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Homework Help: Magnetic Field?

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find B at the center of the 4.3 cm radius semicircle in the figure . The straight wires extend a great distance outward to the left and carry a current I = 6.6 A.


    That is kind of what the figure looks like.

    2. Relevant equations/3. The attempt at a solution

    So, magnetic field = B

    B = μₒI/4R = (4π x 10^-7)(6.6) / 4(0.046) = 4.82 x 10^-5 T

    This was wrong according to the program I use. So since it was a semi-circle I divided in half and also tried 2.4 * 10 ^-5 T. (I also tried 0).

    Any help on what I'm doing wrong?
  2. jcsd
  3. Apr 22, 2010 #2
    I would think the wires cancel each other out so you are just left with the B field of a circular loop divided by 2. I think your equation for a loop is off by 1/2, B = μₒI/2R for a loop. So then B = μₒI/4R would be your final answer but you probably would have noticed that if that was the answer. What program?
  4. Apr 22, 2010 #3
    the program is mastering physics. i have two tries left!
  5. Apr 22, 2010 #4
    Did you try what I said? Nevermind that was your first try.
  6. Apr 22, 2010 #5
    B = μₒI/4R = (4π x 10^-7)(6.6) / 4(0.046) = 4.82 x 10^-5 T

    Isn't it: B = μₒI/4R = (4π x 10^-7)(6.6) / 4(0.043) = 4.82 x 10^-5 T
  7. Apr 22, 2010 #6
    The long wires might not cancel. What is dl x r<hat> for each? (From Biot - Savart Law).
  8. Apr 22, 2010 #7
    http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_feb_28_2003.shtml#hairpin_Biot_Savart_diagrammed [Broken]

    i found this and if you scroll down to the bottom, it is my problem. i plugged in my numbers and got 3.5 * 10^-4. this isn't right either. one try left. =/
    Last edited by a moderator: May 4, 2017
  9. Apr 22, 2010 #8
    The bottom line is that it is either B from a semi circle (which it is not, you tried that) or the sum of 2 wires and the semi circle. From dl x r<hat> the wires do not cancel but add so it is the latter. What is the contribution from one wire? It seems like that page should work. The font doesn't work on my computer so it is hard to read.
  10. Apr 22, 2010 #9
    thank you for the help! i got the right answer. =)
  11. Apr 22, 2010 #10
    Since the B field of a wire is B = μₒI/2pir, the contribution of one wire at r=R is B = μₒI/2piR. Since there are two wires they add B =2*( μₒI/2piR) = μₒI/piR. Adding that with the semicircle would be:
    B = μₒI/piR + μₒI/4R.
  12. Apr 22, 2010 #11
    What was it??
  13. Apr 22, 2010 #12
    More important; have you understood what you have done?
    Do you know how to apply the Biot-Savart equation?
  14. Apr 22, 2010 #13
    I may as well have invented the Biot-Savart Equation.:rofl:
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