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Homework Help: Niels Bohr atom and magnetic fiels

  1. May 13, 2005 #1
    Niels Bohr atom and magnetic fields

    Could anyone help me solve the following problem?

    I am supposed to use Biot-Savart Law. What I tried was:

    The orbit of the electron can be interpreted as a current flowing in a circular spire (on the opposite direction of the electron's velocity). If an element [itex]d \vec{s}[/itex] of the spire produces a field [itex]d \vec{B}[/itex] in the position of the proton, the intensity of [itex] d \vec{B}[/itex] can be written as the following:

    [itex]dB = \frac{\mu_0ids}{4 \pi R^2}[/itex]

    [itex]B = \frac{\mu_0i}{4 \pi R^2} \oint ds[/itex]

    Calculating the integral for the entire circle:

    [itex]B = \frac{\mu_0i2 \pi R}{4 \pi R^2} = \frac{\mu_0i}{2R}[/itex]

    And then I tried to calculate the electical current as a function of the electron's velocity of displacement:

    [itex]i = nq_{e}v_{d}A[/itex]

    Where [itex]n[/itex] is the number of free charged particles, [itex]q_e[/itex] is the charge of an electron, [itex]v_d[/itex] is the velocity of displacement of the charge and [itex]A[/itex] is the area of section of the current conductor. I am not sure what [itex]A[/itex] could be in the original problem. Am I making any sense?


    Last edited: May 13, 2005
  2. jcsd
  3. May 13, 2005 #2


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    Staff: Mentor

    That's the formula for the current in a conductor of cross-section A, in terms of the motion of the electrons in it. In the Bohr atom, you don't have a conductor, you don't have a cross-section, and you don't have a number density of electrons (n).

    What you do have is a single electron traveling in a circular path. Suppose you're standing next to a point on that circular path (and that you're very tiny :smile: ). How many times per second does that electron pass you? How many coulombs per second does that work out to, counting each "pass" as a separate chunk of charge?
  4. May 13, 2005 #3
    Thanks so much! I think I got it now. Just to make sure my ideas are correct:

    In each period [itex]T[/itex], the electron passes through the point "where I am standing" once. That means that there is a charge of [itex]q_e[/itex] every [itex]T[/itex] seconds:

    [itex]i = \frac{q_e}{T} = \frac{q_e}{\frac{2 \pi R}{v}} = \frac{q_e v}{2 \pi R}[/itex]

    Sustituting [itex]i[/itex] in Biot-Savart Law:

    [itex]B = \frac{\mu_0}{2R} \frac{q_e v}{2 \pi R} = \frac{\mu_0 q_e v}{4 \pi R^2} = k_m \frac{q_e v}{R^2}[/itex]

    I think that's it!

    Thanks again
    Last edited: May 13, 2005
  5. May 14, 2005 #4


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    Staff: Mentor

    Yep, that's it. You even did it algebraically, without plugging in any numbers until the end! :!!) My students always calculate every intermediate number along the way unless I whack them by taking points off for it. :yuck:
  6. May 14, 2005 #5


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    Staff Emeritus
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    Gold Member

    Having done a classical calculation of the field at the proton, it is just one step further to calculate the interaction energy of the proton spin with this field. That, at least might tell you something. I don't see what physical idea comes out of calculating the field alone...perhaps it's just to give you practice with calculations like this?
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