# Cycloid motion of electron in perpendicular E and B field

1. Jun 18, 2012

### bobred

1. The problem statement, all variables and given/known data
An infinite metal plate occupies the xz-plane. The plate is kept at zero potential. Electrons are liberated from the plate at y = 0. The initial velocity of the electrons is negligible. A uniform magnetic field B is maintained parallel to the plate in the positive z-direction and a uniform electric field E is maintained perpendicular to the plate in the negative y-direction. The electric field is produced by a second infinite plate parallel to the first plate, maintained at a constant positive voltage $V_{0}$ with respect to the first plate. The separation of the plates is $d$. Show that the electron will miss the plate at $V_{0}$ if

$d>\sqrt{\frac{2mV_{0}}{eB^2}}$

2. Relevant equations
$v_{x}=\frac{E}{B}\left(1-\cos\left(\frac{qB}{m}t\right)\right)$
$v_{y}=\frac{E}{B}\sin\left(\frac{qB}{m}t\right)$
$v_{z}=0$

3. The attempt at a solution
I know this produces a cycloid travelling in the minus x direction. If $r$ is the radius of a rolling circle then $d>2r$ to miss. I think I should be using conservation of energy but dont know the form of the velocity. I am assuming the perpendicular velocity will be the sum of a transverse and rotational velocity?

2. Jun 18, 2012

### TSny

You might try integrating the expression for vy with respect to time to get an expression for y as a function of time. Choose the constant of integration to match the initial condition for y. Then examine the expression.

I don't see an easy way to use energy conservation.

3. Jun 18, 2012

### collinsmark

Oooh, nice problem.

You can use conservation of energy to solve this problem. Well, that and the work-energy theorem. Conservation of energy makes this problem a lot easier. Here are a few things that are noteworthy (you can call them hints if you like):

1) The magnetic forces always acts in a direction perpendicular to the electron's velocity. In other words, the magnetic force never causes the electron's speed to increase or decrease, it only changes the direction. Still in other words, the magnetic force does no work on the electron.

2) You're going to have to determine the maximum speed of the electron. But there are couple of tricks you can do to make it simpler, if you choose to use them. When the electron is at its maximum speed, which direction is going? What's the maximum value of [1-cos(x)]?

3) You'll need to determine a relationship between E and V0, but that should be pretty simple.

4. Jun 18, 2012

### TSny

Ah, nice. I now see that using conservation of energy is a good way to get the result.

My suggestion of integrating vy to get y as a function of time also gets the answer in short order. But I like the energy approach. Thanks.

5. Jun 19, 2012

### bobred

Hi, thanks for the replies.

Part of the question before asked for the expressions of $x(0)=0$ and $y(0)=0$ giving

$x=\frac{E}{B}t-\frac{Em}{qB^{2}}\sin\left(\frac{qB}{m}t\right)$

$y=\frac{Em}{qB^{2}}\left(1-\cos\left(\frac{qB}{m}t\right)\right)$

The expression $\left(1-\cos\left(\frac{qB}{m}t\right)\right)$ at maximum is 2 so $y$ has a max of

$y=\frac{2Em}{qB^{2}}$ and $E=V_{0}/d$ so

$y=\frac{2mV_{0}}{qdB^{2}}$

I keep going around in circles with this.

6. Jun 19, 2012

### TSny

You're essentially there Just interpret what you got. The electron will barely reach the plate if y-max equals what value? Put this value of y into your result and solve for d.

7. Jun 19, 2012

### collinsmark

I think you're on the right track.

What's the magnitude of the electron's charge q? (As in terms of e)?

The variable y is a measure of length (well, technically displacement in the y direction, but that's still a measure of length). What is the value of y when it is at its maximum? (I.e. what's the significance of ymax = d?)

[Edit: TSny beat me to the hint.]

Last edited: Jun 19, 2012
8. Jun 19, 2012

### bobred

Hi

Sorry, went back to the start and had a look at the Lorentz force equations and worked forward from there and using conservation of energy to get the result.
Thanks again.