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How can I derive a formula for change in B field strength?

  1. May 14, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    I'm a bit stuck on this question (which is homework so hints are more welcome than outright answers). The question is:

    A very long wire carrying a current I is moving with speed v towards a small circular wire loop of radius r. The long wire is in the plane of the loop and is too long to be entirely shown in the diagram.

    lQAjR.png

    The strength of a magnetic field a distance x from a long wire is
    $$|B|=\frac{μ_{0}I}{2πr}$$
    What is the equation for the rate of change of the strength of the magnetic field at the centre of the loop?

    As a hint we are given that $$V = \frac{dx}{dt}$$

    Now, I can see that the equation for the strength of the field comes from amperes law, and is essentially the magnetic field along a loop around the wire, divided by the circumference of that loop.

    So it makes sense to me that given everything else in the equation is constant, delta B should come straight from the change in the circumference of a circle as the radius shrinks. Now since

    $$\frac{dC}{dt}=2\pi\frac{dr}{dt}$$

    and since in this case $\displaystyle{\frac{dr}{dt}}$ is simply the velocity of the wire, it seems to me that the change in the field strength at the center of the loop should simply be

    $$\frac{d|B|}{dt}=\frac{μ_{0}I}{2πV}$$

    however I'm getting that this answer is wrong. Can anyone explain where I've made a mistake?

    Thanks for your help.
     
  2. jcsd
  3. May 14, 2016 #2

    blue_leaf77

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    Just from the dimensional analysis, one can tell that it is incorrect.
    The rate of change of the strength of the B field at a fixed point is
    $$
    \frac{d}{dt}|B|
    $$
    Now, plug in the equation for ##|B|## into the time derivative.
     
  4. May 14, 2016 #3
    So, would you expect to get $$\frac{μ_{0}Ir^{2}-2\pi V}{2\pi r^{2}}$$ ?
     
  5. May 14, 2016 #4

    blue_leaf77

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    Is that what you get after differentiating ##|B|## with respect to ##t##?
     
  6. May 14, 2016 #5
    That's what 'I' got, whether or not it is correct is what I was asking? should it be a plus instead of a minus?

    as in, I used implicit differentiation to turn 1/r into -1/r^2 * dr/dt. since in this case dr/dt = -V that gives V/r^2 etc etc
     
  7. May 14, 2016 #6
    ok i think i got it, i forgot that the constants I pulled out the front are being multiplied back in, so it should be $$\frac{μ_{0}IV}{2\pi r^{2}}$$?
     
  8. May 14, 2016 #7

    blue_leaf77

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    Yes.
     
  9. May 14, 2016 #8
    ok, Thank you so much.
    I am way too stressed right now, and you were a big help :)
     
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