Cycloid motion of electron in perpendicular E and B field

[bobred]

Homework Statement

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}}

Homework 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

The Attempt at a Solution

I know this produces a cycloid traveling 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 don't know the form of the velocity. I am assuming the perpendicular velocity will be the sum of a transverse and rotational velocity?

 

[TSny]

You might try integrating the expression for v_y 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.

 

[collinsmark]

Homework Statement

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}}

Homework 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

The Attempt at a Solution

I know this produces a cycloid traveling 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 don't know the form of the velocity. I am assuming the perpendicular velocity will be the sum of a transverse and rotational velocity?

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.

 

[TSny]

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.

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

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

 

[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.

 

[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.

 

[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 y_max = d?)

[Edit: TSny beat me to the hint.]

 

[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.