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Homework Help: Biot-Savart Law problem

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    dB = (μI/4π)(dLsinθ/r^2)

    3. The attempt at a solution

    the flat edges of the loop will not contribute to the magnetic field because sinθ = . Only the curved outer loop with radius I will call r2 and length L2 and inner loop with radius r1 and length L1 will contribute.

    [itex] B = \int_{0}^{L_2} \frac{μI(dL)}{4π(r_2)^2} + \int_{L_1}^{0} \frac{μI(dL)}{4π(r_1)^2}[/itex]

    [itex] B = \int_{0}^{270°} \frac{μI(r_2dθ)}{4π(r_2)^2} + \int_{270°}^{0} \frac{μI(r_1dθ)}{4π(r_1)^2}[/itex]

    [itex] B = \frac{μI}{4π}(\int_{0}^{270°} \frac{dθ}{r_2} + \int_{270°}^{0} \frac{dθ}{r_1})[/itex]

    [itex] B = \frac{μI}{4π}(\frac{270°}{r_2} + (-\frac{270°}{r_1}))[/itex]

    [itex] B = \frac{μI(270°)}{4π}(\frac{1}{r_2} - \frac{1}{r_1})[/itex]

    [itex] B = \frac{μI(270°)}{4π}(\frac{1}{4} - \frac{1}{2})[/itex]

    [itex] B = \frac{μ(.2)(270°)}{4π}( - \frac{1}{4})[/itex]

    [itex] B = -\frac{μ(.2)(270°)}{16π}[/itex]

    and by the right hand rule I think the direction would be into the page. Is my method correct?

    Attached Files:

  2. jcsd
  3. Apr 22, 2015 #2


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    Yes, the method and the direction seem correct. I have not inspected the calculations.

    ( 270° = 3/2⋅π )
  4. Apr 23, 2015 #3
    Thank you! so not converting to radians will give me an incorrect answer?
  5. Apr 23, 2015 #4


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    You wrote three quarters of the circumference of a circle as ##\int r dL##, right ? Would you settle for 270 r or do you think it should be ##3/2\; \pi r## ?
  6. Apr 23, 2015 #5
    I am not sure. I am guessing it should be in radians?
  7. Apr 23, 2015 #6


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    Why guess ?
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