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Force on a dipole moving through a magnetic field

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

    A dipole of moment ##\vec{p}## where p is fixed, moves with velocity ##\vec{v}## though a magnetic field ##\vec{B}##. Show that the force on the dipole is ##\vec{v}\times(\vec{p}\cdot\vec{\nabla})\vec{B}+\dot{\vec{p}}\times\vec{B}##.

    2. Relevant equations

    ##\vec{F}=e(\vec{E}+\vec{v}\times\vec{B})##

    3. The attempt at a solution
    This is how I am picturing the situation:
    Dipole.png

    ##\vec{F}=\sum_i\vec{F_i}##
    ##~~~=e\vec{v}\times\vec{B_1}-e\vec{v}\times\vec{B_2}##
    ##~~~=e\vec{v}\times\vec{B}(\vec{a}+\frac{1}{2}\vec{d})-e\vec{v}\times\vec{B}(\vec{a}-\frac{1}{2}\vec{d})##, Now I taylor expand ##\vec{B}## and disregard high order terms:
    ##~~~\approx e\vec{v}\times[\vec{B}(\vec{a})+\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]-e\vec{v}\times[\vec{B}(\vec{a})-\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]##
    ##~~~=e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})-e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
    ##~~~=e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
    ##~~~=\vec{v}\times(\vec{p}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
    Using the identity ##\vec{\nabla}\times(\phi\vec{B})=(\vec{\nabla}\phi)\times\vec{B}+\phi\vec{\nabla}\times\vec{B}## and replacing ##\vec{\nabla}## with ##\vec{v}##, also replacing ##\phi## with ##(\vec{p}\cdot\vec{\nabla})## we get:

    ##~~~=\vec{v}(\vec{p}\cdot\vec{\nabla})\times\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}##
    ##~~~=\vec{v}\times (\vec{p}\cdot\vec{\nabla})\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}##

    I think the above derivation is correct so far, so all that remains is to show that ##(\vec{p}\cdot\vec{\nabla})\vec{v}\equiv \dot{\vec{p}}## But I am stuck with this, any help will be appreciated.
     
  2. jcsd
  3. Mar 1, 2016 #2

    haruspex

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    The first of those does not look like a valid substitution to me, and I'm not sure about the second. The formula depends on the differential nature of ##\nabla##.
    I would certainly not expect to see a ##\dot v## term arise, and it would not be related to ##\dot p##. You are given that the dipole is constant in magnitude. What does variation in the vector over time therefore represent physically?
     
  4. Mar 1, 2016 #3
    Thanks for the reply haruspex.

    Yes I was quite iffy about making this substitution, do you know of any identity I can apply in order to expand this expression?

    I think it is supposed to represent any rotation the dipole might be experiencing in the field. Although i'm not sure what bearing p being fixed has on the outcome of this question.

    So my last two lines of working are incorrect which is due to the dodgy substitution I made?
     
  5. Mar 1, 2016 #4

    haruspex

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    Yes, I agree with all of that. Not sure of the correct path though... will try to spend some more time on it later.
     
  6. Mar 1, 2016 #5
    Thank you, I will post if I get any farther.
     
  7. Mar 1, 2016 #6

    haruspex

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    Ok.
    I tried thinking of the dipole as two charges, so a rotation of the dipole looks like two parallel currents in the same direction. When the dipole is aligned with the magnetic field, I can see that this produces a net force; but when the dipole, the field, and the rotation vector of the dipole are mutually orthogonal the two currents are aligned with the field, so I don't see why there is a force.
     
  8. Mar 1, 2016 #7

    TSny

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    I think you should let each charge have its own velocity: ## \vec{v_1}## and ##\vec{v_2}##.
    Shouldn't ##\vec{B}(\vec{a}+\frac{1}{2}\vec{d}) = \vec{B}(\vec{a}) + \frac{1}{2} ( \vec{d}\cdot\vec{\nabla} ) \vec{B}|_{\vec{a}}##? (with parentheses)
     
  9. Mar 1, 2016 #8
    But won't the current be in the direction in which the charges are moving,##\vec{v}## ? which would be perpendicular to the magnetic field, so it would feel a force?


    Ok and then would it be the case that ##\vec{v_1}=-\vec{v_2}## ?


    Yeah that should be the case, but does this change my answer (prior to the last two lines)?
     
  10. Mar 1, 2016 #9

    haruspex

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    Say the field is along the z axis, the dipole is along the x axis, and the rotation of the dipole along the y axis. Ignoring the linear movement of the dipole, the charges are moving parallel to the z axis, i.e. the currents are parallel to the field.
    Anyway, since TSny has joined the thread you are in much more capable hands, so I shall bow out.
     
  11. Mar 1, 2016 #10

    TSny

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    No. Their sum is related to ##\dot{\vec{a}}##.

    I don't think you need to invoke the vector identity that you used in the first post. Yes, you should get your final result plus a term that comes from the particles having different velocities.
     
  12. Mar 1, 2016 #11

    TSny

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    Please don't. Excuse my jumping in. :redface:
    I just happened to find that the result follows once you assign different velocities to the two particles.
     
  13. Mar 1, 2016 #12

    TSny

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    I'm probably doing something wrong since I get the stated result without assuming that ##|\vec{p}|## is fixed. I'll need to check over my work.
     
  14. Mar 1, 2016 #13
    Ok I think I have got it. Please excuse my laziness but I would prefer not typesetting it and just posting a picture.

    12790008_798083910296570_1485818228_o.jpg 12810320_798083953629899_1832171667_o.jpg
     
  15. Mar 1, 2016 #14

    TSny

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    That looks good. Note that the condition "p is fixed" is not used.
     
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