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King Group Title

2 Identical wires A and B have the same length L and carry the same current I.Wire A is bent to form a square of side a.B1 and B2 are the values of magnetic induction at the center of the circle and center of the square respectively.Find B1/B2

  • 2 years ago
  • 2 years ago

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  1. King Group Title
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    @amistre64 @Callisto please help...

    • 2 years ago
  2. King Group Title
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    @Mimi_x3 @pratu043 @Diyadiya help

    • 2 years ago
  3. King Group Title
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    @UnkleRhaukus @experimentX help..

    • 2 years ago
  4. King Group Title
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    @Mani_Jha

    • 2 years ago
  5. King Group Title
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    @amistre64 @Callisto

    • 2 years ago
  6. King Group Title
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    @.Sam. @apoorvk

    • 2 years ago
  7. King Group Title
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    ???????

    • 2 years ago
  8. experimentX Group Title
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    can you draw the picture??

    • 2 years ago
  9. King Group Title
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    they havent given one but wait i'll try...

    • 2 years ago
  10. apoorvk Group Title
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    So, okay, I got wire A 'square' and wire B 'circle' right?|dw:1342104894718:dw|

    • 2 years ago
  11. King Group Title
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    yeah...

    • 2 years ago
  12. experimentX Group Title
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    are those independent??

    • 2 years ago
  13. King Group Title
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    and R=L/2pi ?rite?

    • 2 years ago
  14. apoorvk Group Title
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    Now, for wire A, magnetic field at 'O' will be the combined effect due to the four current carrying sides of the square loop.

    • 2 years ago
  15. King Group Title
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    yeah....

    • 2 years ago
  16. apoorvk Group Title
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    yes, radius 'r' for loop 'B' = L/2pi

    • 2 years ago
  17. apoorvk Group Title
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    Now, magnetic field for A at O = 4 x \(\mu_oI/(L/8) \)

    • 2 years ago
  18. King Group Title
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    isnt it mu i /2pir

    • 2 years ago
  19. apoorvk Group Title
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    That's the square loop, (according to the formula -> B = \(\mu_oI/2\pi l\)

    • 2 years ago
  20. apoorvk Group Title
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    Oh yes I missed the '2pi', am sorry please account for that.

    • 2 years ago
  21. apoorvk Group Title
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    Now for the circular loop, \[B= \mu_oI/2r\ = \mu_oI/2(L/2\pi) = \pi\mu_oI/L\]

    • 2 years ago
  22. apoorvk Group Title
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    And for the square loop, we found it out to be --> 4 x \(\mu_oI/2\pi(L/8)\) = 16\(\mu_oI/\pi L\)

    • 2 years ago
  23. King Group Title
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    so we get pi^2/16 but answer is pi ^2 /8sqrt2

    • 2 years ago
  24. apoorvk Group Title
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    Now you can find out the ratio! \[\frac{B_A}{B_B} =\frac{16 μ_oI/πL}{πμ_oI/L} = \pi^2/16\] Now where are we going wrong hmmm?...

    • 2 years ago
  25. King Group Title
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    even i got this answer i was wondering hw this is wrong...... :(

    • 2 years ago
  26. King Group Title
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    :__(

    • 2 years ago
  27. apoorvk Group Title
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    Lol there may just be a printing error in the book. Or. Did you copy out the question correctly?

    • 2 years ago
  28. King Group Title
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    yeah its not a book its a worksheet..okay so thnx didi!!

    • 2 years ago
  29. apoorvk Group Title
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    DIDI??? o.O *facepalm*

    • 2 years ago
  30. King Group Title
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    :D

    • 2 years ago
  31. apoorvk Group Title
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    Arey Sirji, I am a guy! (ab Bhaiya bolne ki zarurat nahi hai lekin :P )

    • 2 years ago
  32. apoorvk Group Title
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    And whatever worksheet or book, please do check out what's wrong and let us know. :]

    • 2 years ago
  33. experimentX Group Title
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    what's the answer anyway??

    • 2 years ago
  34. apoorvk Group Title
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    @experimentX I seem to get (pi^2)/16, but his book says (pi^2)/(8sqrt2).

    • 2 years ago
  35. experimentX Group Title
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    magnetic field due to circular loop is \[ \vec B_1 = {\mu_0I \pi \over L}\] Magnetic field due to square loop is \[ \vec B_2 = {8\sqrt 2\mu_0 I \over L \pi}\] \[ B_1/B_2 = \pi^2/8 \sqrt 2\] http://www.scribd.com/doc/56083415/3/Example-9-1-Magnetic-Field-due-to-a-Finite-Straight-Wire

    • 2 years ago
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