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Charged particles in magnetic field

  1. May 6, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Two particles have equal masses m and electric charges of equal magnitude and opposite sign (+q and –q). The particles are held at rest in uniform magnetic field B. The direction of the field is perpendicular to the line connecting the charges. The particles are released simultaneously. What is the minimum initial separation L that allows the particles not to collide after they are released? Neglect the effects of gravity.


    3. The attempt at a solution

    Let the distance of closest approach be d.
    Applying conservation of energy,
    [itex]\dfrac{-kq^2}{L} = mv^2 - \dfrac{kq^2}{d} [/itex]

    At distance of closest approach,
    Fe=Fb
    [itex]vd^2=kq/B [/itex]

    But it looks like I will need one more equation. If the trajectory of the particles were circular, I could have applied the formula r=mv/qB but this isn't the case here.
     

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  3. May 6, 2014 #2

    BvU

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    Something weird here: if a charged particle is at rest in a magnetic field, it does not experience any force. So releasing it should not bring about any motion. Are we missing an electric field in the exercise formulation ?

    (I do see an Fe in the picture...)

    [edit] total miss. The E field is from the particles themselves. Sorry :redface:. Have to rethink this one.
     
  4. May 6, 2014 #3

    utkarshakash

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    Pay close attention to this line
    "The particles are held at rest in uniform magnetic field B".
    They are being held at rest initially and when they are released, the particles start moving towards each other due to interaction force and thus, they also experience magnetic force. Here Fe denotes the force due to mutual interaction of the particles.
     
  5. May 6, 2014 #4

    BvU

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    Ah, I've had some coffee and now I wanted to ask: what is the limiting case ?
    But you already described it, adding you need another eqn. But: [itex]\dfrac{-kq^2}{L} = mv^2 - \dfrac{kq^2}{d} [/itex] and [itex]vd^2=kq/B [/itex] have how many unknowns ?

    [edit] More coffee: 3 unknowns. Have to rethink some more.
     
    Last edited: May 6, 2014
  6. May 6, 2014 #5

    utkarshakash

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    3 unknowns- L,v and d.
     
  7. May 6, 2014 #6

    BvU

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    ##vd^2=kq/B## should probably be ##vd^2 \ge kq/B## ? But in the limiting case Fe=Fb, as you (also!) already describe niicely.

    Anyway, I never see d=0 appearing: v goes to infinity if d goes to 0. We end up in relativistic quantum electrodynamics!

    We must be doing something wrong: for every L there comes an equation in d that looks solvable (albeit not easily, but we don't need d): ##km/B^2 = d^3(1-d/L)##.

    If I could have my undemocratic way, I would say: there is no lower limit to L !

    (maybe someone more qualified could bring in the Maxwell equations here and
     
  8. May 6, 2014 #7

    TSny

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    See if you can use Newton's 2nd law in the y-direction to find a general relation between the y-component of velocity of one of the particles and its x coordinate. Then see what you can deduce if you use this relation in the energy equation at the point of closest approach.

    Edit: There is no reason that the electric force magnitude Fe should equal the magnetic force magnitude FB at the point of closest approach, in general. But I think it does turn out that for the minimum value of L for which the particles avoid collision, then the two forces do have equal magnitudes at the closest approach. The motion is then interesting.
     
    Last edited: May 6, 2014
  9. May 7, 2014 #8

    utkarshakash

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    Let at any instant velocity of the charged particle on the left side be v and it makes an angle θ with the vertical.
    In Y- direction

    [itex] F_b \sin \theta = mv \cos \theta \dfrac{d(v \cos \theta)}{dy} \\
    F_e - F_b \cos \theta = mv \sin \theta \dfrac{d(v \sin \theta)}{dx}[/itex]

    The expression does not seem nice, though :uhh:. I don't see how does this help.
     
  10. May 7, 2014 #9

    TSny

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    Express Cartesian components of the positively charged particle as ##x##, ##\dot{x}##, ##\ddot{x}##, and similarly for ##y##.

    Can you use the 2nd law to relate ##\ddot{y}## to ##\dot{x}##? Can you integrate it once to relate ##\dot{y}## to ##x##?
     
  11. May 7, 2014 #10

    utkarshakash

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    The direction of Fb changes continuously. How do I express its components along X and Y axis? Without this, it isn't possible to apply Newton's second law.
     
  12. May 7, 2014 #11

    TSny

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    ##\vec{F} = q\vec{V}\times \vec{B}##

    Note ##\vec{V} = \dot{x} \, \hat{i} + \dot{y} \, \hat{j}## and ##\vec{B} = B\, \hat{k}##.

    How do you write the y-component of the cross product in terms of ##\dot{x}## and ##B \;##?
     
  13. May 8, 2014 #12

    utkarshakash

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    Fnet = ma

    [itex] \left( qB \dot{y} + \dfrac{kq^2}{x^2} \right) \hat{i} -qB\dot{x} \hat{j}=m(\ddot{x}\hat{i}+\ddot{y}\hat{j})[/itex]

    Equating respective components

    [itex] qB \dot{y} + \dfrac{kq^2}{x^2}=m\ddot{x} \\
    -qB\dot{x} =m\ddot{y} [/itex]

    Integrating the last equation I get
    [itex]m\dot{y}=-qBx[/itex]
     
  14. May 8, 2014 #13

    TSny

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    Note that x is not the distance between the two charges. It's the x coordinate of the positively charged particle. So, your expression for the electrostatic force in terms of x needs to be modified.

    What about a constant of integration? You need to satisfy the initial conditions.
     
  15. May 8, 2014 #14

    utkarshakash

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    At x=L/2, vy = 0

    [itex]m\dot{y}=-qBx+L/2 [/itex]

    For the electrostatic force, the x should be replaced by 2x.
     
  16. May 8, 2014 #15

    TSny

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    OK. [EDIT: But you need some parentheses to make this correct. Note that qBx does not have the same dimensions as L/2.]

    Can you use this to get an expression for [itex]\dot{y} [/itex] when the particles are at their closest approach distance d?

    What is [itex]\dot{x}[/itex] at the closest approach?

    What does your energy equation in your first post look like at the closest approach?

    Good.
     
    Last edited: May 8, 2014
  17. May 8, 2014 #16

    utkarshakash

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    I get this equation
    [itex]2m\dot{y} = -qB(d+L) [/itex]
    At distance of closest approach [itex]\dot{x}=0[/itex]. So I can replace v with [itex]\dot{y}[/itex] in my energy equation. But the two equations aren't easy to solve.
     
  18. May 8, 2014 #17

    TSny

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    That's not quite correct. Check all of your signs on the right hand side.

    Good.

    It's not too bad if you note that 1/d - 1/L = (L-d)/(Ld). Look for a common factor of (L-d) in the equation. See if you can end up with a quadratic equation for d.
     
  19. May 8, 2014 #18

    utkarshakash

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    You are a genius! :smile: Thank you so much for helping. I got the correct answer

    [itex]L^3 = \dfrac{16mk}{B^2} [/itex]
     
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