Electromagnetism - particle moving in magnetic field

[latentcorpse]

A amssive charged particle moves under the influence of a time varying magnetic field \mathbf{B}=B(r,t)\mathbf{\hat{z}}, where r is the distance from the z axis. Show that the particle can move in a circular orbit in a plane perpendicular to the field, accelerating and decelerating under the influence of the electric field induced by the temporal variation of the magnetic field, provided that the average value of the magnetic field inside the orbit is twice the magetic field at the orbit.

[i.e. if a is the radius of the orbit and \Phi(t) the flux through it, \frac{\Phi(t)}{\pi a^2}=2B(a,t)

I can't really get started on this one. any ideas?

 

[quantumdude]

Well if you're going to calculate the path of the electron then Newton's second law would be a good place to start, right? So what's the force on a charged particle that is subjected to both electric and magnetic fields?

 

[latentcorpse]

ok. yep, Lorentz force is \mathbf{F}=q(\mathbf{E}+|mathbf{v \wedge B}). We know B is in the z direction so the magnetic part of the force can't be in the z direction. I'm not sure about tackling the electric part, or how to express any of this argument mathematically though?

 

[Phrak]

2. Go back to Maxwell to get E.

 

[latentcorpse]

do you mean \int_C \mathfb{E \cdot dl}==-\frac{d \Phi}{dt} as i have an expression for \Phi?

 

[gabbagabbahey]

do you mean \int_C \mathfb{E \cdot dl}==-\frac{d \Phi}{dt} as i have an expression for \Phi?

That would be the one...

 

[quantumdude]

However I would have went with the differential form of that equation.

 

[latentcorpse]

ok. so we get

\nabla \wedge \mathbf{E}=-\frac{2}{\pi a^2} \frac{d \Phi(t)}{dt}

how do i solve this for E?

 

[gabbagabbahey]

What is curl(E) in cylindrical coords?

 

[latentcorpse]

ok i have that expression, how does that help me?
also surely in the equation i wrote in post 8, the rhs should have a direction? would it be the \mathbf{\hat{z}} direction? does that mean i just compare the z components of lhs and rhs to get the z component of E?

 

[gabbagabbahey]

ok i have that expression, how does that help me?

Compare components and integrate.

also surely in the equation i wrote in post 8, the rhs should have a direction? would it be the \mathbf{\hat{z}} direction? does that mean i just compare the z components of lhs and rhs to get the z component of E?

Yes, but I don't see any reason to use the flux instead of the magnetic field...

 

[Phrak]

Aren't you lucky! Three mentor types, helping out.

ok. so we get

\nabla \wedge \mathbf{E}=-\frac{2}{\pi a^2} \frac{d \Phi(t)}{dt}

how do i solve this for E?

You might be getting off track. You want the integral form you posted in #5. C, is the closed loop encompasing the magnetic flux, \Phi. The length of the loop is 2 \pi a of course.

(By the way, why do you represent the cross product with a wedge operator? I haven't seen that before. Doesn't that give you a 2-form?)

 

[latentcorpse]

ok,
1, as to the wedge product. my lecturer uses it, so i do. But yes, I'm also confused about that because I'm taking a course ind ifferential geometry and seeing as \vec{E} is a 1-form and we can consider \vec{\nabla} a 1-form, there wedge product should be a 2-form.

I have a couple of questions about this point:
(i) can we consider \vec{\nabla} as a vector and hence a 1-form?
(ii) is it true that a 2-form is different from a vector? if so, i will ask my lecturer why he's using this notation on Monday?

2, so |\vec{E}|2 \pi a=-2 \pi a^2 \frac{\partial{\vec{B}(\vec{r},t)}}{\partial{t}}

(i) \vec{dl} is an infinitesimal vector round the loop, correct? i.e. its in the \mathbf{\hat{\phi}} direction. How do we know \vec{E} is also in this direction? Surely if the electric field is induced by the magnetic field the only requirement is that it be in the plane perpendicular to B i.e. surely it could just as easily be in the \mathbf{\hat{r}} direction, in which case the dot product in the integral form of Maxwell equation would be 0 and hence game over?
(ii)I still can't see how to proceed without knowing what B is?

 

[gabbagabbahey]

2, so |\vec{E}|2 \pi a=-2 \pi a^2 \frac{\partial{\vec{B}(\vec{r},t)}}{\partial{t}}

This assumes that \vec{E} is uniform over the loop and points in the \mathbf{\hat{\phi}} direction (otherwise you can't pull it out of the integral). If you can't think of any justification for this assumption, then don't make it Try solving the PDE instead

 

[latentcorpse]

can you advise me on that question about the wedge product and differential forms at all?

also what PDE?

 

[gabbagabbahey]

I'm not sure about the wedge product, it's not notation that I'm familiar with.

As for the PDE, that is your curl(E) equation.

 

[Phrak]

Don't get too distracted by this business, but yes, \nabla can be, and is used as a one-form.

\nabla \wedge V = 2 \partial_{[i} V_{j]} = \partial_i V_j - \partial_j V_i = T_{ij}
U_k = (1/2) \epsilon_{ijk} T_{ij}

 

[latentcorpse]

ok. can i just ask where the factor of 2 comes from in \nabla \wedge V=2 \partial_i V_j

also, using this definition we get

\partial_i E_j - \partial_j E_j = - \frac{2}{\pi a^2} \frac{d \Phi}{dt}

how do i solve that? do i need to pick a coordinate system?

 

[Phrak]

If you're still interested in wedge products n stuf later, ask again.

But gabbagabbahey is right! you need the differential form. Phi just gets in the way. What was I thinking?

The only non zero electric field is in the tangential direction. What is E expressed in terms of B?

 

[gabbagabbahey]

\partial_i E_j - \partial_j E_j = - \frac{2}{\pi a^2} \frac{d \Phi}{dt}

how do i solve that? do i need to pick a coordinate system?

Try cylindrical coordinates instead. Compare the components of curl(E) with the components of dB/dt.

 

[gabbagabbahey]

The only non zero electric field is in the tangential direction.

If that were the case, the particle would not move in an orbit. The magnetic field would exert a force in the radial direction out to infinity. An radial component that cancels the magnetic force is necessary.

EDIT Actually there will be a non-zero radial acceleration, so E may or may not have a radial component. However, it will certainly not completely cancel the magnetic force.

 

[Phrak]

If that were the case, the particle would not move in an orbit. The magnetic field would push in the axial direction out to infinity. An axial component that cancels the magnetic force is necessary.

??

In cylindrical coordinates

Z -- axial
R -- radial
Theta -- tangential, or what would you call the theta direction?

 

[gabbagabbahey]

??

In cylindrical coordinates

Z -- axial
R -- radial
Theta -- tangential, or what would you call the theta direction?

Sorry, I meant radial not axial.

 

[Phrak]

Sorry, I meant radial not axial.

No problem. We know the radial component of the electric field must be zero without some centrally located charge. Gauss's Law.

\nabla \cdot E = 0

I'm assuming this all must work out somehow, without actually having solved this problem yet, gabbagabbahey, for no other reason than that betatrons have been built--and somehow have worked, so I hear.

 

[gabbagabbahey]

No problem. We know the radial component of the electric field must be zero without some centrally located charge. Gauss's Law.

\nabla \cdot E = 0

Hmmm, just because \vec{\nabla} \cdot \vec{E} = 0 doesn't necessarily mean that the radial component of E is zero does it?...What if E_r \propto \frac{1}{r}

I'm assuming this all must work out somehow, without actually having solved this problem yet, gabbagabbahey, for no other reason than that betatrons have been built--and somehow have worked, so I hear.

Indeed, I think I even have an old action figure of Betatron somewhere...or was that Galvatron?

 

[Phrak]

Hmmm, just because \vec{\nabla} \cdot \vec{E} = 0 doesn't necessarily mean that the radial component of E is zero does it?...What if E_r \propto \frac{1}{r}

Are you sure your not mixing up r and theta? E_theta is proportional to 1/r where the contained magnetic flux is the same independent of r.

So, you've driven me to look up the integral form, to make sure, which is

\oint_{S} E \cdot dA = Q / \epsilon

Any flux passing through the cylindrical surface sums to zero when Q is zero. By symmetry the flux is everwhere equal in magnitude, and therefore zero.

I have an unfair disadvantage having gone through something similar in regards to solenoids on this forum. I know I have a text on this matter that would settle the problem, but that would be no fun at all.

 

[latentcorpse]

\left(\frac{1}{r} \frac{\partial{E_z}}{\partial{\theta}} - \frac{\partial{E_{\theta}}}{\partial{z}}\right)\mathbf{\hat{r}} + \left(\frac{\partial{E_r}}{\partial{z}} - \frac{\partial{E_z}}{\partial{r}}\right) \mathbf{\hat{\theta}} + \frac{1}{r} \left[ \frac{\partial}{\partial{r}}\left(rE_{\theta}\right) - \frac{\partial{E_{r}}}{\partial{\theta}}\right] \mathbf{\hat{z}} = -\frac{\partial{B}}{\partial{t}}

who can we compare components though? we don't have an expression for B?

 

[gabbagabbahey]

Are you sure your not mixing up r and theta? E_theta is proportional to 1/r where the contained magnetic flux is the same independent of r.[/itex]

I'm sure. As an example, suppose we had \vec{E}=\frac{k\cos\theta}{r}\hat{r}+f(r)\hat{\theta}, would the divergence of that be non-zero?

So, you've driven me to look up the integral form, to make sure, which is

\oint_{S} E \cdot dA = Q / \epsilon

Any flux passing through the cylindrical surface sums to zero when Q is zero. By symmetry the flux is everwhere equal in magnitude, and therefore zero.

By what symmetry exactly? Just because the flux is zero (and it is), doesn't mean the fields can't have a non-zero radial component. Calculate for my above example if you don't believe me.

To be clear, I'm not saying that there has to be a non-zero radial field, I'm saying that there could be. Gauss' Law alone does not exclude this possibility.

 

[gabbagabbahey]

\left(\frac{1}{r} \frac{\partial{E_z}}{\partial{\theta}} - \frac{\partial{E_{\theta}}}{\partial{z}}\right)\mathbf{\hat{r}} + \left(\frac{\partial{E_r}}{\partial{z}} - \frac{\partial{E_z}}{\partial{r}}\right) \mathbf{\hat{\theta}} + \frac{1}{r} \left[ \frac{\partial}{\partial{r}}\left(rE_{\theta}\right) - \frac{\partial{E_{r}}}{\partial{\theta}}\right] \mathbf{\hat{z}} = -\frac{\partial{B}}{\partial{t}}

who can we compare components though? we don't have an expression for B?

Your told B points in the z-direction. \vec{B}=B(r,t)\hat{z}

 

[latentcorpse]

so, if we take, say the \phi component to start with, we get

\partial{E_r} \partial{r} = \partial{E_z} \partial{z}

we don't know what E_r,E_z are though so we can't carry out any integration can we?

 

[gabbagabbahey]

so, if we take, say the \phi component to start with, we get

\partial{E_r} \partial{r} = \partial{E_z} \partial{z}

we don't know what E_r,E_z are though so we can't carry out any integration can we?

Careful, \partial{E_r} by itself has no meaning. Best to write it as \frac{\partial{E_r}}{\partial{z}}=\frac{\partial{E_z}}{\partial{z}}

This alone is insufficient to determine E_z and/or E_r...You need to use all that you know about E; that is you need to use curl(E)=-dB/dt, div(E)=0 and the boundary condition E->0 as r->infinity.

However (!), there is an easier way to find a solution than to actually solve the PDEs. The uniqueness theorem can be used here: if you can by any means find a solution that satisfies the conditions curl(E)=-dB/dt, div(E)=0 and the boundary condition E->0 as r->infinity, then you are guarenteed it is the solution.

So, use what you know about circular motion to infer what E might be and then simply check that it satisfies those 3 conditions.

Hint go back to Newton's 2nd law and assume that you do indeed get a circular orbit with variable speed.

 

[latentcorpse]

ok. first of all, \nabla \cdot E = \frac{\rho}{\epsilon_0} by definition - why have you made it equal to zero, are you assuming there is no charge density (surely when we get an induced electric field we get a current and hence a charge density?)

secondly we would require that the force be purely radial and vary with time in order to get a circular motion with variable speed. how can we relate this to the field - using Coulomb's law or the Lorentx force prehaps?

 

[gabbagabbahey]

(surely when we get an induced electric field we get a current and hence a charge density?)

Why would you think that? An AC power line will produce a time varying magnetic field and hence an electric field, are you saying that the charge density in the power line is non-zero? How about around the power line?

secondly we would require that the force be purely radial and vary with time in order to get a circular motion with variable speed.

Careful, the tangential component of the force won't be zero if the speed is varying.

Equate your expected form of the force with the lorentz force law and compare components.

 

[Phrak]

The cylindrical symmetry of the problem precludes dependency on theta such as

\vec{E}=\frac{k\cos\theta}{r}\hat{r}+f(r)\hat{\theta}

 

[gabbagabbahey]

Cylindrical symmetry precludes dependency on theta such as in this

\vec{E}=\frac{k\cos\theta}{r}\hat{r}+f(r)\hat{\the ta}

How do you know that E is cylindrically symmetric?

 

[latentcorpse]

\vec{F}=a \mathbf{\hat{r}} + c(t) \mathbf{\hat{\phi}} where a is a constant and c(t) is some function of time

the constant radial part of the force gives a circle and the time varying tangential bit varies the speed - i reckon this is wrong though because when i picture it, surely the tangential force I've written down would distort the path - meaning it would no longer be a circle?

i didn't know there was no charge density in a power line? is this general knoweldge? lol, i must appear to be a right idiot!

 

[gabbagabbahey]

\vec{F}=a \mathbf{\hat{r}} + c(t) \mathbf{\hat{\phi}} where a is a constant and c(t) is some function of time

the constant radial part of the force gives a circle and the time varying tangential bit varies the speed - i reckon this is wrong though because when i picture it, surely the tangential force I've written down would distort the path - meaning it would no longer be a circle?

Why do you think 'a' is a constant? Isn't 'a' just centripetal acceleration times mass...what is centripetal acceleration in circular motion? Also, what physical quantity must c(t) represent?

 

[latentcorpse]

ok so \vec{F}=\frac{m \vec{v_r}^2}{r} \mathbf{\hat{r}} + c(t) \mathbf{\hat{\phi}}

im gussing c(t) is something to do with magnetic field or flux but not sure why?

 

[gabbagabbahey]

ok so \vec{F}=\frac{m \vec{v_r}^2}{r} \mathbf{\hat{r}} + c(t) \mathbf{\hat{\phi}}

im gussing c(t) is something to do with magnetic field or flux but not sure why?

Careful, the centripetal force is inwards, not outwards. If v(t) were a constant would there be any tangential acceleration? Remember, for circular motion v(t) is entirely tangential:

\vec{v}(t)=v(t)\hat{\phi}\implies \vec{a}(t)=\frac{d}{dt}\left(v(t)\hat{\phi}\right)=\dot{v}(t)\hat{\phi}+v(t)\frac{d\hat{\phi}}{dt}=\dot{v}(t)\hat{\phi}-\frac{v(t)^2}{r}\hat{r}

This is something you should have learned in your first vector calculus course.

Now compare this to what you get from the Lorentz force law...

 

[Phrak]

How do you know that E is cylindrically symmetric?

B = B(r,t), E = f(B).

 

[latentcorpse]

q(\vec{E}+\vec{v} \cross \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}}

do i just rearrange for E now?

\vec{E}= qm \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}} - \vec{v} \cross \vec{B}
that's a solution is it not? therefore it must be the only solution?

do i now substitute in for B or is this all wrong?

 

[gabbagabbahey]

B = B(r,t), E = f(B).

True, but that information is not contained in Gauss' Law, it comes from Faraday's law. My point is that your earlier post seemed to suggest that just because div(E)=0, there can be no radial component of the field. It is only once you take into account Faraday's Law (and the condition that E goes to zero far from all charge and currents) that you can make this statement. That was my point.

 

[gabbagabbahey]

q(\vec{E}+\vec{v} \cross \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}}

Surely you mean q(\vec{E}+\vec{v} \times \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}}

What is the direction of v for circular motion? what is the direction of B in this case? What does that make v x B?

\vec{E}= qm \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}} - \vec{v} \cross \vec{B}
that's a solution is it not? therefore it must be the only solution?

Huh? It's only a solution if it satisfies the 3 conditions mentioned earlier...you are going to have a difficult time checking if it does in it's current form...

 

[Phrak]

True, but that information is not contained in Gauss' Law, it comes from Faraday's law. My point is that your earlier post seemed to suggest that just because div(E)=0, there can be no radial component of the field. It is only once you take into account Faraday's Law (and the condition that E goes to zero far from all charge and currents) that you can make this statement. That was my point.

Well, that is all contained in the problem statement---or implied. I hunted down the solution in a text. And it's somewhat erronious. It's a first order solution, without any excuses made for the swindle.

 

[latentcorpse]

right well \vec{v} is tangential and B is in z direction so \vec{v} \cross \vec{B} will be in radial direction and for cylindrical coordinates (r,\phi,z), we have

\mathbf{\hat{\phi}} \cross \mathbf{\hat{z}} = \mathbf{\hat{r}}, so i reckon the magnetic force will be radially outwards.

Now, can we show \vec{E} to be in the tangential direction because then we could just compare components and identify

|\vec{E}|=\frac{m \dot{v}(t)}{q}, |\vec{B}|=-\frac{v(t)^2}{qr}

or is this me going off in the wrong direction?
also what are the 3 conditions i wrote down? i don't think i wrote down any boundary conditions, did i?

 

[gabbagabbahey]

right well \vec{v} is tangential and B is in z direction

Right, so \vec{B}=B(r,t)\hat{z} and \vec{v}=v(t)\hat{\phi}...so in terms of v(t) and B(r,t), vxB=____?

\mathbf{\hat{\phi}} \cross \mathbf{\hat{z}} = \mathbf{\hat{r}}, so i reckon the magnetic force will be radially outwards.

\cross is not a valid LaTeX command, use \times instead.

Now, can we show \vec{E} to be in the tangential direction because then we could just compare components and identify

|\vec{E}|=\frac{m \dot{v}(t)}{q}, |\vec{B}|=-\frac{v(t)^2}{qr}

or is this me going off in the wrong direction?

It's going in the right direction, but you need to be more careful with your algebra...

also what are the 3 conditions i wrote down? i don't think i wrote down any boundary conditions, did i?

Your only boundary condition is that E goes to zero far from all sources (i.e. at r-->infinity).

The other two "conditions" are Gauss' law and Faraday's law...first correct your solution, and then check to make sure it satisfies all 3 of these conditions...It is only a valid solution if it does.

 

[latentcorpse]

\vec{v} \times \vec{B} = B(r,t) v(t) \mathbf{\hat{r}}

then q(\vec{E}+\vec{v} \times \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}}

rearranging gives \vec{E}=\frac{m \dot{v}(t)}{q} \mathbf{\hat{\phi}} - \frac{v^2-Bvqr}{qr} \mathbf{\hat{r}}

however as r \rightarrow \infty not all the terms of E dissappear?

 

[gabbagabbahey]

\vec{v} \times \vec{B} = B(r,t) v(t) \mathbf{\hat{r}}

then q(\vec{E}+\vec{v} \times \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{v(t)^2}{r} \mathbf{\hat{r}}

You seem to be missing an m in there still

rearranging gives \vec{E}=\frac{m \dot{v}(t)}{q} \mathbf{\hat{\phi}} - \frac{v^2-Bvqr}{qr} \mathbf{\hat{r}}

however as r \rightarrow \infty not all the terms of E dissappear?

Doesn't that depend on what v(t) is?...Why not take a guess at what v(t) could be...choose one that makes the radial part of E dissappear...what does that make the phi component of E?

 

[latentcorpse]

ok. q(\vec{E} + \vec{v} \times \vec{B}) = m \dot{v}(t) \mathbf{\hat{\phi}} - \frac{m v(t)^2}{r} \mathbf{\hat{r}}

then
\vec{E}=\frac{m \dot{v}}{q} \mathbf{\hat{\phi}} - \frac{mv^2-Bvqr}{qr} \mathbf{\hat{r}}

i want a v(t) with 1/r dependence but i don't know how to pick it or why i want the radial part of E to disappear?

what would you recommend for choosing as v?

 

[gabbagabbahey]

but i don't know how to pick it or why i want the radial part of E to disappear?

There are several reasons why you might want to choose v(t) in a way that makes the radial component of E disappear:

(1) the obvious one is that it makes for a much simpler E and since you are essentially trying to guess the correct E why not make your first guess as simple as possible? (of course that's not a very physical argument, but the point is that if you can luck out and choose v(t) in a way that makes E satisfy the 3 'conditions' stated earlier, then you are guaranteed it is the correct E and hence the correct v(t) and orbital path...so why not choose a v(t) that makes the radial component of E disappear and check if the resulting E satisfies the 3 conditions? If it does, then you've chosen correctly)

(2) A more physical reason is the symmetry mentioned by phrak...Faraday's law essentially tells you that in the absence of non-zero charge densities, E takes on the symmetries of the time-varying B field that creates it. In this case, B is cylindrically symmetric, so it makes sense that E is as-well. The only way a cylindrically symmetric E can be divergence free is if it is entirely tangential.

So, when is the radial component of E zero?