# AC Generator With Repelling Magnets

• benofer90

#### benofer90

Hi Everyone,

I have 4 scenarios in question . Will The EMF(Voltage) be the same in all of the scenarios?
- the difference between the two photos are the polls: one repel one attract .
- i understand that the current(amps) will change due to the wire length assuming same gauge

1) Photo 1: d1 = 1cm, d2=20cm, wire length=240cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
2) Photo 1: d1 = 1cm, d2=10cm, wire length=200cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
3) Photo 2: d1 = 1cm, d2=20cm, wire length=240cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
4) Photo 2: d1 = 1cm, d2=10cm, wire length=200cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second

Photo1:

Photo2

Thank you

Hello Ben, welcome to PF !

Do you have an expression for the EMF ? The F in EMF stands for force; forces are vectors: they have a magnitude and a direction, so you'll need to think about both of these.

hi Bvu, yea I'm aware of it . the EMF(V) will be generated in the oppose direction . but still I'm looking for an answer to the above question. and when i ask if the EMF(Voltage) be the same i mean in magnitude

Do you have an expression for the magnitude of the EMF ?

In your drawings #1, no current is generated in the loops.

Do you have an expression for the magnitude of the EMF ?
what do you mean ?

In your drawings #1, no current is generated in the loops.
Please explain why you think that ?

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Please explain why you think that ?

Use the right hand rule on the horizontal top section of a loop, and again, on the bottom section. The EMF of one section opposes the other. Symmetry says they are equal and opposite.

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Use the right hand rule on the horizontal top section of a loop, and again, on the bottom section. The EMF of one section opposes the other. Symmetry says they are equal and opposite.
then how is it any difference from this ?

Blue and red move in opposite directions

Blue and red move in opposite directions
So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?

Use the right hand rule on the horizontal top section of a loop, and again, on the bottom section. The EMF of one section opposes the other. Symmetry says they are equal and opposite.
So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?

So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?

any thoughts ?

So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?
any thoughts ?

Plenty of thoughts. One of them: the situation isn't all that clear to you yet. What can be done to improve the situation ?

Are you familiar with the Lorentz force on charge carriers ? $$\vec F = q\, \left (\vec E + \vec v \times \vec B \right )$$
There's a picture (bottom one: motion ##\Rightarrow ## emf) in post #16 in this thread

Plenty of thoughts. One of them: the situation isn't all that clear to you yet. What can be done to improve the situation ?

Are you familiar with the Lorentz force on charge carriers ? $$\vec F = q\, \left (\vec E + \vec v \times \vec B \right )$$
There's a picture (bottom one: motion ##\Rightarrow ## emf) in post #16 in this thread

all new to me. making sure I am getting it correctly .

In you picture, e.g. at the lower end, you have M pointing to the left.
M is the magnetic field.
At the lower "horizontal" section of the wire frame, the magnetic field is pointing up.
The velocity vector is pointing to the left. So the Lorentz force on the charge carriers in the wire is pointing away from the viewer.

Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame !

In you picture, e.g. at the lower end, you have M pointing to the left.
M is the magnetic field.
At the lower "horizontal" section of the wire frame, the magnetic field is pointing up.
The velocity vector is pointing to the left. So the Lorentz force on the charge carriers in the wire is pointing away from the viewer.

Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame !
Hold on, when it say magnetic field "right hand rule" little picture I added on top of the drawing. it is referring to the permanent magnet field? and the force is referring to in my case the direction of the coil.

if that is the case then i believe this is the correct drawing .

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One out of three. Still for the bottom "horizontal" section of the wire frame:

The magnetic field thumb is indeed the field from the magnet.
So yes, the middle finger is up (no pun intended ).

The current in this situation is the motion of the charge carriers in the frame. The ##q\vec v## in the Lorentz force expression.
So index finger points to the left.

That leaves the thumb pointing into the screen. Thus representing the direction of the force that these charge carriers "feel". Corresponding to an emf that wants to move these charges.

Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame !

And photo 1, where the emf allegedly oppose each other.

One out of three. Still for the bottom "horizontal" section of the wire frame:

The magnetic field thumb is indeed the field from the magnet.
So yes, the middle finger is up (no pun intended ).

The current in this situation is the motion of the charge carriers in the frame. The ##q\vec v## in the Lorentz force expression.
So index finger points to the left.

That leaves the thumb pointing into the screen. Thus representing the direction of the force that these charge carriers "feel". Corresponding to an emf that wants to move these charges.

Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame !

And photo 1, where the emf allegedly oppose each other.

But, if i look at "fleming's right hand rule" then:
1) the thumb is pointing to the movement(motion of the coil in my case and we can also call it force)
2) First finger (also called Index Finger) is the magnetic field (in my case the magnetic field due to the permeant magnet)
3) second finger (also called middle Finger) is the direction of the current (EMF,V)

no pun was taken

so where am I getting it wrong in photo #2?

Magnetic field is vertical so first finger should point vertically.
Motion is to the left so thumb should point to the left.
Last one left is current.

But, if i look at "fleming's right hand rule" then:
1) the thumb is pointing to the movement(motion of the coil in my case and we can also call it force)
2) First finger (also called Index Finger) is the magnetic field (in my case the magnetic field due to the permeant magnet)
3) second finger (also called middle Finger) is the direction of the current (EMF,V)
Yeah, it works: in the scenario I described: thumb left, index up and middle points way from the viewer.

But if you follow the wiki picture the result of the cross product goes with the thumb.
Must say I think I remember (from 40+ years ago) that in ##\vec c = \vec a \times \vec b## the righthand thumb is a, index is b and result c is middle finger.

Something to do with the properties of the cross product with respect to permutations: swap two -- the sign changes. Swap twice and then it's a plus sign again (even permutations).​

Learn yourself one and only one right hand rule for the cross product. Otherwise you'll end up with a whole life of only 50% chance you're indeed right ! For over 21's the corkscrew rule is perhaps even better...​

But in your picture, at the bottom I see an m pointing up, and an f to the left. That's not good. You can't "also call it force". That's confusing.
And i is not the current. The "cause" of the emf is the motion of the charge carriers (because the wire frame moves). That there is a current is a consequence of the F . There doesn't have to be a current (see e.g. photo 1!). But there is an emf.

so where am I getting it wrong in photo #2?
On the face of it, I don't think you are. You are reversing the field direction and, with the motion in the same direction, I should expect the resultant emf to be in the opposite direction at top and bottom of the coil - giving you a current circulating round the coil. Your logic certainly makes sense to me.
This isn't a situation that's commonly met and I can't see it as a basis for a 'good' generator design. There is the problem that the field will not be as strong as you could get with the normal configuration ( with the iron core that you normally find in alternators) and the fact that you would need a reciprocating motion for the coil, which is not as convenient as a rotation in the normal style of alternator. If you wanted a (DC) Dynamo, the commutator would be pretty hard to implement, too.

Yeah, it works: in the scenario I described: thumb left, index up and middle points way from the viewer.

But if you follow the wiki picture the result of the cross product goes with the thumb.
Must say I think I remember (from 40+ years ago) that in ##\vec c = \vec a \times \vec b## the righthand thumb is a, index is b and result c is middle finger.

Something to do with the properties of the cross product with respect to permutations: swap two -- the sign changes. Swap twice and then it's a plus sign again (even permutations).​

Learn yourself one and only one right hand rule for the cross product. Otherwise you'll end up with a whole life of only 50% chance you're indeed right ! For over 21's the corkscrew rule is perhaps even better...​

But in your picture, at the bottom I see an m pointing up, and an f to the left. That's not good. You can't "also call it force". That's confusing.
And i is not the current. The "cause" of the emf is the motion of the charge carriers (because the wire frame moves). That there is a current is a consequence of the F . There doesn't have to be a current (see e.g. photo 1!). But there is an emf.

Can we simplify your answer . Yes there will be EMF i photo #2 and yes there will be current flowing . Please confirm .

Magnetic field is vertical so first finger should point vertically.
Motion is to the left so thumb should point to the left.
Last one left is current.

HI Cwatters, thank you for your reply . can we agree that Photo number 2 will generate current and EMF . plus that my last right hand rule drawing in the last picture is correct ?

On the face of it, I don't think you are. You are reversing the field direction and, with the motion in the same direction, I should expect the resultant emf to be in the opposite direction at top and bottom of the coil - giving you a current circulating round the coil. Your logic certainly makes sense to me.
This isn't a situation that's commonly met and I can't see it as a basis for a 'good' generator design. There is the problem that the field will not be as strong as you could get with the normal configuration ( with the iron core that you normally find in alternators) and the fact that you would need a reciprocating motion for the coil, which is not as convenient as a rotation in the normal style of alternator. If you wanted a (DC) Dynamo, the commutator would be pretty hard to implement, too.

Hi sophiecentaur,

thank you for your reply . I think i agree with everything you said. but the wonders of life might prove all roads leads to rome (meaning it could be very efficient). Can we agree on the face of it that a current will flow and an EMF will exist in case Photo #2?

Hi sophiecentaur,

thank you for your reply . I think i agree with everything you said. but the wonders of life might prove all roads leads to rome (meaning it could be very efficient). Can we agree on the face of it that a current will flow and an EMF will exist in case Photo #2?
There is no reason, imo that this form of alternator would be 'efficient'. You can get a much stronger field between two opposite poles, separated by a small gap, (conventional alternator and motor design) than the field on the surface of a magnet pole, on its own. There are often several ways of achieving an effect 'in principle' but they won't necessarily all be as effective or efficient. I think this is the case here.
But well done to spot the basic idea. Shows a bit of lateral thinking, I think.

PS If there's a net emf round the circuit then, of course, a current will flow. It's just a matter of how much emf.

benofer90