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Magnetic Flux through conductor coil

  1. Aug 7, 2017 #1

    Luk

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    1. The problem statement, all variables and given/known data
    Consider a conductor loop which rotates about the y-axis with angular velocity w. The positive y-axis is directed into the paper, the x-axis to the right and the z-axis upwards. There is an inhomogeneous magnetic field along the positive x-axis. Calculate the magnetic flux through the conductor loop as a function of \Phi, which is the angle between the loop and the z-axis.

    2. Relevant equations
    The inhomogeneus magnetic field can be described by: B = B_0 * (1 - z^2/a^2) * e_x

    3. The attempt at a solution
    Φ = ∫∫ B dA

    So how do I figure out dA ? I was thinking about dA = l dz cos(φ)

    Φ = ∫ B l cos(φ) dz with the integral limits -a and a.
     
  2. jcsd
  3. Aug 7, 2017 #2

    TSny

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    Looks good. Is the length l a function of z?
     
  4. Aug 7, 2017 #3

    BvU

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    Hello Luk, :welcome:


    This isn't really a relevant equation in the PF sense: it is given in the exercise statement.

    Your integral limits need reconsidering: ##a## ? You didn't tell but it seems to be a parameter of the field, not of the geometry of the loop !
    And your choice of ##\phi## gives you another expression for ##\vec B\cdot d\vec A##
    Other than that, your battle plan seems just fine to me ! Go ahead !
     
  5. Aug 7, 2017 #4

    Luk

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    no, the length l is fixed and directed along the y-axis.
    So, I am a little confused, because I have a solution to this problem which is different from mine. In the solution, the dA is expressed in terms of cylindrical coordinates.

    I think it's helpful and since it is available online, it should be fine to provide the link to the question:
    http://www.tet.tuhh.de/downloads/education/pruefungen/TET-II-Klausursammlung.pdf

    You can see a figure of the problem on page 26 and the solution on page 30. It's not in english, but still useful. I don't understand why dA can be expressed the way it is. Also I don't understand the first manipulation of the equation.
     
  6. Aug 7, 2017 #5

    TSny

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    Good point. I presumed that ##a## is the radius of the loop so that B would go to zero at the circumference of the loop. But, that should have been stated in the problem.

    I thought so, too, at first. But I think it's OK, unless I'm still overlooking something.
     
  7. Aug 7, 2017 #6

    Luk

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  8. Aug 7, 2017 #7

    TSny

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    Ugh! I was assuming a circular loop. Bad form on my part. Sorry.

    Thanks for the picture.
     
  9. Aug 7, 2017 #8

    TSny

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    The picture shows that ##\phi## is the angle the loop makes to the x axis, not the z axis. So, you need to be careful with whether or not you want ##\cos \phi## or ##\sin \phi##.
     
  10. Aug 7, 2017 #9

    Luk

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    So, here's my calculation (I hope formatting is fine):

    Φ = ∫ l * B_0 * (1- (z^2/a^2) ) * cos(φ) * e_x dz , with limits -a to a

    Φ = l * B_0 * [(z - z^3/(3a^2)) * sin(φ)] * e_x

    Φ = l * B_0 * [(a - a^3/(3a^2)) * sin(φ) - (-a + a^3(/(3a^2)) * sin(φ)] * e_x

    Φ = l * B_0 * 4/3 a * sin(φ) * e_x


    I don't know what the e_x is supposed to tell me. It is the unitary vector along the x-axis. But do I need it for the calculation?
     
  11. Aug 7, 2017 #10

    TSny

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    In the picture in the statement of the problem, they show the loop when it is parallel to the yz plane. Draw the picture for some other time, such as when the plane of the loop is at a 60 degree angle from the x-axis. If you take a strip of area of the loop of length l and width dr, note that dr is not parallel to the z axis. So, dr ≠ dz.
     
  12. Aug 7, 2017 #11

    TSny

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    ##\hat{e}_x## is a unit vector giving the direction of B. ##\hat{e}_{\phi}## gives the direction of the area vector. So, when you do the scalar product of the field with the area, you get a scalar product ##\hat{e}_x \cdot \hat{e}_{\phi}##. How can you express this in terms of ##\phi##?
     
  13. Aug 7, 2017 #12

    TSny

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    When the loop is not parallel to the yz plane, you need to distinguish between r and z. r measures distance in the plane of the loop. The solution expresses z as a function of r and uses r as the integration variable.
     
  14. Aug 7, 2017 #13

    Luk

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    allright. I'm not too comfortable with cylindrical coordinates. I mean, I .. well given the problem I wouldn't come up with the idea of using cylindrical coordinates. And I think I'm close to solving it.

    So, again, with the help of your comments I got this far:

    dr ≠ dz
    dz = dr * sin(φ)

    Φ = ∫∫ vec(B) * d vec(A) ( I know, this looks weird, I am using vec() to indicate B and dA are vectors. So in this formula I should consider the scalar product of B and the orthogonal vector of A. I didn't do that in the following calculation, so I assume I have to include some vector in the end)

    Φ = ∫ l * B_0 (1 - (z^2/a^2) ) * sin(φ) dr , with limits -a to a for the integral

    now, I replace z with r * sin(φ)

    and when I perform the calculation I end up with
    Φ = 2 l B_0 * a * (sin(φ) - ⅓ sin^3(φ))

    and this turns out to be the solution (except they get a negative sign and I don't). So last question: how do I include the scalar product of the unit vectors? I mean x and φ doesn't seem to go well together. x is carthesian and φ polar, isn't it?
     
  15. Aug 7, 2017 #14

    TSny

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    The scalar product just produces a number, not a vector. So, you would not include some vector in the end. To see where the negative sign comes from, note the directions of the unit vectors ##\hat{e}_x## and ##\hat{e}_{\phi}##. What is the scalar product of these unit vectors?

    upload_2017-8-7_12-57-56.png

    The solution takes the area vector of the loop to be in the ##\hat{e}_{\phi}## direction. You might be taking the area vector to be in the opposite direction, which would be OK. If so, your flux will differ in sign from their flux.
     
  16. Aug 7, 2017 #15

    Luk

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    The scalarproduct gives the projection of one vector onto the other, am i right?
    e_Φ * e_x = cos(90° + φ)

    I think from your sketch it can be seen that, since unit vectors have a length of 1, the scalarproduct is -1.
    Thank you so much for your help btw, I really wouldn't have got this far without it!
     
  17. Aug 7, 2017 #16

    TSny

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    Yes. A trig identity gives cos(90° + φ) = -sinφ. This is where the -sinφ comes from in the solution:

    upload_2017-8-7_20-56-7.png
     
  18. Aug 16, 2017 #17

    Luk

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    Thanks again for your help, TSny!

    I just took another glance at the question and there is something that I still don't understand:
    I want to calculate the induced electric field in the part labeled I.
    E = v x B
    since B is given the crucial part is to find out v. The velocity should be equal to the angular velocity times the distance: wa. And the direction is e_φ
    And that's a problem for me, because I don't know how to calculate the cross product of a vector with component in e_φ direction and one pointing in e_x direction. Can you help me with this ?
     
  19. Aug 16, 2017 #18

    TSny

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    How do you find the cross product of two vectors in general? Can you find the angle between the velocity vector and ##\hat e_x##?
    upload_2017-8-16_12-50-47.png
     
  20. Aug 16, 2017 #19

    Luk

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    let's see. There are different ways of how to compute the cross product, here's one:
    $$\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

    But for unit vectors - which I consider - I always go with the notion that the cross product will result in a vector which is orthogonal to the vectors involved in the cross product. So i x j = k, for example. Since e_φ and e_x are part of different coordinate systems, however, I don't know how to apply this thought.

    In the picture that you uploaded, the angle between the velocity vector and e_x should be equal to 90° + θ. In the first part of the question I have used this angle: The scalar product between two vectors equals the product of their lengths times cos(α). I think.
     
  21. Aug 16, 2017 #20

    TSny

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    Yes. Use this method to find the scalar product. You know the magnitude of each vector and the angle between them.
     
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