of the coil does not overlap. If we start at the top and wind all the way round once, then that will be 4 cm of the wire wrapped around. How do we know how much area this would cover (clearly 2cm along, but how much down?)

 

[rude man]

I presume we are also meant to assume that the winding of the coil does not overlap. If we start at the top and wind all the way round once, then that will be 4 cm of the wire wrapped around. How do we know how much area this would cover (clearly 2cm along, but how much down?)

See my previous post.

You have a coil of dimensions 2 cm x 2 cm. That is a perimeter of 8 cm. So I was wrong too - you will have 80/8 = 10 infinitely thin turns of wire. So N = 10 and A = 4 cm^2. Change that to m^2 if I were you.

 

[CAF123]

If the coil is a planar square, then will it not be 20 turns? (I can see that if the wire was wrapped around a central core for example then it would be 10).

 

[rude man]

The wire is wrapped about the periphery. That makes it 10 turns.

Assuming you get past this nuance, have you decided on a max. torque orientation yet?

 

[CWatters]

What Rude_man said. The perimeter is 8cm, wire is 80cm so 10 turns. Assume wire is very thin so all turns occupy the same space.

More here..

http://www.schoolphysics.co.uk/age1...tromagnetism/text/Torque_on_a_coil/index.html

I remember the formulae as Tmax = BAIN.

 

[CAF123]

The wire is wrapped about the periphery. That makes it 10 turns.

Assuming you get past this nuance, have you decided on a max. torque orientation yet?

How can you tell it is wrapped around the perimeter? I was assuming it was wrapped lengthways. The maximal torque is when the the coil and B field are parallel.

What Rude_man said. The perimeter is 8cm, wire is 80cm so 10 turns. Assume wire is very thin so all turns occupy the same space.

More here..

http://www.schoolphysics.co.uk/age1...tromagnetism/text/Torque_on_a_coil/index.html

I remember the formulae as Tmax = BAIN.

So $\tau = 1.6 \times 10^{-5} N m$. Do you have any ideas about my problem regarding the electric motor? I get the idea behind it, but I had a question about it. (see my post 2 for the sketch)

 

[rude man]

. The maximal torque is when the the coil and B field are parallel.

You need to be specific about the orientation of the coil. It's best to talk of the normal to the coil's area. So where does the normal vector point with respect to the B field ?

So $\tau = 1.6 \times 10^{-5} N m$. Do you have any ideas about my problem regarding the electric motor? I get the idea behind it, but I had a question about it. (see my post 2 for the sketch)

Take it one step at a time to determine the torque:

1. what is the force on one side of the coil?
i = 0.2A
l = 2cm = 0.02m
B = 0.02T
10 turns
F = i l x B = ?

2. Torque = F x d, what is d here?

3. Repeat for the other side.

 

[CAF123]

You need to be specific about the orientation of the coil. It's best to talk of the normal to the coil's area. So where does the normal vector point with respect to the B field ?

If the coil is wrapped around the perimeter of the square then when the square is in the plane of the B field the torque is maximal (i.e the normal vector of the square points in the upwards direction, perpendicular to the B field.)

Take it one step at a time to determine the torque:

1. what is the force on one side of the coil?
i = 0.2A
l = 2cm = 0.02m
B = 0.02T
10 turns
F = i l x B = ?

Yes, so since B and L are perpendicular, F = BIL(N) = 0.0008N

2. Torque = F x d, what is d here?

There will be a torque with respect to the centre of the square, so d is 1cm. This means $\tau$ = d(BILN) = 8 x 10^-6N m

3. Repeat for the other side.

On the other side, the same but the force will point in the opposite direction.
So total torque: $\tau$ = F(1cm) + F(1cm) = 2(0.01)F = 1.6 x 10^-5N m

 

[rude man]

If the coil is wrapped around the perimeter of the square then when the square is in the plane of the B field the torque is maximal (i.e the normal vector of the square points in the upwards direction, perpendicular to the B field.)

Yes, so since B and L are perpendicular, F = BIL(N) = 0.0008N

There will be a torque with respect to the centre of the square, so d is 1cm. This means $\tau$ = d(BILN) = 8 x 10^-6N m

On the other side, the same but the force will point in the opposite direction.
So total torque: $\tau$ = F(1cm) + F(1cm) = 2(0.01)F = 1.6 x 10^-5N m

Excellent! good work.

Now, as to the motor: what happens when the normal to the coil rotates until it's in line with the B field & overshoots a small amount due to inertia? What is the direction of the torque now? Same as before or opposite?

 

[CAF123]

Excellent! good work.

Thanks, any ideas about my problem regarding the second part of the problem? (I posted the sketch in post 2).

 

[rude man]

Thanks, any ideas about my problem regarding the second part of the problem? (I posted the sketch in post 2).

See my edited last post.

 

[CAF123]

Now, as to the motor: what happens when the normal to the coil rotates until it's in line with the B field & overshoots a small amount due to inertia? What is the direction of the torque now? Same as before or opposite?

It will be in the opposite direction.

 

[rude man]

It will be in the opposite direction.

Right. So what would be a way to continue making it go in the same direction?

 

[CAF123]

Reverse the direction of the B field or the current every 90 degrees.

 

[CWatters]

http://en.wikipedia.org/wiki/Commutator_(electric )

 

[rude man]

Reverse the direction of the B field or the current every 90 degrees.

Yes and no.

Yes, 90 deg after the max. torque position you would want to do that. But what about the other positions of the normal? Should you switch again 90 deg after that?

Consider the direction of torque at every orientation of the coil and you will change your mind about reversing the current or field every 90 deg.

 

[CAF123]

I have seen an online lecture and there it appears that the current should reverse every 180 degrees. But I have considered the forces at numerous orientations of the coil and I don't yet see it. For example, start the coil at -90 degrees. Put the B field to the left and current inwards. Then the force is upwards. So the coil will rotate up to 0 degrees mark. Then after this, with nothing changed, the force will still be upwards but on the other side of the coil so ths tends to push the coil back to 0 degrees. How then, is it possible for the coil to get to +90 degrees?

 

[rude man]

I have seen an online lecture and there it appears that the current should reverse every 180 degrees. But I have considered the forces at numerous orientations of the coil and I don't yet see it. For example, start the coil at -90 degrees. Put the B field to the left and current inwards. Then the force is upwards. So the coil will rotate up to 0 degrees mark. Then after this, with nothing changed, the force will still be upwards but on the other side of the coil so ths tends to push the coil back to 0 degrees. How then, is it possible for the coil to get to +90 degrees?

Look again at what I said.

Yes, the current or field needs to be reversed at 0 deg, just as you say. But ask yourself this: when should it be reversed again? What happens at say + 135 deg if you reversed it again at +90?

You are using the correct reasoning and should be able to see why the reversal has to take place every 180 deg, not every 90. And at what points the reversal should occur.