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