# Velocity of a beam

1. Oct 5, 2005

### jena

Hi,

My Question:

What is the velocity of a beam of electrons taht go undeflected and magnetic fields of magnitude 8.8 x 10^3 V/m and 3.5 x 10^.9 T, respectively? what is the radius of the electron orbit if the electric field is turned off?

My Work:

V= E/B
V=(8.8 x 10^3 V/m)/(3.5 x 10^.9 T)
V=2.51 x 10^-6 m/s

r=mv/qB
r=(9.11 X 10^-31 kg)(2.51 x 10^-6 m/s)/(1.6 x 10^-19 coul)(3.5 x 10^-3 T)
r= .004m

Are these correct

Thank You

2. Oct 6, 2005

### Andrew Mason

If B = 3.5 x 10^9 T (not 10^.9), and there is an electric field of 8.8e3 V/m, both perpendicular to the direction of v, and to each other, your answer is correct, although that seems like a very slow electron beam (it is very strong magnetic field). I would check the given values again. (You are using 3.5 x 10^-3 T in the next part, which works out to v = 2.51 x 10^6 m/sec). You should show your reasoning as well: e.g.
For a straight line path, force = 0 so $q\vec{v} \times \vec{B} = q\vec {E} => v =E/B$

Where do you get 3.5 x 10^-3 T?. Again, show your reasoning and you won't be confused:
Centripetal force is supplied by the Lorentz force:$F_c = mv^2/r = qvB$ so $r = mv/qB$.
AM

3. Oct 6, 2005

### mukundpa

going right, (calculation part not seen.)

4. Oct 8, 2005

### jena

Hi,

I'm sorry for replying so late but I recalculated the answer for the question as asked by Andrew Mason

and came up with 2.51 x 10^12 m/s, is this answer resonable.

Also for the second portion of the question:

I can still use this equation

r=mv/qB

and come up with the right answer?

Thank You

5. Oct 9, 2005

### Andrew Mason

Unfortunately, this is faster than the speed of light i.e. it is not reasonable. You should provide us with the magnetic field value. I suspect it is 3.5 x 10^(-3) T. based on your answer in the second part. If that is the case, the speed should be 2.51 x10^6 m/sec which is a reasonable value.

Of course. But you have to use the correct value for v. If you substitute v = E/B,

$$r = mE/qB^2$$

AM