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Magnetic Field of Coils

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Two thin coils of radius R = 3 cm are d = 13 cm apart and concentric with a common axis. Both coils contain 10 turns of wire with a conventional current of I = 3 amperes that runs counter-clockwise as viewed from the right side (see the figure).

    18-089-HW_two_coils_sym.jpg

    (a) What is the magnitude and direction of the magnetic field on the axis, halfway between the two loops, without making the approximation z >> r? (For comparison, remember that the horizontal component of magnetic field in the United States is about 2 ✕ 10-5 tesla).
    |B| = _ T
    direction: c

    (b) In this situation, the observation location is not very far from either coil. Calculate the magnitude of the magnetic field at the same location, using the 1/z3 approximation.
    |Bapprox| = _ T

    The percent error of an approximate result can be found by symimage.cgi?expr=100%2Aabs%281-%28approx%29%2F%28exact%29%29.gif . What percentage error results if you calculate the magnetic field using the approximate formula for a current loop instead of the exact formula?
    percent error: _ %

    (c) What is the magnitude and direction of the magnetic field midway between the two coils if the current in the right loop is reversed to run clockwise?
    magnitude: _
    direction: _


    2. Relevant equations
    |B| = (mu_0 / 4pi) * (I * R^2 * 2pi) / (R^2 + z^2)^(3/2)
    |B_approx| = (mu_0 / 4pi) * (2 * I) / r

    mu_0 / 4pi = 1e-7
    R = radius
    r = distance from ring
    z = distance from center of ring along z-axis

    3. The attempt at a solution
    I tried using the |B| formula with I = 30 (3A * 10 turns of wire), R = 0.03 m, z = d/2 = 0.065 m. I got 4.6238e-5 T as my answer and it was wrong. This seems too large. Do I need to include the other wire loop, by adding? Do I need to keep I = 3 A?
     
  2. jcsd
  3. Oct 6, 2016 #2

    kuruman

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    Homework Helper
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    Hi cranen and welcome to PF.:welcome:

    Yes you need to consider the superposition from both loops. Please invest some time to learn how to use LaTeX and use it to enter your equations. It will help us help you much more efficiently.
     
  4. Oct 6, 2016 #3

    gneill

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    Staff: Mentor

    Yes you need to include both loops. Yes the current is 3 A for each loop.

    It looks to me that your result for one loop is off by a factor of ten (too large). Check your exponent calculations.
     
  5. Oct 7, 2016 #4
    I tried ## |B| = 1e-7*\frac {3A*0.03m^2*2π}{(0.03m^2+0.065m^2)^{3/2}} ## and then multiplied it by 2 to incorporate both rings and got 9.247e-6. It was wrong
     
  6. Oct 7, 2016 #5

    kuruman

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    Thank you for using LaTeX. The way you have written your equation with numbers makes it difficult to figure out what's what. Try using symbols and substitute the numbers at the very end. This will make it much easier for us to diagnose your mistakes. I don't see where the number 7 comes from in your answer and I don't see the 10 (number of loops) anywhere either. Also, there should be no factors of pi in your expression.

    On edit: I just realized that the "7" is an exponent.
     
  7. Oct 7, 2016 #6
    ## |B| = \frac μ {4π} \frac {I*R^2*2π} {(R^2 + z^2)^(3/2)} ##
    R = 0.03 m
    z = d/2 = .065 m
    I = 3 A
    ## \frac μ {4π} = 1e-7 ##
     
  8. Oct 7, 2016 #7

    kuruman

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    Great! You are a fast LaTeX learner. In the expression that you have, note that 2π in the numerator and 4π in the numerator is just 1/2. This expression is for the B-field from one coil. You need to multiply this by the number of coils that you have contributing to the B-field at the point of interest. What is that number?
     
  9. Oct 7, 2016 #8
    So it would be ## |B| = \frac μ 2 \frac {IR^2} {(R^2+z^2)^{3/2}} ## and then since there are two coils I would use just have ## |B| = μ \frac {IR^2} {(R^2+z^2)^{3/2}} ##
     
  10. Oct 7, 2016 #9

    kuruman

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    Each coil has 10 turns and you have an expression for the B-field due to 1 turn, So ...
     
  11. Oct 7, 2016 #10
    So I would need 20 total, or 10 if I used the very last equation I typed?
     
  12. Oct 7, 2016 #11

    kuruman

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    Yes and yes.
     
  13. Oct 7, 2016 #12
    Thanks so much for all the help!
     
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