Homework Help: Electromagnetism - particle moving in magnetic field

1. Feb 25, 2009

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?

2. Feb 25, 2009

Tom Mattson

Staff Emeritus
Re: Electromagnetism

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?

3. Feb 25, 2009

latentcorpse

Re: Electromagnetism

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?

4. Feb 25, 2009

Phrak

Re: Electromagnetism

2. Go back to Maxwell to get E.

5. Feb 26, 2009

latentcorpse

Re: Electromagnetism

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

6. Feb 26, 2009

gabbagabbahey

Re: Electromagnetism

That would be the one...

7. Feb 26, 2009

Tom Mattson

Staff Emeritus
Re: Electromagnetism

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

8. Feb 26, 2009

latentcorpse

Re: Electromagnetism

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?

9. Feb 26, 2009

gabbagabbahey

Re: Electromagnetism

What is curl(E) in cylindrical coords?

10. Feb 26, 2009

latentcorpse

Re: Electromagnetism

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?

11. Feb 26, 2009

gabbagabbahey

Re: Electromagnetism

Compare components and integrate.

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

12. Feb 27, 2009

Phrak

Re: Electromagnetism

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

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?)

Last edited: Feb 27, 2009
13. Feb 28, 2009

latentcorpse

Re: Electromagnetism

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) 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?

14. Feb 28, 2009

gabbagabbahey

Re: Electromagnetism

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

15. Feb 28, 2009

latentcorpse

Re: Electromagnetism

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

also what PDE?

16. Feb 28, 2009

gabbagabbahey

Re: Electromagnetism

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.

17. Feb 28, 2009

Phrak

Re: Electromagnetism

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

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

Last edited: Feb 28, 2009
18. Feb 28, 2009

latentcorpse

Re: Electromagnetism

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?

19. Feb 28, 2009

Phrak

Re: Electromagnetism

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?

Last edited: Feb 28, 2009
20. Mar 1, 2009

gabbagabbahey

Re: Electromagnetism

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