# Homework Help: Cathode Ray - No Electric Field

1. Feb 2, 2008

### dachikid

[SOLVED] Cathode Ray - No Electric Field

1. The problem statement, all variables and given/known data
Consider J. J. Thomson’s 2nd experiment, the discovery of the electron. Turn on a magnetic field, but turn off the electric field. If the electrons enter a region of uniform magnetic field B and length l, show that the electrons are deflected through an angle θ ≈ (e l B)/(mvl) for small values of θ, where e is the absolute value of the electron’s charge, m is the mass of the electron, and v is the electron’s horizontal speed. (You may need to use the approximation θ ≈ tan θ ≈ sin θ, which applies for small θ in radians.)

2. Relevant equations

3. The attempt at a solution

So I get how Thomson figured out the ratio of q/m with the whole cathode ray tube experiment. I can certainly show the equation for the ratio as q/m=(tan(θ)E)/(l*B^2)
I know by adjusting the magnetic field and the electrical potential, he was able to eliminate the deflection of the electron, and determine the velocity of the electron.

However, I'm clueless at answering the above question. I know that a force due to the magnetic field will act on the electron with F=qvB, , I'm having a hard time figuring out where to start Any Suggestions?

2. Feb 3, 2008

### Shooting Star

The force is actually q(vXB), so the initial direction of v is very important.

Can you tell us if B points in the +z dircn, and a charge is traveling in the +y dircn, what sort of trajectory the charge will have? That'll solve your problem.

3. Feb 3, 2008

### dachikid

so if I'm looking at a three dimensional coordinate system, +y is to the right, +x is coming out towards me and +z is up. So the electron travels in the +y direction and the magnetic field is in the -z direction. By the right-hand rule, the force on the electron causes it to travel towards the -z direction, or downward.

So the trajecotory of the electron is moving in the -z direction, so that will give me tan(theta)=vz/vy and vz=az*t
Using l=vy*t and solving for t, I get vz=az*l/vy so now tan(theta)=(az*l)/(vy^2)
Now working to solve for the unknown quantity az and using Fb=m*az=qvB
az = qvB/m, now tan(theta)=qvBl/m*vy^2
If I clean it up by cancelling the v in the numerator with a vy in the denominator, and equating q with e. I get tan(theta)=eBl/m*vy

I'm missing that l in the denominator - frustrating!

Thanks for the help!

4. Feb 4, 2008

### Shooting Star

The relation you have derived is correct, and if theta is small, then theta = eBl/mv, where v is the speed with which the electron had entered the region with the magnetic field.

However, I feel that you have not understood certain salient points about the trajectory which I had asked about. You have taken the force to be qvB, not only at the point where it enters the region with B, but also afterward. Why? Will the speed not change, if the particle is acted upon by a force? In this particular case, with the given directions of B and v, what sort of curve the trajectory will be?

5. Feb 4, 2008

### dachikid

The force acts on the electron the entire time it is traveling through the magnetic field. The trajectory will follow the right hand rule. If I may, using an imaginary CRT tube, looking at the tube from the front, as sitting watching a television, the electron would make contact on the left side of the screen. This would be caused by the force caused by the B field.

Sorry for the poor analogy, but it's all I could think of.

Thanks again for the help!

6. Feb 4, 2008

### Shooting Star

The force on the electron at the initial point will be in the +x dircn. Think of the cross product. If the initial velocity v and B are mutually perpendicular, then the force qbv acts in the direction perpendicular to both B and v. This means that the force is always perp to v and constant in magnitude, and so it is a centripetal force of constant magnitude. As a result, the charge moves in a circle with constant speed. In this case, the plane of the circle is parallel to the x-y plane. The force cannot change the speed of the charge, but can change only the direction of the velocity. (The Cyclotron is based on this phenomenon.)

The relation that you had derived was correct, but the derivation is totally wrong. I’m giving a simple derivation.

The radius r of the circular arc can be obtained from the equation for centripetal acceleration. Equating the centripetal and magnetic forces we have,

mv^2/r = evB => r = mv/eB.

The magnetic field ends after a dist l, and suppose the electron has been deflected by an angle theta. Then theta is also the angle subtended at the centre by the arc. The length of the circular arc is C, say, so that C = r*theta. If the arc C is very small, then C~l. So,

theta = C/r = l/(mv/eB) = elB/mv.

7. Feb 4, 2008

### dachikid

Thanks for the help Shooting star, your explanation has helped tremendously and makes sense.

This sounds incredibly interesting, given the time, I will be looking into it.

Thanks again!

8. Oct 16, 2008

### JoshuaR

Re: [SOLVED] Cathode Ray - No Electric Field

Um, I don't know if it's terrible to reply to a thread from so long ago, but I ran into the exact same problem and a different derivation (incorrect).

Idea as follows, in the distance through the magnetic field, l, the electron moves a height h=ltan$$\theta$$.
This same distance can be related through kinematics as 1/2at2. Since t=l/v, tan$$\theta$$=1/2al/v2.
With F=mv2/R=qvB, tan$$\theta$$=qBl/(2mv). With small angles, this finishes as...

angle=qBl/(2mv).

So what's wrong with this derivation that it's off by a factor of two? Is it that the derivation does not account for F=q(vxB), and the velocity of the electron changes as it travels through length l?

9. Oct 16, 2008

### Shooting Star

Re: [SOLVED] Cathode Ray - No Electric Field

I think you have answered it yourself. The formula that you have used is valid for constant acceleration in a certain direction, which is not the case here. Here, the centripetal acceleration is constant in magnitude and points towards the centre of the arc.