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Particle spiraling in a magnetic field

  1. Dec 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle has specific charge α=q/m=108 C/kg. Then it enters a bubble chamber with magnetic field of induction B=10-2 T (the field is perpendicular to particle's velocity). Particle experiences drag force at the same time F = -kv, and as a consequence the particle spirals inward. After 2 full rotations the radius decreaces by 2%. Then after the magnetic field was switched off, the particle travels L=30 cm before coming to a stop? What was particle's initial velocity?



    2. Relevant equations

    FLorentz = Bvq

    F=ma = m dv/dt

    Fdrag=-kv



    3. The attempt at a solution

    I began by writing down the following equation at time t=0

    Bvq = mv2 /2, where v is the inital velocity of the particle (just before drag force started acting on it).

    Then, I balanced 2 expressions involving drag force and m dv/dt.

    kv = m dv/dt , k ds/dt = m dv/dt ==> k ds = m dv, which upon integrating gives ks=mv.

    Now, my question is what to do next. I know that these two forces, drag force and Lorentz force are not acting in the same direction. That implies that one component of the velocitiy decreases while other stays the same. Bvyq = mvy2/R. Moreover, we have a fact that radius decreases as well. I would like to know how can I combine these equations to get the initial velocity of the particle. Thank you in advance! :D
     
  2. jcsd
  3. Dec 23, 2013 #2

    TSny

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    Hello ajdin and welcome to PF!

    You have force on the left and energy on the right.

    Think about the direction of the drag force and how that would affect the signs in this equation.
     
  4. Dec 23, 2013 #3
    Thank you very much! :D

    In the first part, I ment Bvq = mv2/R, which after simplyfing gives v = Bqr/m=B[itex]\alpha[/itex]m.

    In the second part, the quation becomes, -kv = m dv/dt, since drag force acts in the opposite direction (and since the the particle is spiraling, is it right to say that the drag force is tagential?). When I integrate both sides of the equation -kv = m dv/dt, I get

    [itex]\int[/itex] dv/v = [itex]\int[/itex] -k/m dt, and after evaluating this integral from v0 to v(t) on the left side, and from 0 to t on te right side I get:

    lnv(t) - lnv0 = -k/m t

    lnv(t)/v0 = -k/m t

    v(t) = v0 exp (-k/m t)

    Could you tell me if I derived these expressions correctly? Now, I think that only one component of this velocity matters, when we want to balance Lorentz force with centripetal force.

    Bvyq = mvy2/R. But then, I have to find an angle between the drag force and Lorentz force. Drawing a diagram, I end up having this equation:

    sin[itex]\varphi[/itex] = Fdrag/(Fdrag2+Florentz2)1/2

    I would like to ask you if I am on a right track, and what should I do next? I have that piece of information that radius decreases after 2 complete circles, that is after 4[itex]\pi[/itex] rad. Thank you very much :D
     
  5. Dec 23, 2013 #4

    TSny

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    OK. (This assumes the particle may be considered as moving in circular motion, which is a good approximation for this problem since the radius only decreases by about 1% per revolution.)

    I'm not sure what the last expression means here.

    Good

    That looks correct.

    I don't think you need to worry about the angle ##\varphi## since it will be very small.

    By what percentage must the speed decrease in order for the radius to decrease by 2%?
     
  6. Dec 23, 2013 #5
    Thank you for your answer!

    In this part, I made a stupid error, writing v=B[itex]\alpha[/itex]m instead of v=B[itex]\alpha[/itex]R.

    Since, we've already derived the relation between velocity and radius, I would say that velocity has to decrease by 2% in order for the radius to decrease by 2% as well. So using our equation where the velocity decreases exponentially, we have:

    v(t) = (1-0.02)v0 = 0.98v0

    v(t) = v0exp(-k/m t)

    0.98v0 = v0 exp(-k/m t)

    exp(-k/m t) = 0.98

    -k/m t = ln0.98 <==> k/m t = -ln0.98 = 0.02020271

    After the magnetic field was turned off, we'll have a particle whose velocity is 0.98v0 and on which the drag force is acting. Neglecting gravity, after magnetic field was turned off, the equation becomes:

    kv = ma

    k ds/dt = m dv/dt ==> k ds = m dv, integrating this I end up having:

    kL = m(v(t) - 0), since particle comes to a stop.

    kl = mv(t) = 0.98v0

    Now, I have a problem involving t.

    I have this relation k/m t = -ln0.98 = 0.02020271 and if I write this equation kv = m dv/dt which after integrating gives

    k dt = m dv/v, and here I have a problem, since integral goes from v(t) to 0. I think this is the crucial step in this problem, and I know that I have to combine these expressions to get the answer, but apparently I am not creative enough. Thank you once again for your help :D
     
  7. Dec 23, 2013 #6

    TSny

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    Your work looks good!

    A typo here with a missing m.

    Can you derive an expression for the period T of one orbit in the magnetic field?
     
  8. Dec 23, 2013 #7
    Yes, again a typo here. It should be kl=mv(t) = 0.98mv0

    If we refer to the first equation, mv02/R = Bqv0, we see that

    v=BqR/m = BR[itex]\alpha[/itex] , where [itex]\alpha[/itex] = q/m

    v = 2[itex]\pi[/itex]R/T

    2[itex]\pi[/itex]R/T = BR[itex]\alpha[/itex]

    T = 2[itex]\pi[/itex]/B[itex]\alpha[/itex]

    Then the total time will be 2T

    t = 2T = 4[itex]\pi[/itex]/B[itex]\alpha[/itex]

    In my previous post, I posted this equation:

    0.98mv0 = kl, and k/m t = -ln 0.98

    Combinig these two equations we have:

    v0= kl/0.98m = l/0.98 -ln0.98/t

    v0=-ln 0.98/0.98 l/2T = -ln 0.98/0.98 B[itex]\alpha[/itex]l/4[itex]\pi[/itex]

    v0=492.3966752 m/s

    I hope that I haven't made any errors and I hope that this result sounds reasonable. I really don't know how I can thank you for your amazing tips and hints on how to approach this problem. Thank you very much :D
     
  9. Dec 23, 2013 #8

    TSny

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    Your work looks good to me and I got the same answer. (No need for all those non-significant digits.)

    Nice work!
     
  10. Dec 23, 2013 #9
    So I will write 492.4 m/s as a final answer. Thank you very much once again.
     
  11. Dec 23, 2013 #10

    TSny

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    OK. The initial data is only given to 1 or 2 sig figs. Don't know how picky your instructor is.

    The answer seems small for a particle in a bubble chamber. That speed yields an orbit radius of about 0.5 mm which also seems small. But, anyway, I don't see any errors.

    If you're curious, here's a paper that deals with the general case where the radius might decrease a lot per orbit.

    http://www.nim.nankai.edu.cn/upload/yuyinben/chenjl/circle_jpa_v2.pdf

    http://mathworld.wolfram.com/Involute.html

    http://mathworld.wolfram.com/LogarithmicSpiral.html

    http://mathworld.wolfram.com/LogarithmicSpiralInvolute.html
     
    Last edited: Dec 23, 2013
  12. Dec 23, 2013 #11
    This reminds me of a problem I saw in Irodov's book. There, we had no drag force, but still, it was very interesting to work on that problem. Thank you very much for your help. I feel that I've learnt something new and important, something I will definitely use in the future :D
     
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