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Why does the speed remain the same?

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    image.jpg



    2. Relevant equations
    F= qvB sin theta
    F=ma



    3. The attempt at a solution
    A) 6.2x10^18N
    B)9.5x10^8m/s^2
    C) this is the problem when I searched lot of solutions they say that speed remains the same
    Could someone explain this to me
     
  2. jcsd
  3. Mar 19, 2016 #2

    Orodruin

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    How is the force direction related to the particle velocity?
     
  4. Mar 19, 2016 #3
    Using the right hand rule, v and b are at 52 degrees and the force should be 90 degrees to the velocity. Am I right?
     
  5. Mar 20, 2016 #4

    Orodruin

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    Yes, but what conclusion can you draw from this? (The magnetic force is always at a right angle to both the field and the velocity!)
     
  6. Mar 20, 2016 #5
    I don't understand that yet
     
  7. Mar 20, 2016 #6

    Orodruin

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    So let me ask you this: how do you compute the speed?
     
  8. Mar 20, 2016 #7
    It's already given in the question 550m/s
     
  9. Mar 20, 2016 #8
    Think about a circular motion: The velocity can change in direction but the speed remains the same..
     
  10. Mar 20, 2016 #9

    Orodruin

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    I mean in general, not necessarily in this problem.
     
  11. Mar 20, 2016 #10

    ehild

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    Orudruin asks, how do you get the speed (scalar) if you know the velocity ( a vector) .
     
  12. Mar 20, 2016 #11
    Speed = distance/ time
    If you know the velocity the magnitude is the speed right?
     
  13. Mar 20, 2016 #12

    Orodruin

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    Yes, so how do you find out the magnitude of a vector? Or, more useful, how do you find the squared magnitude of a vector? (The magnitude is constant if the magnitude squared is constant!)
     
  14. Mar 20, 2016 #13
    I didn't quite get that but I ll give it a shot
    Magnitude squared in this case 550^2
     
    Last edited: Mar 20, 2016
  15. Mar 20, 2016 #14
    I didn't understand the magnitude square thing

    But I think this is what happened
    An alpha particle with a velocity of 550m/s goes in a magnetic field in an angle of 52 degrees
    So like in the right hand rule the alpha particle is deflected 90 degrees to the field so the acceleration is to change the velocity by changing the direction but the speed remains the same.

    If that is right how would I know if the speed did change?
     
  16. Mar 20, 2016 #15

    Orodruin

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    This is something you will have to understand to understand the solution of the problem.
    We are not interested in this particular case at the moment, we are interested in the general case when you are given a velocity vector ##\vec v##. If I give you a general velocity vector ##\vec v = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z##, what is the speed of the object?
    This is what we are trying to guide you to! This is why you should follow the steps I am giving you.
     
  17. Mar 20, 2016 #16
    I'm really trying to get this in my head please continue. I'll follow the steps
    As for the question with the 3D vector, I don't know but if it was just x and y only v would be found using Pythagoras theorem. I know that much. I don't know about the 3 D one
     
  18. Mar 20, 2016 #17

    Orodruin

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    Pythagoras' theorem holds in an arbitrary number of dimensions.
     
  19. Mar 20, 2016 #18

    ehild

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    Have you learnt about the scalar product (dot product) of two vectors?
     
  20. Mar 20, 2016 #19
    I think I got it tell me if this is right
    V= square root Vx^2+Vy^2+Vz^2

    Is this the speed?
     
  21. Mar 21, 2016 #20

    Orodruin

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    Yes, so what is the derivative of the speed with respect to time? You may use that dVx/dt = ax (x-component of acceleration) and so on.
     
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