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Generators generate anger :( (1 Viewer)

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I want to clear this up...

1) Let's assume there is a magnetic field into the page and there is current (conventional current?) pointing down and there is a force directed to the right.
There is a motion arrow pointing left, opposite to the force arrow.
How does it actually affect the wire and what is the difference between those two quantities ?

2) In a motor, when it turns clockwise I just acknowledged that the force on the left wire turns the armature up then clockwise (N-S pole)*, now what the hell does motion do there ? Oppose the force ?


3) If I turned an AC generator by hand clockwise (N-S pole), where is the motion and force arrows directed and where does conventional current flow?


4) In all generators, no torque means no voltage right ?


* - North pole on the left and South pole on the right
(Permanent magnets for convenience please)
 

alcalder

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f3nr15 said:
I want to clear this up...

1) Let's assume there is a magnetic field into the page and there is current (conventional current?) pointing down and there is a force directed to the right.
There is a motion arrow pointing left, opposite to the force arrow.
How does it actually affect the wire and what is the difference between those two quantities ?
This sounds like it is describing Back EMF. If the wire is moving to the left this will produce a conventional current down the wire. However, this now produces a back EMF to oppose the change that caused it. THUS, the conventional current down the wire in that magnetic field produces a force opposing the motion that caused it. So there is, if you like, a drag on the rotor of the generator.

2) In a motor, when it turns clockwise I just acknowledged that the force on the left wire turns the armature up then clockwise (N-S pole)*, now what the hell does motion do there ? Oppose the force ?
See above. Back EMF again. In a motor, we are providing the current, which is providing the motion. Thus the back emf will produce a current to produce a force to oppose the motion we are creating.

3) If I turned an AC generator by hand clockwise (N-S pole), where is the motion and force arrows directed and where does conventional current flow?
When you are trying to work out the current generated you look at an individual electron. Orientate your hand according to the motion of this electron - motion of electron (stationary in the wire) is current (obviously opposite direction is conventional current), the magnetic field (as per normal) and the force/palm of hand indicates the direction the electron will move up or down the wire. This then tells you which direction the current will flow in the wire. Make sense? It's hard to explain without a diagram.

4) In all generators, no torque means no voltage right ?
In a motor, no torque = sum of all rotational forces on the wires in the rotor = 0

So you might have the voltage turned on but the rotor may be stuck so all the forces up and the forces down add to zero. Therefore there is no torque BUT the voltage is on. You would see this in a DC motor with one coil for example.

In a generator, if there is no torque, the rotor is not rotating so it would not be producing any voltage/current. So, yes, in all generators no torque means no voltage.


Hope I wasn't too dyslexic with my grammar ;)
 

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Thanks, those responses were somewhat more accurate than my teachers ... No problem with grammar, very fine actually.


1) The force from EMF is like "drag" ?

2) The force is simply generated by back EMF which will be overcome by the motion of the motor eventually ?

3) If I turn clockwise, conventional current flows from left to right relative to the magnetic poles (N-S)?

My right-hand rule is based on the Longman though.
I is parallel to my arm ahead, F is upwards and perpendicular to my arm (which is my thumb) and B faces left (instead of F)

One more thing, do you teach one bit of calculus to your Physics students ?
I kind of feel sorry for the General Maths student who has trouble comprehending derivatives ... like Induced Voltage is equal to the change in flux per time unit, Vind = -d(phi)/dt
 

alcalder

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f3nr15 said:
1) The force from EMF is like "drag" ?
Yes, back EMF is like drag.

2) The force is simply generated by back EMF which will be overcome by the motion of the motor eventually ?
Yes. Back EMF is not greater than the current producing it, so there is at least some current that flows. Back EMF kind of provides an increasing resistance in the turn of the loop.

3) If I turn clockwise, conventional current flows from left to right relative to the magnetic poles (N-S)?
If you turn the loop clockwise (let me see if I can describe this) the wire on the left side of the loop has electrons moving up (kind of like conventional current down). Magnetic field N-S means the electron experiences a force towards you.

THe wire on the right side of the loop has electrons moving down (conventional current up). Magnetic field N-S means the electron experiences a force away from you.

If we now look at the loop from the top, the electrons move around the wire loop in an anti-clockwise direction so the conventional current generated is clockwise (so left-to right). Correct!

My right-hand rule is based on the Longman though.
I is parallel to my arm ahead, F is upwards and perpendicular to my arm (which is my thumb) and B faces left (instead of F)
I like to use the one where the thumb is the current (because there is one wire), the fingers are the B field (because we always draw it wil mutiple x's or o's and Force is then the palm of the hand because we push with our palm. It is easier to remember.

BUT always remember to put your pen down to do it. LOL

One more thing, do you teach one bit of calculus to your Physics students ?
I kind of feel sorry for the General Maths student who has trouble comprehending derivatives ... like Induced Voltage is equal to the change in flux per time unit, Vind = -d(phi)/dt
I don't believe I have taught calculus because it can kind of be explained without it. However, I have said to Extension maths students that it is not wrong to use your 3u and 4u maths to attempt and complete the questions. Sometimes 4u students feel like they are cheating when they use the derivation of motion equations method to complete projectile motion questions.

I may have once shown how finding the gradient of a graph can help find a quantity but not gone into the calculus (but have mentioned calculus can be used and anyone interested could see me later).

But I truely feel that students who are doing General Maths are likely to struggle with many aspects of Physics, especially manipulation of equations. When I was at school they told us you could not do Physics unless you were doing 3u maths. But I believe they can't say that anymore.
 

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Do you still teach your students E = vlB because I don't recall that formula being in the syllabus, and I thought because formulas are provided in the Physics HSC you don't need any more Maths skills than in Year 10.
I chose 3U Maths because of Physics, otherwise I'd be doing General Maths.


Also there is something else I need to know.

This is a screeshot of Q7 in last year's HSC Physics paper.
http://img137.imageshack.us/img137/3742/06q7nc7.gif

F = LIBsinx

F = (0.4)(3.0)(0.5)(sin45) = 0.424264068 ... = 0.424N (INCORRECT ANSWER)
F = (0.4)(3.0)(0.5)(sin90) = 0.6N (CORRECT ANSWER)

My teacher was referring to this question which fooled many of his students.
He said something about magnetic fields are always perpendicular, why is it sin 90 instead of sin 45 ? Because I would have been fooled as well.
 

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f3nr15 said:
Do you still teach your students E = vlB because I don't recall that formula being in the syllabus, and I thought because formulas are provided in the Physics HSC you don't need any more Maths skills than in Year 10.
I chose 3U Maths because of Physics, otherwise I'd be doing General Maths.


Also there is something else I need to know.

This is a screeshot of Q7 in last year's HSC Physics paper.
http://img137.imageshack.us/img137/3742/06q7nc7.gif

F = LIBsinx

F = (0.4)(3.0)(0.5)(sin45) = 0.424264068 ... = 0.424N (INCORRECT ANSWER)
F = (0.4)(3.0)(0.5)(sin90) = 0.6N (CORRECT ANSWER)
Sadly incorrect as well

l is the length of the wire in the magnetic field.
0.4/l=sin45
l=0.4/sin45

F=BILsin(theta)
=0.5*3*0.4/sin45*sin90
=0.849N
My teacher was referring to this question which fooled many of his students.
He said something about magnetic fields are always perpendicular, why is it sin 90 instead of sin 45 ? Because I would have been fooled as well.
The sin 45 is used to find the length of the wire.

The angle theta is the angle between the wire and magnetic field. The wire is in the plane of the page and the magnetic field is perpendicular into the page, so they are perpendicular to each other and the angle theta=90
 

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