Electromagnetism - particle moving in magnetic field

[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 disappear?

 

[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 disappear?

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 disappear...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?

 

[latentcorpse]

v(t)=\frac{D(t)}{r} where D(t) is some polynomial in t?

 

[gabbagabbahey]

If the radial part is zero, doesn't that mean \frac{mv^2-Bvqr}{qr}=0...solve that for v

 

[kof9595995]

refer to the feyman lectures on physics Vol 2 chapter17-3

 

[latentcorpse]

for a non trivial solution v(t)=\frac{B(r,t)q^2r^2}{m}

i still don't really see what to do next though?

 

[gabbagabbahey]

for a non trivial solution v(t)=\frac{B(r,t)q^2r^2}{m}

i still don't really see what to do next though?

First, check your algebra because this result is a little off...second calculate dv/dt and hence find the tangential component of E...then check to see if E satisfies Gauss' Law (use the differential form) and Faraday's law (use the integral form).

 

[latentcorpse]

v(t)=\frac{B(r,t)qr}{m}

\Rightarrow \frac{d v(t)}{dt}=\frac{qr}{m} \frac{\partial{B(r,t)}}{\partial{t}}

therefore the tangential component of E is E_{\phi}=r\frac{\partial{B(r,t)}}{\partial{t}}


shall i just take a gaussian pillbox to test both Gauss and Faraday?

 

[gabbagabbahey]

shall i just take a gaussian pillbox to test both Gauss and Faraday?

Why use the integral form of Gauss' law? Since you now know E (or at least have made a guess at what E is), just take the divergence of it.

And Faraday's law requires a loop, not a surface.