Electron speed due to magnetic field.

[sarahaha288]

Homework Statement

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?

Homework 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.

The Attempt at a Solution

I attempted to use the equation Fm=q(velocity/c x B)... but I am not sure what to use for the magnetic force.

 

[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

 

[sArGe99]

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

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

 

[thebigstar25]

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

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

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

 

[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??

 

[thebigstar25]

In your question, it said that "uniform magnetic field whose strength is 10MN/C" but for me that doesn't 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 ?

 

[sarahaha288]

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

 

[thebigstar25]

I don't really get you! what B-Bar supposed to mean is it the same as B?? and what is M(in 10 MN/c) ? , I am sorry if I can't be helpful but if you could clarify these questions then I may can help ..

 

[RoyalCat]

Are you working in CGS units by any chance?
In that case, refer to this wiki page:
http://en.wikipedia.org/wiki/Lorentz_force#Lorentz_force_in_cgs_units

This makes sense in terms of magnitude too, since B fields are usually only a fraction of a Tesla in strength, whilst here the numerical value for the field is on the order of 10^7!

 

[thebigstar25]

Are you working in CGS units by any chance?
In that case, refer to this wiki page:
http://en.wikipedia.org/wiki/Lorentz_force#Lorentz_force_in_cgs_units

This makes sense in terms of magnitude too, since B fields are usually only a fraction of a Tesla in strength, whilst here the numerical value for the field is on the order of 10^7!

I have checked the link you provided, but I can't see where it mentions that both E and B can be expressed in N/C unit? and I don't 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

 

[RoyalCat]

I have checked the link you provided, but I can't see where it mentions that both E and B can be expressed in N/C unit? and I don't 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

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.

 

[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 ..

 

[sarahaha288]

thanks so much! that helped a lot.