Determine the ratio of the charge to the mass

[Vladi]

Homework Statement

A beam of electrons passes undeflected through two mutually perpendicular electric and magnetic fields. If the electric field is cut off and the same magnetic field maintained, the electrons move in the magnetic field in a circular path of radius 1.14 cm. Determine the ratio of the electronic charge to the electron mass if E = 8.00 kV/ m and the magnetic field has flux density 2.00 mT.

Homework Equations

F=(q)(v)(b)*sin(theta)
F=(m(v)^2)/R
F=q*e

The Attempt at a Solution

The magnetic force must be equal to the centripetal force; thus...
q*v*b*sin(90)=((m)(v)^2)/r
-->q/m=v/(r*B)
If I have velocity, I'll be able to solve this problem. I'm not sure how the electric field ties into all of this; my gut tells me that I need to use it to solve for the velocity, but I have no clue how. Some additional work has been attached. Any tips are appreciated.

 

[BvU]

Hello,

beam of electrons passes undeflected through two mutually perpendicular electric and magnetic fields

I don't see this being used in your calculations. Am I missing something ?

 

[haruspex]

View attachment 210117

Homework Statement

A beam of electrons passes undeflected through two mutually perpendicular electric and magnetic fields. If the electric field is cut off and the same magnetic field maintained, the electrons move in the magnetic field in a circular path of radius 1.14 cm. Determine the ratio of the electronic charge to the electron mass if E = 8.00 kV/ m and the magnetic field has flux density 2.00 mT.

Homework Equations

F=(q)(v)(b)*sin(theta)
F=(m(v)^2)/R
F=q*e

The Attempt at a Solution

The magnetic force must be equal to the centripetal force; thus...
q*v*b*sin(90)=((m)(v)^2)/r
-->q/m=v/(r*B)
If I have velocity, I'll be able to solve this problem. I'm not sure how the electric field ties into all of this; my gut tells me that I need to use it to solve for the velocity, but I have no clue how. Some additional work has been attached. Any tips are appreciated.

What about the force from the electric field? You have not used this. What equation can youwrite involving that?

 

[Vladi]

What about the force from the electric field? You have not used this. What equation can youwrite involving that?

If the speed of the particle is properly chosen, the particle will not be deflected by these crossed electric and magnetic fields. Does this imply that the magnetic force is equal to the force of the electric field? If so, this is what I come up with.

 

[haruspex]

If the speed of the particle is properly chosen, the particle will not be deflected by these crossed electric and magnetic fields. Does this imply that the magnetic force is equal to the force of the electric field? If so, this is what I come up with. View attachment 210156

Ok.
Compare with the value at https://en.m.wikipedia.org/wiki/Mass-to-charge_ratio.

 

[Vladi]

It looks like my answer is correct. This is what I learned: If the particle will not be deflected by these crossed electric and magnetic fields, this implies that the magnetic force is equal to the force of the electric field. Thank you for your help. It is much appreciated.