Electron traveling through two parallel charged plate. What is Magnetic Field?

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An electron traveling at 1.0 X 10^7 m/s between two parallel charged plates, separated by 1.0 cm and charged by a 200V battery, requires a specific magnetic field to avoid deflection. The electric field strength is calculated to be 20,000 N/C, and the magnetic field must be oriented to counterbalance the electric force acting on the electron. The relationship between electric and magnetic forces is established through the equation F = qE = qvBsin(x), where the angle x can affect the resulting magnetic field strength. It is noted that if the electron travels parallel to the plates, the magnetic field must be perpendicular to the electric field to maintain equilibrium. Ultimately, the discussion emphasizes the need for proper alignment and magnitude of the magnetic field to achieve a net force of zero on the electron.
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An electron travels with speed 1.0 X 10^7 m/s between two parallel charged plates. The plates are separated by 1.0 cm and are charged by a 200V battery. What magnetic field strength and direction will allow the electron to pass between the plates without being deflected.?

F= qvBsin\theta
V=Ed

E=V/d = 200V/(1X10^-2)= 20,000. That is the strength of the E-field. Would this also be the force acting on the electron?

If so i could set the top equation(force equation) equal to this. When i did this i came out with a HUGE number for the Magnetic field... umm for the the magnetic field to not have any impact(cause deflection) on the traveling electron it would have to be going in the same direction as the electron. in which case sin\theta where it would be sin(0) is 0.. thus the force would be zero. I may just be confusing myself. Help please

link for a diagram http://img81.imageshack.us/img81/7625/platediagramkl3.jpg
 
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Force due to mag field should balance electric force, so should be in oppo dir.
 
So E=B? but in the opposite direction. What exactly is producing the magnetic field? the plates right?
 
Xaspire88 said:
So E=B?
How is it possible, check the dimensions of E and B.

Xaspire88 said:
What exactly is producing the magnetic field? the plates right?
The mag field under consideration is by someone else, and we are supposed to find it.
 
What do you mean by dimensions?
 
k. See this way : F = q*E = q*v*B*sin(x)

E and B can't have same units in SI. So you can't equate them.
 
but i can relate the force F= E*q and F=q*v*B
in which case

B= (E*q)/(q*v)

B= E/v

is that a correct path to follow?
 
Except the case of misinterpretation, the question seems incomplete coz
q*E = q*v*B*sin(x) will have various soln for various sin(x).
 
isn't it safe to assume that the velocity and the magnetic field are at 90 degrees to each other?
 
  • #10
:-) . Its dangerous to assume so, I may lead you to wrong soln.
 
  • #11
how else would you suggest i go about it? Would you say that the electron is traveling along a path parallel to the plates? if so the E Field would be going perpendicular to that and thus the Magnetic field would also be perpendicular to the path of the electron.
 
  • #12
You will get a soln for every B outside the plane of v and E (ofcourse the value of B will change for diff positions), thus perpendicularity betw v and B is not an essential condition for this situation.
 
  • #13
being that you say it is not essential, i am confused as to how i would then arrive at a value for the b-field that does not contain some variable. This is from a non calculus based class so try not to make it to complicated for me :)
 
  • #14
Sorry, but looks as if info given is not enough. Though you can find the component of B perp to v and E.
 
  • #15
This is a couple years after this was posted but to those reading and yet to read it, they have been on the correct path. Solve for E which in this case is 20000. Then set the two force equations(F=qE and F=qvBsinx) eqaul to each other and solve for B. sinx will be eqaul to 90 degrees since it has to oppose the electric field force. Its the same concept of hovering, just enough force has to be supplied in the opposite direction of gravity so the net force will be zero.
 
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