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Find the magnetic field at the origin

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    2niy2ph.jpg



    2. Relevant equations

    B = u0*I / 2pi * r
    that's the only equation relevant in this problem

    U
    0 is a constant wth a value of : 4pi x10^-7

    3. The attempt at a solution

    okay so what i did was use the equation above and find the magnetic field (B) for I1 I3 and the x and y component of I2.Also, i used the right hand rule to determine the direction of the magnetic field for each one and I get :

    I1 = 3.57x10^-5 T in the -y direction
    I2 x-comp = 2,17x10^-5 T in the -y direction
    I2 y-comp = 2.17x10^-5 T in the +x direction
    I3 = 3.57x10^-5 in the -x direction

    I tried subtracting -x direction value from the +x to find the x-component at the origin and i tried adding the 2 -y values to find the y-comonent of the magnetic field at the origin I get :

    x-component : -1.4x10^-5 T
    y-component : 5.74x10^-5 T

    but that is said to be wrong , so please any help is appreciated
     
  2. jcsd
  3. Nov 23, 2016 #2

    TSny

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    Check the values for the x and y components of the field due to I2.
     
  4. Nov 23, 2016 #3
    i still get 2.17 x10 ^ -5 for both. the "r" in the x and y directions is the same which is 0.0387. and the current if I2 is 4.20A. so thats :

    B= [4pi x 10^-7 * 4.20A]/[2pi*0.00387] = 2.17 x 10^-5 T for both of them

    I don't see where i made a mistake
     
  5. Nov 23, 2016 #4

    TSny

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    Did you use the correct value of r for I2? How did you calculate the x and y components of B2 from the magnitude of B2?
     
  6. Nov 23, 2016 #5

    TSny

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    For I2 it appears that you might be trying to get Bx by doing something like Bx = μ0I2/ (2πy), where y is the y-component of r. But this is not correct.
     
  7. Nov 23, 2016 #6
    should i then use sqrt[x^2+y^2] to find the distance between I2 and the origin and then use the formula to find the magnetic field of that and then using the tangent of the angle tan[x] = 3.87cm/3.87cm and use that to determine the x and y components ?

    In fact i did that and i get 4.19x10^-7 for Bx and 1.53x10^-5 , but Bx seem to be way too small so i am not sure this is right
     
  8. Nov 23, 2016 #7

    TSny

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    Yes, although I would need to see your work to make sure that you are doing it correctly.

    What did you get for the magnitude of B2?
     
  9. Nov 23, 2016 #8
    i got 1.54x10^-5 , but if this is right everything else is correct ? i just follow the same steps using these new values or...
     
  10. Nov 23, 2016 #9

    TSny

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    OK. Then what did you get for B2x and B2y?
    Your results for B1 and B3 look right.
     
  11. Nov 23, 2016 #10
    what i got for B2x and B2y is what i said above , the stuff using the tangent and all that and the 1.54x10^-5 is the magnitude of B2

    4.19x10^-7 for Bx and 1.53x10^-5 for By
     
  12. Nov 23, 2016 #11

    TSny

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    These values aren't correct. Show how you got them.
     
  13. Nov 23, 2016 #12
    i did what is mentioned above and then i used these fomulas Bx = Bsin(x) and By = Bcos(x)
     
  14. Nov 23, 2016 #13

    TSny

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    You have the right method but you must be making a numerical error somewhere. It's a little confusing to use the symbol x for x-component and also for an angle.

    I can't identify the error without seeing your explicit calculation. What did you get for the angle x?
     
  15. Nov 23, 2016 #14
    i'm sorry but my calculations are a real mess and if i take a picture and post it here it won't help u in the slightest but here , i got 1.56 for the angle which is oddly small which is why i tried again and got 45 for the angle and now my values are :

    Bx = 1.09 x 10^-5
    By = 1.09 x 10^-5
     
  16. Nov 24, 2016 #15

    TSny

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    OK, except you need to check if you have the correct signs for these components.
     
  17. Nov 24, 2016 #16
    so By should be negative and Bx should be positive ?
     
  18. Nov 24, 2016 #17

    TSny

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    Yes.
     
  19. Nov 24, 2016 #18

    gneill

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    Staff: Mentor

    You need to be able to present your work if you expect helpers to be able to help you. If your handwritten efforts are indecipherable then please take the time to type your work into the post. The editor provides tools for creating reasonably well formatted math equations, and LaTeX syntax is also available for presenting more advanced math.
     
  20. Nov 26, 2016 #19
    so now i do the same steps which subtracting the -x value from the +x value and adding the 2 -y values together ?

    if that was correct i get : -4.66x10^-5 T in the i direction and 4.66x10^-5 in the j direction

    sorry for the late reply i had a cal final exam so i didn't have the time for physics
     
    Last edited: Nov 26, 2016
  21. Nov 26, 2016 #20
    yeah , sorry about that i'll do it next time
     
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