# Force on a charge centered around a rotating magnet

• hemalpansuriya
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

#### hemalpansuriya

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

• Delta2
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 ?
what are your thoughts? Which of Maxwell's equations probably applies here? Also, is this question for schoolwork?

what are your thoughts? Which of Maxwell's equations probably applies here? Also, is this question for schoolwork?
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.

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. • Delta2
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.

Last edited:
On a side note, one interesting mechanics problem, is what should be the spin angular frequency of the magnet (around its COM) such that its pole looks always toward the center of the circle, while the magnet rotates around the center of that circle.
I am not sure if this spin of the magnet should be neglected or not.

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.
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.

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

• Delta2
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.
Please checkout solution that I have tried,
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.

• Delta2
Sorry I can't follow your solution, equation (4) is a possible condition but not necessary. Also ##E_x=E_y=0## is another possible condition but not necessary.
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.

On second thought, maybe equations (5) and (6) are not completely wrong but you have to add integration constants that are not constants actually I think in (5) you have to add a function ##C_1(x)## and in (6) a function ##C_2(y)##. And then find a method to determine ##C_1(x),C_2(y)##

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.

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.

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

• Delta2
So, Ez is zero at the contre ( because x and y = 0 ). Right ?
Yes.

Yes.
Ok. Thank you

• Delta2
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/

Last edited:
• Delta2
I think all the replies above are well analyzed.

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.

• Delta2