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The strength of induced electric field

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    A thin and straight metallic rod of length L is rotating about its middle point with angular velocity ##\omega## in a uniform magnetic field B. The axis of rotation is perpendicular to the length of the rod and parallel to the magnetic field. The strength of the induced electric field is?
    a) Constant throughout the rod
    b) maximum at the two points halfway between the centre and the two ends
    c)minimum at the two points halfway between the centre and the two ends
    d) minimum at the ends and maximum at the centre
    e) maximum at the ends and minimum at the centre
    2. Relevant equations
    ##\nabla \times E = -\frac{dB}{dt}##
    Emf,## \epsilon = \int f_b.dl##
    3. The attempt at a solution
    According to the equation shown above and Faraday's law a change in magnetic field causes an induced electric field and in this question the magnetic field is said to be uniform. Hence from the equation, dB/dt is zero and therefore E becomes zero. So is it right to conclude that E remains constant throughout the rod?.
     
  2. jcsd
  3. May 4, 2017 #2
    why is it wrong to use the equation ##\nabla \times E = -\frac{dB}{dt}##?
    Motional emf is ## \epsilon = \int f_b .dl##
    ##f_b = vB##
    ## \epsilon = \int vB .dl##
    ## \epsilon =BLv##
     
  4. May 4, 2017 #3

    cnh1995

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    I read the problem wrong. I was thinking the rod is fixed at one of its ends. That gives a non-zero emf.
    But since the rod is moving about its centre, the emf should be zero.
    It isn't wrong. But if the rod were moving about one of its ends, you could tell, by looking at the motional emf equation whether the emf is constant throughout its length.
     
  5. May 4, 2017 #4
    Okay so the induced electric field is going to be constant throughout the length right? Also could you elaborate how emf will be zero using the equation?
     
  6. May 4, 2017 #5

    cnh1995

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    Using the motional emf equation, you can see at any given instant, the two halves of the rod move in the opposite directions. That gives zero emf in the rod.
    Zero emf is not any of the options, so I guess that could be the answer. However, I think the word 'constant' implies the quantity is non-zero.
     
  7. May 4, 2017 #6
    I tried finding it with the equation and I am not getting zero as the answer for emf. Its possible that I have made a mistake. I tried to do it in two ways and I end up getting the same non zero answer. This is how I did it:
    Method 1:
    ## \epsilon = \int f_b .dl##
    ## v=L\omega/2##
    ## f_b = BL\omega/2 ##
    integrating it from -L/2 to L/2
    ## \epsilon = \frac{BL^2 \omega}{8}##

    Method 2:
    A small area swept by one half of the rod ## dA= \frac{L^2 d\theta}{8}## (Area of a sector = ##1/2 r^2 d\theta##)
    ## \omega = d\theta/dt##
    therefore ##dA= \frac{L^2\omega dt}{8}##
    differential flux ## d\phi = B.dA##
    substituting dA ## d\phi = \frac{BL^2\omega dt}{8}##
    and emf ##\epsilon = d\phi/dt = \frac{BL^2\omega}{8}##

    So I can't see where I have made the mistake.
     
  8. May 4, 2017 #7

    cnh1995

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    Can you do that in this case? At any instant, the portions of the rod between -L/2 to 0 and 0 to +L/2 are having opposite linear velocities. The direction of the B-field remains the same. Only the velocities are opposite. Apply the right hand rule for each half and see what the emfs do to each other.
     
  9. May 4, 2017 #8

    TSny

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    This gives you the potential difference between the center of the rod and one end. But you are asked about the magnitude of the electric field at different points inside the rod.

    Consider an arbitrary point inside the rod and imagine a free charge carrier of charge q located at that point. Identify the forces that act on q.
     
  10. May 4, 2017 #9
    There would be magnetic force given by ##F_b = q(v\times B)## but it has opposite direction depending on the rotation of the rod.
     
  11. May 4, 2017 #10

    TSny

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    OK. But there is another force also. When the rod begins to rotate, free charge moves relative to the rod (due to the magnetic force) until charge accumulates in certain regions of the rod. The charge accumulations produce the induced electric field that you are interested in. It might help to review the simpler example of a rod moving with pure translation in a uniform B field. http://www.kshitij-iitjee.com/motional-emf
     
  12. May 4, 2017 #11
    I am not sure how is it right to say that charges move due to magnetic force because I have read that magnetic forces do no work. So in this case, the induced electric field will be qE = qvB on one half of the rod and the directions will be opposite for the respective forces on the other half. Hence the net effect of induced EF will be zero?
     
  13. May 4, 2017 #12

    TSny

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    You want to know how E varies inside the rod. So, use qE = qvB to help you.
     
  14. May 4, 2017 #13
    Yes right, so the EF will be maximum at the ends and minimum at the centre? because ##v=r\omega## velocity is maximum at the ends.
    On a side note, I did not get how emf will be zero and integrating the rod from -L/2 to L/2 is incorrect as cnh1995 said could you please give an insight on that especially using the equations?
     
  15. May 4, 2017 #14

    TSny

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    Right, the magnetic force does no work. But without the magnetic force, a free charge in the rod would not move significantly with respect to the rod as the rod begins to spin. To account for the work done on a free charge as the rod begins to move in the B field, you have to take into account additional forces of interaction with the atoms in the rod.

    But once things settle down and there is no further movement of charge relative to the rod, you only need to take into account the magnetic force and the electric force due to the induced E field.
     
  16. May 4, 2017 #15

    TSny

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    Yes. By symmetry you can see that E is in fact 0 at the center of the rod.
    I'm not sure exactly what is being discussed here. The potential difference between two points of the rod is (minus) the integral of the induced electric field between the points. If you integrate from one end of the rod to the other, you would get zero potential difference. This makes sense because the potential must be the same at each end due to symmetry. If you integrate from the middle to one end, I think you will get a potential difference of ##\frac{B \omega L^2}{8}##.
     
  17. May 4, 2017 #16
    I understood that emf will be zero at the centre and earlier I didn't take into account the opposite sign for the force while doing the integration. Now however Integrating from middle to end gets me a different answer and this is how.
    ##\epsilon = - \int f_b.dl##
    ##f_b = Bv##
    since the rod is rotated at its centre, ##v=L \omega/2##
    substituting this into the integral for the limits 0 to L/2
    ##\epsilon = - \int f_b.dL##
    ##\epsilon = - \int_{0}^{L/2} BL\omega/2 dL##
    Here I get ##\epsilon = - \frac {B\omega L^2}{16}## and this is for one half of the rod. Can't see where the error is. The answer is off by 1/2
     
  18. May 4, 2017 #17

    TSny

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    ##f_b## should represent the electric force per unit charge at an arbitrary point a distance ##l## from the center when you do ##\int_0^{L/2} f_b \, dl##. ##v \neq \omega L/2## except at the end of the rod.
     
  19. May 4, 2017 #18
    Yes I think I realized where the mistake is, I took ##v = L\omega /2## and not an arbitrary distance. Should have just been ##L\omega## while integrating or like you said f_b = E to avoid the confusion.
     
  20. May 4, 2017 #19

    TSny

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    OK, but the symbol ##L## is reserved for the total length of the rod. So, it is confusing to say that ##L \omega## is the speed at an arbitrary point. Might be better to use ##x## for an arbitrary distance from the center and integrate with respect to ##x##.
     
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