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Question about Biot-Savart Law (B-field)

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data
    擷取.PNG
    擷取1.PNG
    I=10A,L=0.5m,a=0.3m,x=0.2m
    2. Relevant equations
    upload_2016-4-19_16-38-1.png

    3. The attempt at a solution
    by the Biot-Savart Law
    ds cross r=-dxcosθ k(direction k)
    dB= μI/4π∫-dxcosθ/r^2
    B=μI/4π∫-dxcosθ/r^2 (from ?? to ??) in magnitude
    question:
    1) what is ?? to ??
    2)how dx and r respect to dθ????


     

    Attached Files:

  2. jcsd
  3. Apr 19, 2016 #2

    BvU

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    Look up the magnetic field from a current carrying infinitely long straight wire, an application of the BS law.
     
  4. Apr 19, 2016 #3
    where is the start and end point of integration? 0 to inf?
    then change cosθ and r respect to x? also the direction B-field is -k?
     
  5. Apr 19, 2016 #4

    BvU

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    Are you aware of what you are doing ? What BS entails ? You write dx for an integration in the y direction, you write dB and on the next line B for the same expression on the right. You want to use the Biot Savart Law, which is fine, but somehat over the top: Ampere is already fine here!
    No, ##-\infty## to ##\infty##: everywhere where there is a contribution to the B field.
     
  6. Apr 19, 2016 #5
    OK thx
     
  7. Apr 19, 2016 #6

    BvU

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    Any idea what ##\theta ## is ?
     
  8. Apr 19, 2016 #7
    Yes after check with my lecture notes and upload_2016-4-19_18-31-7.png for Magnetic Field for an Infinite Long, Straight Conductor
     
  9. Apr 19, 2016 #8

    BvU

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    Don't see no ##\theta## :smile:

    Can you derive ##B={\mu_0I\over 2\pi a}## from BS :rolleyes: ?

    But I spy an ##a## and that's not the ##a## in 'your' problem statement :nb) !
     
  10. Apr 19, 2016 #9
    upload_2016-4-19_18-57-3.png upload_2016-4-19_18-57-22.png
    upload_2016-4-19_18-57-16.png
    θ1=-π/2,θ2=π/2,If the conductor is an infinitely long, straight wire
    hence upload_2016-4-19_18-59-22.png
    i think a is r when i used in this question
     
  11. Apr 19, 2016 #10
    can i ask you another question?
    擷取4.PNG 5.PNG
    using faraday law and lens law
    e.m.f=-Nd(flux)/dt
    =-200A(dB/dt)
    =-200(aL)(dB/dt) how i link dB/dt with velocity dv/dt=5ms^-1???
     
  12. Apr 19, 2016 #11

    BvU

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    Reassuring you found it; especially the transition from ##dx## (your ##dy##) to ##d\theta## isn't all that trivial.
    The end result is Ampere's law for the straight long wire: ##B \, 2\pi a = \mu_0 I##.
    Good.

    And yes, ##a=r##.
     
  13. Apr 19, 2016 #12

    BvU

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    You use the result of part i.
    And Flux is not ##AB## but ##\int \vec B(r)\cdot d\vec A\ \ ## !
     
  14. Apr 19, 2016 #13
    i feel confuse that what is the changing time in this question, since v=5ms^-1 with constant velocity,the question just asking the e.m.f at x=0.2m,but at the beginning of the figure ,it alreay in x=0.2m, so how can i cal?
     
  15. Apr 19, 2016 #14
    6.PNG
     
  16. Apr 19, 2016 #15

    BvU

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    Do the integral for the flux. Should give you something that depends on x: ##\Phi(x)## . Then ##{d\over dt} \Phi = {d\over dx} \Phi \ {dx\over dt} = \Phi' v, \ \ ## to be evaluated at x = 0.2
     
  17. Apr 19, 2016 #16
    do you mean dr/dt in this case? since B=μI/2πr (dB/dr)(dr/dt)=(dr/dt)(-μI/2πr^2)=5(-μI/2π(0.2)^2) for x=0.2m
     
  18. Apr 19, 2016 #17

    BvU

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    No. For the flux, r is just an integration variable. ##v## is ##{dx\over dt} = 5 ## m/s.
     
  19. Apr 19, 2016 #18
    i know the concept but i do not know how to solve
    e.m.f=-Nd/dx(∫B dA) (5) then how can i solve it????
     
  20. Apr 19, 2016 #19

    BvU

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    Try to work out ##\int B \, dA ##

    Note: e.m.f=-Nd/dt (∫B dA)
     
  21. Apr 19, 2016 #20
    B=
    ∫B dA=∫μI/2πr^2 dA <----how A line with r dA=2πrdr?
     
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