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Induced electric field

  1. Mar 8, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    In the below figure, magnetic field is decreasing at a rate of 1.5T/s in a circular region of radius 2m. EMF developed in the rod will be?
    Untitled.png

    2. Relevant equations
    $$V_{\text{induced}}=-A\frac{\mathrm{d}B}{\mathrm{d}t}$$

    3. The attempt at a solution
    I tried using induced electric fields. They will be concentric circles and its magnitude is given by $$E=-\frac{r^2}{R^2}\frac{\mathrm{d}B}{\mathrm{d}t}$$ (where ##r## is the radius of the circular region and ##R## is the radius of the circular electric field) I took the components of this electric field along the rod and then tried to used $$V=\int E_rdr$$ But finding ##E_r## is difficult. There will be two points on the rod which cut the circular electric field. What should be my approach?
     
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  3. Mar 8, 2016 #2

    BvU

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    Hi Ti,

    There is something wrong with ##
    E=-\frac{r^2}{R^2}\frac{\mathrm{d}B}{\mathrm{d}t}\quad ## -- it doesn't fit dimensionally.
    Where does it come from ?

    I liked the expose here (although it ends abruptly) : it tells you magnitude and direction of the induced ##\vec E##, so all you need to do is take the component along the rod and integrate
     
  4. Mar 8, 2016 #3

    Titan97

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    Aw snap. It's $$E=\frac{r^2}{2R}\frac{dB}{dt}$$

    It comes from $$\int \vec{E}\cdot\mathrm{d}\vec{l}=-A\frac{\mathrm{d}B}{\mathrm{d}t}$$
     
  5. Mar 9, 2016 #4

    BvU

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    And then some trigonometry and an integral and Bob's your uncle.
     
  6. Mar 9, 2016 #5

    Titan97

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    The solution says that 1/5th of flux says through OAB. They solved the question in a single step while I am still stuck. How did they get 1/5?
     
  7. Mar 9, 2016 #6

    BvU

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    A ?
     
  8. Mar 9, 2016 #7

    Titan97

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    The ends of rods are A and B
     
  9. Mar 9, 2016 #8

    BvU

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    Very practical, with an A and a B for Area and magnetic field. Where is B ?
    Well, at least that should pin down the position of B (end of rod).
    Interesting that there is a one step solution. I figured we were heading for some real work. Can you post it ? (This time I don't mind a picture...)
     
    Last edited: Mar 9, 2016
  10. Mar 9, 2016 #9

    Titan97

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    @BvU I will post the picture soon. I don't have my Lumia with me now. My tab does not have a camera.
     
  11. Mar 9, 2016 #10
  12. Mar 9, 2016 #11

    Titan97

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    @gracy I will post the pictures after 7:00 IST
     
  13. Mar 9, 2016 #12

    Titan97

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  14. Mar 9, 2016 #13
    If we can somehow prove that triangle BOX = 36 degrees.
    Then yes, flux passing through AOB will be 1/5th of the total flux as 72 × 5= 360
    (I am not sure though)
     
    Last edited: Mar 9, 2016
  15. Mar 9, 2016 #14

    cnh1995

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    We can't prove that. It should be shown in the diagram. I guess there's a misprint?
     
  16. Mar 10, 2016 #15

    BvU

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    We assume ##\angle OBA = 54^\circ## as well and proceed.

    The bright idea (that I missed too o:) ) is to construct a Faraday loop BOAB where you know the induced emf and where you know the contributions from BO and AO are zero (why ? :smile: ).

    It would still be interesting to carry out the integral as mentioned in post #4 -- a lot of work that should give the same result.


    ( I use the term Faraday loop because I resent the word Faradian )

    Then:
    worried me. Is it clear that there is an induced E field as (for R > r) described in post #3 all over the place ?
     
  17. Mar 10, 2016 #16

    Titan97

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  18. Mar 10, 2016 #17

    Charles Link

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    The electric field can be determined for each r>R using Stokes law. (Draw a separate circle of radius r and let the circle and compute E which is tangent to the circle at each point.) Then you can find r as a function of ## \theta ## where each circle of radius r crosses the line. I think the trigonometric factors in computing the integral of E*ds cancel with a simple ## d\theta ## result. This longer integral approach can be shortened just by applying Stoke's law as mentioned above. It appears it doesn't even need to be a straight line path and Stokes law will give the same answer for the EMF for any path from A to B that stays outside of R.
     
    Last edited: Mar 10, 2016
  19. Mar 10, 2016 #18

    cnh1995

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    The electric field lines cutting OA and OB will be circular in nature, hence the field direction at each point will be perpendicular to OA and OB. So, there will be no emf in OA and OB. Hence, emf is developed only in rod AB.
     
    Last edited: Mar 10, 2016
  20. Mar 10, 2016 #19

    Charles Link

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    In post #17, I should have called the electric field E or E(r) instead of Er. (I was able to edit it.)
     
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