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Magnetism (2 Different Ones)

  1. Mar 7, 2005 #1
    Problem 1:
    A proton is shot with a speed of 8x10^6 m/s at 30 degrees to an x-directed field with B=.15 T. Completely describe the path followed by the proton. Describe the path in words and also give the radius, the period, and the pitch.

    The proton is going to take a spiral - screwdriver sort of descent I believe. I know somehow to get the radius, period etc you use the speed and angle its going at, but not sure how.

    Problem 2:
    A beam of particles of charge q enters a region where an elextric field is uniform and directed downward. Its value is 80,000 V/m. Perpendicular to E and directed into the page is a magnetic field B=.4 T. If the speed of the particles is properly chosen, he particles will not be deflected by these cross fields. What should this speed be?

    Completely lost on this. Two perpendicular fields of different values cross, and you need the speed the beam should be going to prevent it being deflected - havent seen one like this before so any help is appreciated
     
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  3. Mar 7, 2005 #2

    dextercioby

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    For the second,it's much simpler than the first.What forces act on the particle and what's the condition for zero deviation??

    For the first,u need to decompose the magnetic field into 2 components [itex] B_{parallel} [/itex] and [itex] B_{perpendicular} [/itex].

    What's the equation of motion...?And why is this decomposition useful...?

    Daniel.
     
  4. Mar 7, 2005 #3
    For the second - the forces acting on the particle are the two fields and its speed, gravity is ignored i believe...I feel like im going in circles on it, i dont know the conditions for 0 deviation, that may be the issue.

    For the first - the equation for a moving particle in a field is:
    Force on the particle = qvB sin(angle)....V = 8x10^6. B=.15T Angle is 30 degrees, charge on a proton is 1.6x10^-16 if I remember correctly. By finding the force on the particle I can find out the other parts using rotational motion? Is that the right approach?
     
  5. Mar 7, 2005 #4

    dextercioby

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    Partly to the first.The charge of the proton is [tex] 1.6\cdot 10^{-19} C [/tex].And for getting the step of the helicoid,use the fact that the:
    [tex] Step=normal \ component \ of \ velocity \ \times \ period [/tex]

    As for the second,simply write down the second law of Newton.

    Daniel.
     
  6. Mar 7, 2005 #5
    First: What do you mean by step, sorry I seem to be farmiliar with different termanology..Im guessing step means the pitch because pitch is the distance traveled in one revolution. To do that I would need the period.

    Second: F=MA...MA is unknown, forces are the results of the field..dont see how that helps

    I appreciate all the help by the way
     
  7. Mar 7, 2005 #6

    dextercioby

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    No,sorry,it was a mot-à-mot translation of the corresponding term from Romanian.Yes,the period is simple to know.You'll just need the radius of the projection cirle and the tangential velocity.

    Do you agree that:
    [tex] m\vec{a}=\vec{F}_{el}+\vec{F}_{magn} [/tex] ?

    If so,the answer the 2 questions
    1.What's the condition for zero deviation?
    2.How do the 2 forces look like...?

    Daniel.
     
  8. Mar 7, 2005 #7
    First: Period = Ok so the period is the velocity * period and the period is 1/frequency
    I have the velocity but none of the other parts

    Second: Yes MA = Force of the electric + Force of the magnetic.
    0 deviation would occur I believe if the acceleration is 0?
     
  9. Mar 7, 2005 #8

    dextercioby

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    Second:Correct.Now carry on with the problem.

    First:Pay attention as you have 2 distinct components of the velocity:eek:ne ALONG THE FIELD AND ONE PERPENDICULAR...So write the formulas as to specify exacly which are u talking about.

    Daniel.
     
  10. Mar 7, 2005 #9
    Ok I see what you mean on the first one, part of the velocity is going in the x direction and some is going down or to the perpendicular. I get the components of that using the sin and cos functions.

    Sin(30) * V = Perpendicular Speed = 4x10^6 M/s
    Cos(30) * V = X Direction Speed = 6,928,203 M/s

    So every second it will go down 4x10^6m and over 6,928,203m. Those are two points on the circle so with that I can find the radius and thus find the frequency and pitch.

    correct?
     
    Last edited: Mar 7, 2005
  11. Mar 7, 2005 #10

    dextercioby

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    With the tangential one u find the radius and the period,with the normal one,the pitch.

    Daniel.
     
  12. Mar 7, 2005 #11
    Right, thanks a lot, its all good now
     
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