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!

EMF induced in coil by magnetic field

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=63835&stc=1&d=1384141892.png


    2. Relevant equations



    3. The attempt at a solution
    Well I tried this (I thought it'd work, but it's not the right answer and I'm not sure how to do it now) Anyway this is my attempt:

    [tex]\Phi = \mu_{o}\frac{NIA}{\ell} \\ \varepsilon=N\frac{d\Phi}{dt}[/tex]
    So then
    [tex] \varepsilon=-N(I_{max} \omega cos(\omega t+\phi)\frac{\mu_{o}N(Lw)}{w+h} \implies \varepsilon=(-100)(5)(cos(200\pi t+\phi))(200\pi)\frac{\mu_{o}(100)(.2)(.05)}{.1}[/tex]

    (which ended up giving me the wrong answer)
     

    Attached Files:

    • loop.png
      loop.png
      File size:
      16.6 KB
      Views:
      218
  2. jcsd
  3. Nov 10, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    This is an exercize in faraday's Law - but you have to understand the equations.
    The magnetic flux changes with time because the current changes - but it also changes with the distance from the wire. ##d\phi = B(r)dA##
     
  4. Nov 10, 2013 #3
    I'm not understanding why it's $$d\phi = B(r)dA$$ specifically dA, the area wouldn't be changing, only the distance. The B is the magnetic field, but what is this value in this case? And finally, where do you get the r from?

    (Basically I don't understand that equation)
     
  5. Nov 10, 2013 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ##d\phi## is the amount of magnetic flux though the area element ##dA## at a distance ##r## from the wire, in the plane of the coil.

    The magnetic field due to a long straight wire depends on the distance from the wire as well as the current.
     
  6. Nov 11, 2013 #5
    Ok that makes sense, but what do I do with it? If I were to integrate both sides it would just end up ## \phi=BrA##

    Also, wouldn't r therefore be changing too as the current moves around the loop?

    Edit: So ##B=\frac{\mu_{o}I}{2\pi r}## Substituting for the above equation gives ## \phi=\frac{\mu_{o}IrA}{2 \pi r}## I think the r's will cancel so then: ##\phi=\frac{\mu_{o}IA}{2\pi}##. I don't know what to do with this now lol
     
    Last edited: Nov 11, 2013
  7. Nov 11, 2013 #6
    I think I got it...$$d\Phi_{B}=\vec{B} \cdot d\vec{A}=\frac{\mu_{o}IL}{2\pi r}dr$$ Then: $$\Phi_{B}=\int_{h}^{h+w}\frac{\mu_{o}IL}{2\pi r}dr=\frac{\mu_{o}IL}{2\pi}ln(\frac{h+w}{h})$$ Then I can just plug it into: $$\varepsilon=-\frac{d\Phi_{B}}{dt}$$

    Correct?
     
  8. Nov 11, 2013 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No it doesn't. Because dA is made up of a component of dr ... you need to integrate over the length and width of the loop.

    i.e. if you divide the area into strips length L and width dr then the area element at distance r is dA=Ldr ... Now you have ##d\phi = B(r)Ldr## ... notice that B is a function of r?
    Have you never constructed integrals like this before?

    The current induced in the loop obeys Faraday's Law. It depends on the changes in the total flux through the loop.
     
  9. Nov 11, 2013 #8
    Yes I see, I was just confused mainly because of B as a function of r I thought you ment B multiplied by r, but I think I got the correct way of doing it now (above)

    Thanks.
     
  10. Nov 11, 2013 #9

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yeah: You posted while I was typing ... hang on...
     
  11. Nov 11, 2013 #10

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK - that looks good. Well done :)
    It was just a matter of getting your head in the right place huh?

    Formatting notes: the natural logarithm is \ln in LaTeX.
    In general, standard function are formatted by putting a backslash in front of their name ;)
    You can make the brackets go around big arguments by using \left( \right) instead of ( ).

    So - yeah: $$\varepsilon = \frac{\mu_0LI_{max}}{2\pi}\ln\left(\frac{h+w}{h}\right)\frac{d}{dt}\sin( \omega t +\phi )$$
     
  12. Nov 11, 2013 #11
    I just had to slow down and analyze what I was doing and I saw an example similar to this in the book that made it clear.
    Yeah that's what I got and the derivative is just ωcos(ωt+ϕ) then plug in numbers
     
  13. Nov 12, 2013 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Of course the acid test is if this gives the answer that was expected.
    If you feel keen you can try looking at the relationship in terms of the reactance of the coil.
     
  14. Nov 12, 2013 #13
    Can you explain this further?
     
  15. Nov 13, 2013 #14

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Explain which bit?

    If the treatment does not get the marks, then it means we misunderstood the question (or the model answer is wrong.)

    A coil of wire is an electrical component called an inductor - it is a reactive load so it has a reactance. (Look up "back emf".)

    The coil's presence close to the wire means there is an additional load on whatever is making the current in the wire... which is why you cannot just run a coil next to the power lines and get free electricity.

    But you may prefer to wait until it comes up in the course.
     
  16. Nov 13, 2013 #15
    I don't believe that is part of my E&M physics course. We just had an exam on the magnetism part and now we're starting on waves and optics (more modern stuff). I already took an electrical engineering course and we did use inductors quite a bit, but we never talked about back emf other than in lab.
     
  17. Nov 13, 2013 #16

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well .... that's why I said, "if you feel keen" ;)
    It can help to make connections between different parts of the work you've done.
    But that's OK - you're done with this problem now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted