Magnetic Field, Potential, Velocity

AI Thread Summary
An electron traveling at 1.0 x 10^7 m/s between parallel charged plates separated by 1.0 cm and powered by a 200 V battery requires a specific magnetic field strength to avoid deflection. The electric field (E) is calculated as 20000 N/C, and to maintain zero net force, the magnetic force must counterbalance the electric force. Using the Lorentz force equation, the magnetic field strength (B) is derived as 2 mT, with the direction determined by the right-hand rule, indicating it is out of the page for an electron. The discussion emphasizes the importance of correctly applying the right-hand rule and understanding the forces acting on charged particles. Overall, the calculations and conceptual understanding of forces are crucial for solving this problem.
fredrick08
Messages
374
Reaction score
0

Homework Statement


An electron travels with speed 1.0 x 107
m/s between the two parallel charged
plates shown in the adjacent figure. The
plates are separated by 1.0 cm and are
charged by a 200 V battery. What magnetic
field strength and direction will allow the
electron to pass between the plates without
being deflected?

Homework Equations


B=(\mu/4\pi)(qvsin\theta/r^{2})

The Attempt at a Solution


well I am pretty sure that F=0N, and the the direction the field has to be in the same direction of the velocity, so sin(theta)=0... but i have no idea how find the field strength, anyone have any ideas?
 
Physics news on Phys.org
wat can i do with this Voltage... i need a current don't i?... i can't find the right formulas, and getting really frustrated.. argh
 
The question referred to a picture. It may require a picture or at least an accurate description before we can help. Is the electron traveling in a direction parallel to the plates?
 
yes sorry... the electron is traveling parallel through the parallel plates.. i see if i can upload pic
 
Well if that's the case, no picture is required. Just think in terms of how much force is needed to counter-balance the force exerted on the particle by the E-field. Then use the Lorentz force equation to find the magnetic flux density needed.
 
oh i don't know if we are meant to do it like that, because we havnt done flux or lorentz...
 
velocity.jpg
 
Well, "magnetic flux density" is another word for "magnetic field strength" and "Lorentz force" is a more general term for "force on charged particle due to B-field".
 
oh ok so F=(E+v X B)... i don't understand because doesn't f have to equal 0, for the electron to pass through undisturbed, as soon as there is a force the electron is going to change direction? unless its opposing the velocity, which would make it slow down?
 
  • #10
thus E=V/s=200/.01=2000N/C but how do i find B?
 
  • #11
fredrick08 said:
oh ok so F=(E+v X B)... i don't understand because doesn't f have to equal 0, for the electron to pass through undisturbed, as soon as there is a force the electron is going to change direction? unless its opposing the velocity, which would make it slow down?
You're missing out q here. F has to be zero in order for the electron to pass undeflected. Note that the force exerted by the E-field and that by the B-field is perpendicular to its velocity, and hence does not affect its speed in that direction.

fredrick08 said:
thus E=V/s=200/.01=2000N/C but how do i find B?
Use the equation you stated earlier. Although another similar approach would be to separate the two equations: F=qE and F=Bqv and this two are acting in opposite directions, so you equate them and solve for B. Note that you have to indicate the direction in which the B-field is applied.
 
  • #12
yes sorry i missed q... so qE=Bqv =>E=Bv=>B=E/v=2000/1x10^7=2x10^-4T in the direction of the velocity?
 
  • #13
Check your value for E-field. And note note that magnetic force will always act in a direction perpendicular to both the velocity and direction of the B field. Use the right-hand rule to get the direction.
 
  • #14
ya its 20000 not 2000... oh ok yes i remember now... duh... lol so the direction of B will be out of the page? since the E travels + to -...
 
  • #15
Remember this is an electron, not a positive charged particle.
 
  • #16
na I am confused... what does this mean... won't the electron want to move towards the positive charged plate??
 
  • #17
oh for an electron its the exact opposite, so use left hand, so its going into the page? but is my maths correct, does 2mT sound right?
 
  • #18
no wait before i was using my left hand ad it was saying out... my right says in, thereofre it must be out of the page? please help I am really confused
 
  • #19
No I meant to say that when you apply the F=qv X B vector equation you must note that the resulting direction using the right-hand rule holds for a positive charge. The negative charge goes in the opposite direction.
 
  • #20
Don't use your left hand. Your right hand would do, just reverse the direction.
 
  • #21
yes so if my right hand says the direction is into the page for a +ve charge it is out the page for a -ve charge ie, an electron?
 
  • #22
EDIT: Ok, you're right on this. You can visualise the +ve charged particle as moving from the right to the left. By the way, don't switch it around twice; either consider an equivalent positive charged particle or just reverse the direction at the end of your hand-twisting. Don't do both.

Just consider a positive charge from left to right (same as the electron). Assume the B-field passes into the page. Use the right hand rule and reverse the direction at the end. Is the result the direction you want? If not, assume instead the B-field passes out of the page and do the same.
 
Last edited:
Back
Top