Electron moving in electric field

  • #1
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Homework Statement


<pic>


The Attempt at a Solution



well i tried taking v = vx (i) + vy(j)
then listed forces acting in x, y direction using vx,vy and divided by mass to get acceleration

then u used v = u + at for both x and y axis and surprisingly go vy/vx as √3

so means deflection is 60° viz wrong
and what was surprising about this method was that be it any position .. it always give 60° ... coz it doesn't require any distance traveled by particle or time ... nothing .... whats wrong ... ???

BUT MORE IMPORTANTLY ............

How do i do this question
i have 5 question of nearly this type ...... plz help me
 

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Answers and Replies

  • #2
gneill
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The magnetic field will always be perpendicular to the velocity of the electron. What do you know about the resulting motion?
 
  • #3
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i did the part (a) and (c) .... i know its circular!!!

I just need help with part (b) .... forgot to mention it earlier .... :tongue2:
 
  • #4
gneill
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Hmm. If you've derived a solution that works for part a, you should be able to use the same framework for b and c. Perhaps you should explain how you solved a.
 
  • #5
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Hmm. If you've derived a solution that works for part a, you should be able to use the same framework for b and c. Perhaps you should explain how you solved a.

For (a) & (c) :

as magnetic field is providing centripetal force ... qvB = mv2/r

r = mv/qB

as in (a) d is slightly less that mv/qB .... particle will complete half circle ... deflection wil be 90 degree

for (c) d is slightly less that 2mv/qB ... particle will complete full circle ... deflection will be 180 degree

But i dont know how to do (b)
 
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  • #6
gneill
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Okay. For parts (a) and (c) I think you meant quarter circle (90 degrees) and half circle (180 degrees) respectively.

For part (b), only a smaller portion of the circle will be followed before the electron leaves the field. If you sketch a diagram which shows a complete circle of radius R, and you suppose that the electron "enters" the circle at the bottom (the -90 degree angle point) moving to the right, then the electron would follow the arc of the circle. It moves to the right and up until it covers an x-distance equal to the extent of the field. What angle of arc has it traversed?
 
  • #7
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sinθ = x/r
and in this case (x=r/2),

θ = 30°
 
  • #8
gneill
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sinθ = x/r
and in this case (x=r/2),

θ = 30°

Okay! So through what angle has the electron's trajectory been rotated? Remember, its velocity is always perpendicular to R.
 
  • #9
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if you mean deflection in the velocity then i dont know :(
 
  • #10
gneill
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Draw a little vector representing the velocity as it "enters" the circle at the bottom. Draw another at the end of the radius vector when it just leaves the field (when the angle through which the R vector has moved is 30 degrees). How much has the velocity vector rotated?
 
  • #11
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i expected it to be 30 degree but look at the pic

......


angle 1 would be 30 degree then angle 2 would be 60 degree ........ then i could have easily got deflection as 30
but this was only true if "red line" was perpendicular to radius ... which it is not!!!!

still confused!!! :(
 

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  • #12
gneill
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Go back to the diagram that came with the problem. What angle depicted there is the angle representing the trajectory angle?
 
  • #13
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Woops ....

its the blue one right???

yes that will be 30

Dumb mistake!!!!!! :shy:

Thanks a lot for your help gneill !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :biggrin:
 

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  • #14
gneill
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Right! You're welcome.
 

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