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- Thread starter hemalpansuriya
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In summary, a rotating magnet will generate an electric field at the center. The electric field depends on the angular frequency of the rotation.f

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Mentor

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what are your thoughts? Which of Maxwell's equations probably applies here? Also, is this question for schoolwork?Summary::A magnet is rotating in a circle facing its one of the pole towards the centre with radius R. If we put a charge q at the centre, what will be force on charge?

Charge will experience a rotating magnetic field around it. What will be electric field ( If any ) at the centre, generated by rotation of magnet ?

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We can apply equation, curl E = -dB/dt to find electric field generated at the centre due to rotating magnet facing its one of pole at the centre. The question is not for homework, its my personal curiosity.what are your thoughts? Which of Maxwell's equations probably applies here? Also, is this question for schoolwork?

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Many thanks for your effort.

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Sorry, but I deleted my post #4 as there's a mistake in it. I will post a correct solution - if it comes to me.Many thanks for your effort.

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If ##\omega## is angular frequency of the rotation, then the rotating magnetic field at the center is given in cartesian components (if I am not mistaken) by $$\vec{B}=B_0(\cos\omega t\hat x+\sin\omega t\hat y)$$

Then we have to solve the PDE $$\nabla\times \vec{E}=-\frac{d\vec{B}}{dt}$$ which in fact is a system of three PDEs, because ##\frac{d\vec{B}}{dt}## will have two components (x and y) and we ll also get one more equation by equating the z-component of curl to 0 (I am not sure if we need this last PDE). We will have three unknowns (##E_x,E_y,E_z##) and 3 PDEs.

EDIT: Don't know if @rude man did the same mistake as me, but I completely neglected the generation of additional B-field according to Maxwell-Ampere's law. Maybe if ##\omega## is small we can safely neglect it.

Then we have to solve the PDE $$\nabla\times \vec{E}=-\frac{d\vec{B}}{dt}$$ which in fact is a system of three PDEs, because ##\frac{d\vec{B}}{dt}## will have two components (x and y) and we ll also get one more equation by equating the z-component of curl to 0 (I am not sure if we need this last PDE). We will have three unknowns (##E_x,E_y,E_z##) and 3 PDEs.

EDIT: Don't know if @rude man did the same mistake as me, but I completely neglected the generation of additional B-field according to Maxwell-Ampere's law. Maybe if ##\omega## is small we can safely neglect it.

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I am not sure if this spin of the magnet should be neglected or not.

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No, my mistake was to let ## (\nabla \times \bf E)_\theta = \partial E_r/\partial \theta - \partial E_\theta/\partial r = \omega B_0 ## or whatever I wrote.. Should have been obvious - the ##\theta## component of curl can't have any ##theta## in either derivative. Has to be ## \partial E_r/\partial z - \partial E_z/\partial r = \omega B_0 ##. But now I can't get a grip on the boundary values.

Then we have to solve the PDE $$\nabla\times \vec{E}=-\frac{d\vec{B}}{dt}$$ which in fact is a system of three PDEs, because ##\frac{d\vec{B}}{dt}## will have two components (x and y) and we ll also get one more equation by equating the z-component of curl to 0 (I am not sure if we need this last PDE). We will have three unknowns (##E_x,E_y,E_z##) and 3 PDEs.

EDIT: Don't know if @rude man did the same mistake as me, but I completely neglected the generation of additional B-field according to Maxwell-Ampere's law. Maybe if ##\omega## is small we can safely neglect it.

I'm still looking at this. Good luck on your approach also.

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Please checkout solution that I have tried,

Then we have to solve the PDE $$\nabla\times \vec{E}=-\frac{d\vec{B}}{dt}$$ which in fact is a system of three PDEs, because ##\frac{d\vec{B}}{dt}## will have two components (x and y) and we ll also get one more equation by equating the z-component of curl to 0 (I am not sure if we need this last PDE). We will have three unknowns (##E_x,E_y,E_z##) and 3 PDEs.

EDIT: Don't know if @rude man did the same mistake as me, but I completely neglected the generation of additional B-field according to Maxwell-Ampere's law. Maybe if ##\omega## is small we can safely neglect it.

B = B0cos(wt) î + B0sin(wt) ĵ

∇ x E= -∂B/∂t,

-∂B/∂t = B0wt sin(wt) î - B0wtcos(wt) ĵ

(∂Ez/∂y - ∂Ey/∂z) î - (∂Ez/∂x - ∂Ex/∂z) ĵ +(∂Ey/∂x - ∂Ex/∂y) k̂ = B0wt sin(wt) î - B0wtcos(wt) ĵ

(∂Ez/∂y - ∂Ey/∂z) = B0wt sin(wt) ………… (1),

(∂Ez/∂x - ∂Ex/∂z) = B0wtcos(wt) ………….(2)

And (∂Ey/∂x - ∂Ex/∂y) = 0…………(3)

From equation (3),

» y. Ey = x. Ex

» ∂Ey/∂x = 0 and ∂Ex/∂y = 0 …………(4)

» So, Ey = 0 and Ex = 0

Putting values of equation (4) into equation (1) and (2), we get,

∂Ez/∂y = B0wt sin(wt) and ∂Ez/∂x = B0wtcos(wt)

Integrating both, we get,

Ez = y.B0wt sin(wt) …………(5)

Ez = x.B0wtcos(wt)…………..(6)

Adding (5) and (6), 2Ez = y.B0wt sin(wt) + x.B0wtcos(wt)

Ez = (1/2)*( y.B0wt sin(wt) + x.B0wtcos(wt))

Now, at centre, x = 0 and y = 0,

So Ez = 0

So at the centre, all components of electric field is zero. So there will be no force.

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Also you end up with two functions for ##E_z## but obviously they are not equal (not for every x,y) so the condition that led you to this solution, which is ##E_x=E_y=0## simply is not true.

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So your final conclusion that E(0,0,0)=0 seems correct to me, though I don't agree completely with the method you used to solve the system of PDEs.

BTW this solution holds for a small range of x and y, cause we took the field of the magnet to be homogeneous, which might be true near the pole of the magnet but not very true as we getting far from the magnet.

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So, Ez is zero at the contre ( because x and y = 0 ). Right ?After a bit more thought I end up that $$E_x=E_y=0, E_z=yB_0\omega\sin{\omega t}+xB_0\omega\cos{\omega t}$$ is one possible solution, I think it satisfies the original curl equation.

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Yes.So, Ez is zero at the contre ( because x and y = 0 ). Right ?

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Ok. Thank youYes.

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We may be able to apply the electric vector potential ## W ## where ## \bf E = \nabla \times \bf W ##.to this problem.

To do so we assume a finite thickness ##h## of the B field in the z direction. ##h## can be any real constant.

Then

##\bf W = (1/4\pi) \int -\partial B_0/\partial t ~ dv/s ~~~ \hat \theta ##

with volume element ## dv = 2\pi ~r ~ h ~dr ## and distance ## s = r ##

since we are looking at the center. Also, the B field is assumed to extend ## 0 < r < a ##, ##a## any real constant.

Thus,

## \bf W = (-1/4 \pi) ~\int_0^a \omega ~ B_0 ~~ 2 \pi ~ r ~ h~dr/r ~~~ \hat \theta ##

or ## \bf W = -ah \omega B_0 /2 ## = constant

which of course means ## \bf E = \nabla \times \bf W = 0 ## at the origin.

All this depends on the validity of modeling the B field as I have shown, particularly the assumption that h and a are constants. So there is as they say "an element of doubt". At least the dimensions check. I'll continue looking at it.

https://www.physicsforums.com/insig...ectric-vector-potential-and-its-applications/

To do so we assume a finite thickness ##h## of the B field in the z direction. ##h## can be any real constant.

Then

##\bf W = (1/4\pi) \int -\partial B_0/\partial t ~ dv/s ~~~ \hat \theta ##

with volume element ## dv = 2\pi ~r ~ h ~dr ## and distance ## s = r ##

since we are looking at the center. Also, the B field is assumed to extend ## 0 < r < a ##, ##a## any real constant.

Thus,

## \bf W = (-1/4 \pi) ~\int_0^a \omega ~ B_0 ~~ 2 \pi ~ r ~ h~dr/r ~~~ \hat \theta ##

or ## \bf W = -ah \omega B_0 /2 ## = constant

which of course means ## \bf E = \nabla \times \bf W = 0 ## at the origin.

All this depends on the validity of modeling the B field as I have shown, particularly the assumption that h and a are constants. So there is as they say "an element of doubt". At least the dimensions check. I'll continue looking at it.

https://www.physicsforums.com/insig...ectric-vector-potential-and-its-applications/

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- #17

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I believe the answer to the force can be illustrated by a simple example.

Assuming that the magnetic field at the center point is defined as ##\vec{B}=B_0(\cos\omega t\hat x+\sin\omega t\hat y)## , there is a small sphere with uniformly distributed electric charge at this point, then then according to equation ##\nabla\times \vec{E}=-\frac{d\vec{B}}{dt}~##, this sphere should be subject to varying x-axis torque and varying y-axis torque.

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