# Electron speed due to magnetic field.

1. Feb 9, 2010

### sarahaha288

1. The problem statement, all variables and given/known data

electrons emitted by a certain device are observed to travel in a circular path with a radius of 2.0cm when plaves in a uniform magnetic field whose strength is 10MN/C. What is the speed of the elctrons emitted by this device?

2. Relevant equations
I know that 10MN/C is "b-bar" as it is called in the book. But i'm not sure what equation to use.

3. The attempt at a solution
I attempted to use the equation Fm=q(velocity/c x B).... but im not sure what to use for the magnetic force.

2. Feb 9, 2010

### thebigstar25

im not really sure about it , but i think that since these electrons are moving in a circular path in a uniform magnetic field , then using newton's 2nd Law :

qvB = m v^2/R

3. Feb 9, 2010

### sArGe99

Yes, the magnetic force provides the necessary centripetal force for the circular motion of the electrons.

Fm = q (v X B) = m*v2 / r
Don't really know much about b-bar. The units should be either Tesla or Gauss.

4. Feb 9, 2010

### thebigstar25

I believe that if the given value 10MN/C ( 10 mega newton per coulomb) , then this given value is the electric field (E) ..

5. Feb 9, 2010

### sarahaha288

the B-bar is measured in tesla. and if the electric field vector is the one that is given, how do I find the magnetic field in order to use the equation Fm=mv^2/B if i am trying to solve for v??

6. Feb 9, 2010

### thebigstar25

In your question, it said that "uniform magnetic field whose strength is 10MN/C" but for me that doesnt make sense? how it is a magnetic field with this unit, I know that it should be tesla, weber per meter square or volt.second per meter square but not newton per coulomb ? can you please check the question once more ?

7. Feb 9, 2010

### sarahaha288

The magnetic field for B-Bar is definately 10MN/C
the book says that vector B is in units of tesla which is: (N/C)(m/s^-1)

8. Feb 9, 2010

### thebigstar25

I dont really get you!! what B-Bar supposed to mean is it the same as B?? and what is M(in 10 MN/c) ? , im sorry if I cant be helpful but if you could clarify these questions then I may can help ..

9. Feb 9, 2010

### RoyalCat

10. Feb 9, 2010

### thebigstar25

I have checked the link you provided, but I cant see where it mentions that both E and B can be expressed in N/C unit? and I dont think that is possible for two different quantities to have the same units even if different system ..

If it is really mentioned somewhere then I apologize , and please quote from where it is mentioned .. & thanks

11. Feb 9, 2010

### RoyalCat

That is the whole point of the CGS system. :) $$\vec B$$ is defined differently there, so that $$\vec B$$ and $$\vec E$$ have the same units. The Lorentz force in CGS units would be:

$$\vec F=q(\vec E+\frac{\vec v}{c} \times \vec B)$$

See the link for the exact transformations in going from one unit of measurement to the other.

12. Feb 9, 2010

### thebigstar25

hmmm I have never encounter this before .. thanks for bringing this up :) .. and I hope that sarahaha288 would be able now to solve the problem , it is just converting back B to tesla and then substituting in qvB = m v^2/R to get v ..

13. Feb 10, 2010

### sarahaha288

thanks so much! that helped alot.