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Magnetic Flux, Equator and flying bullets!

  1. Nov 15, 2008 #1


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

    At the equator, the earth’s magnetic field is approximately horizontal, is directed towards the north and has a value of [tex] 8 × 10^{-5} [/tex] T.


    Estimate the EMF induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot east. Assume the bullet has length of 1 cm, a diameter of 0.4 cm and travels at a speed of 300 m/s (for simplicity, one can assume that the bullet has a square cross-section).

    2. Relevant equations

    [tex] \epsilon = -\frac{d\Phi}{dt} [/tex]

    [tex] \Phi = \int B da = BHx [/tex]

    3. The attempt at a solution

    I think I have done this question, but would like to check to see it I have done it correctly...

    [tex] \Phi = BHx [/tex]

    [tex] \epsilon = -\frac{d\Phi}{dt} [/tex]

    [tex] \epsilon = -\frac{dBHx}{dt} [/tex]

    B and H are constants:

    [tex] \epsilon = -BH\frac{dx}{dt} [/tex]


    [tex] \frac{dx}{dt} = v [/tex]

    [tex] \epsilon = -BHv [/tex]

    we know:

    B = [tex] 8 × 10^{-5} [/tex]
    H = 0.004
    v = 300

    thus inserting the values I get:

    [tex] \epsilon = -9.6 \time 10^{-5} [/tex] Volts

    Does this look correct?

  2. jcsd
  3. Nov 15, 2008 #2


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    Does this look like the right solution?

  4. Nov 16, 2008 #3


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    Does it look like I have attempted the question in the correct way? Does it look right?


  5. Nov 16, 2008 #4
    Hi TFM,

    Your answer looks good to me, although I wouldn't use H as a variable when considering B-fields since it could be confused with the auxillary field H. This question is familar to a hall effect question, although the circumstances are different you get the same final equation. I geuss when it comes down to it though you still have moving charges in a B-field, so in essence it's the same. Odd.
  6. Nov 17, 2008 #5


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    Ok that's good. For the second part,


    What is the EMF if the bullet was travelling South?

    I am assuming that the same equation will be used, which should give the same value, since none of the variables appear to have changed. The only real difference is that the bullet is no longer travelling perpendicular, but it does have a perpendicular surface (to the current). Does this sound right?

  7. Nov 17, 2008 #6
    If the bullet is travelling south then all the charges inside it are also travelling south. The velocity of the charges will be parallel to the magnetic field. What is the force on a charge when moving parallel to a magnetic field?
  8. Nov 17, 2008 #7


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    Well the formula is:

    [tex] Q(v \times B) [/tex]

    which is a cross product, so the force will be 0?

  9. Nov 18, 2008 #8


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    Does this look right? If so, how would this force connect to the EMF force, [tex] \epsilon [/tex]?

  10. Nov 18, 2008 #9
    note: emf isn't a force.

    The two methods (Faraday and Lorentz) should lead to the same answer. However I am now confused, since when the bullet moves south there is still a surface parallel to B (so B.da is non zero) which will sweep out an area and lead to an emf. However the Lorentz force law contradicts this, because it says that since all the charges are now moving parallel to B the force on them must be zero.

    Please can someone explain this odd inconsistency? I think I have something obvious mixed up somewhere.
  11. Nov 19, 2008 #10
    I think that as the bullet is now travelling parallel to the B-field there will be no change in magnetic flux through any of it's surfaces or open surfaces, since there will be as much B-field going in as there is coming out and it's not cutting any field lines. Does that make sense?
  12. Nov 20, 2008 #11


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    That does indeed make sense.

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