1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fairly hard induced EMF problem

  1. Aug 2, 2017 #1
    1. The problem statement, all variables and given/known data
    6.gif
    (ignore the suggested problems for test 3.)
    If for whatever reason the image doesn't load the given's are that:
    The area is 25pi(.125m)^2 (A circular coil with 25 turns and a radius of 12.5cm.) The magnetic field varies with time and is: (1.5T/s)t I hat + (3.33T-(2.25(T/s^2)t^2))J hat -(2.10T)k hat The Area lies flat on the xz plane.





    2. Relevant equations
    EMF=Acos(phi)(db/dt) (as only the magnetic field varies with time.





    3. The attempt at a solution
    Area has already been given.
    db/dt=(1.5T/s)I hat -(4.5T/s)t J hat
    The problem arises when I try to calculate phi.
    My first instinct was to use phi as the angle between the normal of the area and db/dt (rather than B) but this was incorrect.
    I know now that we can calculate phi with the following formula:
    B dot A=BAcos(phi)
    As the normal to the area lies purely in the J hat direction BxAx and BzAz are both zero.
    Which leaves us with:
    ByAy=BAcos(phi)
    The problem is that the magnetic field we're given varies with time. How do we remove this dependency (if we even have to) to use this formula to calculate?
    My first instinct is that we turn this:
    (1.5T/s)t I hat + (3.33T-(2.25(T/s^2)t^2))J hat -(2.10T)k hat
    Into this:
    (1.5T/s)s I hat + (3.33T-(2.25(T/s^2)s^2))J hat -(2.10T)k hat
    Leaving us with:
    (1.5T)I hat + (1.08T)J hat -(2.10T)k hat
    But I'm unsure whether or not this is actually correct.
     
  2. jcsd
  3. Aug 2, 2017 #2

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    There is another way to calculate the dot product between two vectors, which you do not need to actually know the value of the angle.
    You can just multiply the components, then add them.
    For example if vector u = ai + bj + ck and vector v = di + ej + fk, then the scalar product of u and v is a*d + b*e + c*f. This will work, even if the vectors are represented by expressions, rather than just numbers. Now you have a time varying expression for A dot dB. A is the vector normal to the xz plane.
     
  4. Aug 2, 2017 #3
    That's literally what I did.
    Regardless we're only doing that dot product to get the angle phi.
     
  5. Aug 2, 2017 #4
    You can determine ##B_y## by examining the expression for ##\vec{B}##.

    ##B_y## is the coefficient of ##\hat{j}##.
     
  6. Aug 2, 2017 #5
    Yeah there's no new information here unless I'm missing something.
    What I am getting from this is that we can have the angle phi be represented by an expression varying with time rather than a concrete number:
    Arccos((3.33T-(2.25(T/s^2)t^2))(25pi(.125m)^2)/sqrt((1.5T/s)t)^2 + (3.33T-(2.25(T/s^2)t^2))^2 +(2.10T)^2))(25pi(.125m)^2)
     
  7. Aug 2, 2017 #6
    Yes...
    The coefficient being...?
    Remember it varies with time and the J hat part has a part with no reliance on time, and one with a reliance.
     
  8. Aug 2, 2017 #7
    I feel like emitting a roar of triumph that a lion would find intimidating.
    This means that phi also varies with time.
    Which means our combined equation is:
    EMF=Acos(phi)(db/dt)+BA-sin(phi)(dphi/dt)
    Where A is 25pi(.125m)^2

    Phi is Arccos((3.33T-(2.25(T/s^2)t^2))(25pi(.125m)^2)/sqrt((1.5T/s)t)^2 + (3.33T-(2.25(T/s^2)t^2))^2 +(2.10T)^2))(25pi(.125m)^2

    db/dt (1.5T/s)I hat -(4.5T/s)t J hat

    B is (1.5T/s)t I hat + (3.33T-(2.25(T/s^2)t^2))J hat -(2.10T)k

    and dphi/dt is whatever the nasty lookin' derivative of
    Arccos((3.33T-(2.25(T/s^2)t^2))(25pi(.125m)^2)/sqrt((1.5T/s)t)^2 + (3.33T-(2.25(T/s^2)t^2))^2 +(2.10T)^2))(25pi(.125m)^2 is
     
  9. Aug 2, 2017 #8
    Can someone see a problem with this math so long as we have a time (an actual number of seconds) to plug in?
    The only thing I'm not sure of is whether or not to use the full B or the j hat part.
     
  10. Aug 2, 2017 #9

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    So one way to find the dot product of two vectors is to find the magnitude of each vector, then multiply by the angle between them, this is what your formula
    EMF=|A| cos(phi)(d|B|/dt) is doing. But you can find the dot product the way, below, which doesn't even need the cosine or the angle.

    {scalar product of u and v is a*d + b*e + c*f., if vector u = ai + bj + ck and vector v = di + ej + fk }

    This link talks about calculating the induced EMF, without mentioning anything about cosines.
    https://webhome.phy.duke.edu/~schol/phy152/howtos/faraday_howto.pdf
    Whoops, the one above, is the one that I intended. The one below is OK too, but it talks about the angle and cosines.
    http://physics.bu.edu/~duffy/PY106/InducedEMF.html
     
    Last edited: Aug 2, 2017
  11. Aug 2, 2017 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You might find it easier to work with a more fundamental formula.
    In terms of the area and field vectors and the number of turns, what is the total flux? What function of that determines the EMF?
     
  12. Aug 2, 2017 #11
    Please use LaTeX to write your equations. Few if any will read them in the format you've written.
     
  13. Aug 2, 2017 #12

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    If you want to use cos(φ), then I suggest you do this. Break it up into 3 different components of B and find 3 angles: φi, φj and φk.
    So find the angle φi between the i component, and the vector normal to xz plane,
    then find the angle φj between the j component, and the vector normal to xz plane,
    and finally the angle φk between the k component, and the vector normal to xz plane. What are the cosines of each of those angles?
     
  14. Aug 3, 2017 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Or if not comfortable with Latex, at least use the superscript, subscript and Greek characters available with the buttons above the text area (X2, X2, Σ).
     
  15. Aug 3, 2017 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I can see that you are making it a deal more complicated than it is. You can leave the vectors behind quite quickly and never need to find the angle.
    Please try my suggestion in post #10.
     
  16. Aug 3, 2017 #15

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Yes I agree with @haruspex . The solution to the problem can be much simpler than you are making it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fairly hard induced EMF problem
  1. Induced emf problem (Replies: 2)

  2. Problem on induced EMF (Replies: 5)

  3. Induced EMF problem (Replies: 4)

Loading...