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!

Help clarifying an answer for magnetics question.

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    http://www.aijaa.com/img/b/00334/3685081.jpg [Broken]

    2. Relevant equations

    [tex]B= \frac{ \mu _0}{2 \pi r}I[/tex]

    3. The attempt at a solution

    I got it wrong the two first times (I got [tex]\frac{ \mu _0I}{2 \pi a}(1- \frac{a}{b})[/tex]), then the assistant teacher showed how it was done, but wasn't able to tell me why. This is the answer I was showed: [tex]B=B_a-B_b= \frac{ \mu _0I}{4}( \frac{1}{a}- \frac{1}{b})= \frac{ \mu _0I}{4a}(1- \frac{a}{b})[/tex] Can anybody clarify what the heck happened between steps 2 and 3, everything else I can understand.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 26, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Just factor out 1/a from within the parentheses:
    (blah) = 1/a(a*blah)
     
  4. Feb 26, 2009 #3
    Uh, if you compare my answer and the correct answer, I'm missing [tex]* \frac { \pi}{2}[/tex]. If you misunderstood my original text [tex]B_a-B_b= \frac {\mu _0I}{4}( \frac{1}{a}- \frac{1}{b})[/tex] is the part where, the black magic happens.
     
  5. Feb 26, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Sorry about that. I wasn't clear on what you meant by "between steps 2 and 3", so I had to guess. (You meant between expressions 2 and 3 in your equation.)

    In any case, why are you using this formula?
    That's the field from a long straight wire, not half of a current loop.
     
  6. Feb 26, 2009 #5
    That's the thing, I don't yet quite get the idea behind line integrals and I tried to derive the expression from Amperé's law. But isn't the magnetic field in the center of half current loop the sum of infinitely small straight wires?
     
    Last edited: Feb 26, 2009
  7. Feb 26, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, infinitely small straight wires. You cannot use the result for an infinitely long straight wire. Instead, integrate the magnetic field from a small current element as given by the Biot-Savart law. (The integration is trivial, by the way.) Give it a shot.
     
  8. Feb 26, 2009 #7
    Ah, okay. I've never heard of that law so we haven't gone through it in our course. I'm going to look it up. I'll make sure the lecturer gets a few praises. :biggrin:
     
  9. Feb 26, 2009 #8
    Now it makes sense. So [tex]B= \frac{ \mu_0I}{4 \pi r^2} \int dl[/tex], from 0 to [tex] \pi r[/tex]. It gives me [tex]B= \frac{ \mu_0I}{4r}[/tex]. Thanks for the help and expertise our teachers obviously lack, Doc Al! :tongue2:
     
  10. Feb 26, 2009 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Now you're cooking. (And you're welcome!)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help clarifying an answer for magnetics question.
Loading...