#### artis

Depends what you mean by linear generator. I say that linear generator in meaning that it is a generator, which all parts are moving together lineary, does not work.
Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply

#### olgerm

Gold Member
Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply
Rotational and linear case may not give the same result.
I am not 100% sure. But I think Faraday disc would generate electricity if you rotated its brushes together with its disc.

#### artis

No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.

#### olgerm

Gold Member
No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.
You are probably right. I cant get intuition of the situation. Would it not depend of shape of the wire from brush to center of disk and shape of B-field?

#### artis

Well I have read in theory that if the brushes and return wire could get magnetically shielded then in theory the current in the loop should be the sum of the current generated in the disc portion but in real life this is not possible at least haven't heard anyone done that.
The best practical way to "shield" the return path from getting any canceling current generated is to keep it still and that is the reason for the sliding contacts

But I think you got it , it's a Lorentz force generator so it's very peculiar to the B field and parts of a single loop moving while others staying still unlike in a conventional generator you simply change the field strength through a loop and get induced current.

#### greswd

because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.
Sorry, I don't understand what you're saying

#### greswd

Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply
If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgerm

#### greswd

You see, the Lorentz force formula uses the relative velocity seen in your rest frame. Not exactly following the principle of relativity.
@artis @jartsa
maybe @olgerm is right and it does depend on the angular velocity. and this would be investigating a rarely mentioned chapter of physics.

#### olgerm

Gold Member
No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.
You are right, it cant produce DC current. EMF summarised over one full rotation in loop that consist of wire disc and brush must be proportional to change in B-field flux compared to last time, that the wire was in that position beacuse of maxwells III equation. And magnetic flux throught the loop must be same every time it gets to that position, because B-field is constant(not changing in time).

#### greswd

You are right, it cant produce DC current. EMF summarised over one full rotation in loop that consist of wire disc and brush must be proportional to change in B-field flux compared to last time, that the wire was in that position beacuse of maxwells III equation. And magnetic flux throught the loop must be same every time it gets to that position, because B-field is constant(not changing in time).

If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgerm
There is a distinct physical effect in the rotational case which cannot be ignored.

#### olgerm

Gold Member
There is a distinct physical effect in the rotational case which cannot be ignored.
This was not about disc breaking scenario, but about (rotational) faraday generator, when brushes rotate together with disc.

#### olgerm

Gold Member
How exactly depends on details (shape of pieces, properties of the material etc), but generally, EMF would smoothly go to zero if the disc breaks.

#### greswd

This was not about disc breaking scenario, but about (rotational) faraday generator, when brushes rotate together with disc.
I wasn't referring to the disc-breaking scenario. Also, I said that brushes or no brushes, the disc gets polarized.

#### greswd

How exactly depends on details (shape of pieces, properties of the material etc), but generally, EMF would smoothly go to zero if the disc breaks.
That's an interesting assumption. You see, as it breaks, its tangential velocity remains unchanged, and by the Lorentz force law, the force should not change.

#### olgerm

Gold Member
That's an interesting assumption. You see, as it breaks, its tangential velocity remains unchanged, and by the Lorentz force law, the force should not change.
• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.
• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.

#### greswd

• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.
Let's say its a huge magnet, with a wide magnetic field.

• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.
Then we assume near distances.

#### olgerm

Gold Member
we assume near distances.
after the disc breaks its pieces must get only füther and füther from each other as time passes. EMF is not directed to center of disc, but crosswise to it if time from breaking approaces infinity. polarizing effects lack to exist as time from breaking approaches infinity.

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#### olgerm

Gold Member
why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?
linear velocity is not important because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.
Sorry, I don't understand what you're saying
E,B,v are different in different frames of reference, but meaningful(frame invariant) claims same in all frames of reference. E,B,v are different in frames of reference, where linear generator is in rest and where it is moving, but whether it is generating power or not is same in both frames of reference.U is EMF.
In frame where linear generator is in rest:
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*(\vec{0}+\vec{0}\times \vec{B}))=0$

In frame where linear generator is moving:
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=$
(because Maxwell's III equation)
$\oint(dl*(\vec{v}\times \vec{B}))-\frac{\partial \oint (dS*\vec{B})}{\partial t}$=
(beacause stokes theorem)
$\oint(dS*(rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}))= \oint(dS*(\vec{v}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))=$(because according to Maxwells II equation $div(B)=0$)
$\oint(dS*(\vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))=$
(because the idea based on a rigid body moving lineary with constant speed $\forall_i(\frac{\partial v}{\partial x_i}=0)$)
$\oint(dS*(-(\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))$
(beacuse we assume that all EM field is created by the magnet and magnetic field is soly determined by position of magnet, $B(t)=f(\vec{X_{magnet}}(t))$ time $\Delta t$ ago was $B(t-\Delta t)=f(\vec{X_{magnet}}(t-\Delta t))=f(\vec{X_{magnet}}(t)-\Delta t*v)$, it must be that $(\vec{v}\cdot \nabla)\vec{B}=\frac{\partial \vec{B}}{\partial t}$)
$-2*\oint(dS*(\vec{v}\cdot \nabla)\vec{B})$

It should be 0 in both frames of reference, but I probably made a sign error somewhere.

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#### artis

if the disc breaks into pieces then each individual piece effectively becomes a small linear faraday generator because even a single electron rotating above a homogeneous B field experiences the Lorentz force and would be deflected sideways and the same thing happens in a conducting element being dragged through a B field at 90 degree angle which is essentially the Faraday disc.

Again the physics to the best of my knowledge does not change whether in the rotational or linear scenario.

#### olgerm

Gold Member
if the disc breaks into pieces then each individual piece effectively becomes a small linear faraday generator because even a single electron rotating above a homogeneous B field experiences the Lorentz force and would be deflected sideways and the same thing happens in a conducting element being dragged through a B field at 90 degree angle which is essentially the Faraday disc.
Again the physics to the best of my knowledge does not change whether in the rotational or linear scenario.
EMF goes smaller and smaller as time passes from the disc break because:
• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.
• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.

#### greswd

after the disc breaks its pieces must get only füther and füther from each other as time passes. EMF is not directed to center of disc, but crosswise to it if time from breaking approaces infinity. polarizing effects lack to exist as time from breaking approaches infinity.
But it becomes linear straightaway when it breaks though. Straightaway.

#### greswd

E,B,v are different in different frames of reference, but meaningful(frame invariant) claims same in all frames of reference. E,B,v are different in frames of reference, where linear generator is in rest and where it is moving, but whether it is generating power or not is same in both frames of reference.
But it doesn't explain why use the angular velocity when the LFL depends on linear velocity.

#### greswd

if the disc breaks into pieces then each individual piece effectively becomes a small linear faraday generator because even a single electron rotating above a homogeneous B field experiences the Lorentz force and would be deflected sideways and the same thing happens in a conducting element being dragged through a B field at 90 degree angle which is essentially the Faraday disc.

Again the physics to the best of my knowledge does not change whether in the rotational or linear scenario.
But let's say the conductor and the magnet are moving together, at the same speed.

In the rest frame, no Lorentz force. But in the moving frame, you might expect a Lorentz force. Some people argue that there is an induced electric field that counters the magnetic force in the moving frame.

But then you might wonder why this doesn't apply to the rotational case.

Hence the request for transitional experiments.

#### olgerm

Gold Member
But it becomes linear straightaway when it breaks though. Straightaway.
Yes, but EMF approxes zero over time. Pieces are moving, in this case, relative to magnet. That is why this may produce EMF, but linear generator, where circut moves with magnet would not.

But it doesn't explain why use the angular velocity when the LFL depends on linear velocity.
What is LFL?

But let's say the conductor and the magnet are moving together, at the same speed.
In the rest frame, no Lorentz force. But in the moving frame, you might expect a Lorentz force. Some people argue that there is an induced electric field that counters the magnetic force in the moving frame.
post #68 explains exactly that.

#### greswd

Yes, but EMF approxes zero over time. Pieces are moving, in this case, relative to magnet. That is why this may produce EMF, but linear generator, where circut moves with magnet would not.
But what about the disc getting polarized when disc and magnet rotate together?

What is LFL?
Lorentz Force Law

post #68 explains exactly that.
But the point I'm making is:
But then you might wonder why this doesn't apply to the rotational case.

Hence the request for transitional experiments.

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