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Ampere's Law and conductor

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data

    A conductor carrying current ‘I’ is in the form of a semicircle AB of radius ‘R’ and lying in the x-y plane with its centre at origin as shown in the figure. Find the magnitude of ∫B.dl for the circle 3x2 +3z2 =R2 in the x-z plane due to curve AB.

    Ans (1-√3/2)μ0I



    2. Relevant equations



    3. The attempt at a solution

    Applying Ampere’s Law ∫B.dl= μ0I for the closed loop i.e the given circle ∫B.dl turns out to be zero since there is no current flowing through the loop .But this is incorrect .

    I would be grateful if somebody could help me with the problem.
     

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    Last edited: Sep 10, 2014
  2. jcsd
  3. Sep 10, 2014 #2

    TSny

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    Hello, Tanya.

    Ampere's law only applies to a current that is part of a complete circuit. The current can't just start at A and end at B. So, you'll have to imagine a way to complete the circuit if you want to use Ampere's law.

    (Do you have a typographical error in your equation for the circular path in the x-z plane?)
     
  4. Sep 10, 2014 #3
    Hi TSny

    Thanks for the response .

    Yes ,there was a typo in the equation for circular path . I have fixed it .

    Should I join the path from B to A to complete the circuit ?
     
  5. Sep 10, 2014 #4

    TSny

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    Maybe. :smile:
     
  6. Sep 10, 2014 #5

    nrqed

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    I think you are expected to ignore any wire beside the one shown because you may take for granted that anything else will not produce any flux through your surface (for example, the remaining wire could extend away along the y axis). Just obtain the magnetic field produced by the section shown and calculate the line integral (or calculate the integral of [itex] \int \vec{B} \cdot \vec{dA} [/itex] and use Stokes theorem if you have seen that).
     
  7. Sep 10, 2014 #6
    I am actually smiling while reading this .

    This gives correct answer :smile: .

    So basically the semicircular conductor carrying current is equivalent to a closed circuit conducting path from A to B ( semi circle ) and B to A ( straight line ) .

    Does that sound right ?
     
    Last edited: Sep 10, 2014
  8. Sep 10, 2014 #7

    TSny

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    Well, the semicircular current alone is not equivalent to a closed current loop which consists of the semicircular part plus a straight current from B to A. Perhaps I'm misinterpreting what you are saying.

    You want to find ∫Bdl for the field of just the semicircle current. This is not the same as ∫Bdl for the complete closed loop consisting of the semicircle and straight line. But you can relate the two ∫Bdl 's using ideas of superposition.
     
  9. Sep 10, 2014 #8

    rude man

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    I don't see the rationale in doing that. (Which means nothing, really.) :tongue2:

    Personally I would have used Biot-Savart to find B(x,z), then integrated B.ds.
     
  10. Sep 10, 2014 #9

    TSny

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    Hi, rude man. You can use Biot-Savart for the semicircular current, but it looks a little messy to me. I think it's easier to use Ampere's law and some judicious superpostion so that you only have to use Biot-Savart for a simpler geometrical current distribution.
     
  11. Sep 10, 2014 #10
    Ok :redface:

    Magnetic field due to straight wire BA at the circular path ##B' = \frac{3μ_0I}{4\pi R}##

    Magnetic field due to semi circular wire AB = ##B##

    ##\oint \vec{B'} \cdot \vec{dl} = \frac{\sqrt{3}μ_0I}{2}##

    ##\oint \vec{B} \cdot \vec{dl} + \oint \vec{B'} \cdot \vec{dl} = μ_0I ##

    ##\oint \vec{B} \cdot \vec{dl} = μ_0I(1-\frac{\sqrt{3}}{2}) ##

    Why is it giving correct answer ? I may have made some calculation mistake .
     
    Last edited: Sep 10, 2014
  12. Sep 10, 2014 #11

    TSny

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    That looks good.

    Not sure why you have some doubt. Looks like you are using the idea of superposition as illustrated in the attached picture. You do want to make sure you understand the signs of the various terms (which depend on which way the currents are flowing and which way you are integrating around the circular path).
     

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  13. Sep 10, 2014 #12

    rude man

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    I will stipulate B' = 3μ0I/4πR as derived by Biot-Savart. (I didn't check it but OK).

    But then why not ∫B' dl = 3μ0I/4πR * √3(2πR) = 3√3μ0I/2 ?

    And then how can ampere be used with the completed semicircle plus straight wire? How can that closed semicircle circuit be interpreted by ∫B.dl = μ0I?
     
  14. Sep 10, 2014 #13

    TSny

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    What is the radius of the circular path of integration (3x2 + 3z2 = R2)?

    Not sure I'm following. Ampere's law can be used for any distribution of steady current and any shape of path of integration. For the completed semicircular current loop, Ampere's law will hold when integrating over the circular path in the x-z plane. The value of B in Ampere's law is the net magnetic field (at a point on the path of integration) due to all current elements in the complete circuit of current.
     
  15. Sep 10, 2014 #14

    rude man

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    The net current due to the complete semicircular wire piercing the x-z circle = 0!
     
  16. Sep 10, 2014 #15

    TSny

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    To get the radius of the circular path, write the equation as x2 + z2 = R2/3.

    The attached picture shows the "completed" path of current (to form a complete circuit). There's a current going through the path of integration. The magnetic field in Ampere's law at points on the path of integration is the net field due to the semicircle and the straight section.
     

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  17. Sep 10, 2014 #16

    rude man

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    Oops, I got the radius wrong. It's 1/sqrt(3). Senior moment .. thanks for your patience!
     
  18. Sep 10, 2014 #17
    I may have used the idea of superposition but this is not what I intended in post#10 :shy:. I didn't take straight conductor AB into consideration or did I :frown: ?

    I thought of the semicircular conductor as equivalent to the closed loop .This means I am still not sure with Ampere's Law and superposition principle.

    Are we applying Ampere's law to only closed loop ? Then what is the role of straight conductor AB (in which current is flowing in opposite direction) ?How are we integrating the two concepts i.e how are we using the superposition principle ? Could you reflect more on the sign issue ?
     
    Last edited: Sep 10, 2014
  19. Sep 10, 2014 #18

    TSny

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    Does this picture help?
     

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    Last edited: Sep 11, 2014
  20. Sep 11, 2014 #19
    I may be having trouble with signs .

    Suppose we consider direction of current I to be positive downwards with B' positive clockwise .

    Then while using Biot Savart Law to calculate ∫B'.dl , I would use ##\oint- \vec{B'} \cdot \vec{dl} = -\frac{\sqrt{3}μ_0I}{2}## ,as B' and I for red current would be in opposite directions to that of blue current .

    The minus sign on both the sides cancel and we are left with ##\oint \vec{B'} \cdot \vec{dl} = \frac{\sqrt{3}μ_0I}{2}##

    We then use this result while applying Ampere's law to the loop .

    Is it correct ?
     
  21. Sep 11, 2014 #20

    TSny

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    You use Biot-Savart to find the magnitude and direction of B' on the path of integration due to the red current alone. You find that the direction of B' is counterclockwise as viewed from above.

    Then ##\small \oint \vec{B\:'} \cdot \vec{dl}## ##\small = \pm |B \: '|\; 2\pi R / \sqrt{3}##, where the sign will depend on whether you choose to integrate counterclockwise or clockwise around the path.

    You then combine this result with the result for ##\small \oint \vec{B \:''} \cdot \vec{dl}## for the D-shaped loop (containing the blue current) which is easily found from Ampere's law. Of course you should integrate around the path in the same direction you chose for the red current. Here I have used the notation B'' for the field produced by the D-loop. (The field produced by just the black semicircular current alone is the vector sum of B' and B''.)
     
    Last edited: Sep 11, 2014
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