1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic Field at the point P

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the magnetic field at the point distance R outside a rectangular loop of sides 2R, 4R carrying current I as shown in the attached diagram
    4bZvS1zgrPJ7p_U1p0n4xRCJrHrjaTgm3GovX7cX1DxrcWbRVlBZjUodEhgqNt8YTq2Q_VxHCZQeZS65Oqk6yLr1JhfWnUbA.png

    2. Relevant equations
    Biot-Savart Law
    iOj7251a0abeL8eOAH_Ykg5QrnyZabcOLgI-r77LlH2q7d3IRXdsar6Gp0IhSnKJvY8tfLzW-TfCvGIgFoC3FF1nz8UUthIA.png

    3. The attempt at a solution
    Could I just break this up into 4 separate lines and then calculate using the Biot-Savart Law for a finite wire? The magnetic field at the point, P, would then be the sum of the magnetic fields due to each line, no?
     
  2. jcsd
  3. Aug 7, 2017 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes.
     
  4. Aug 7, 2017 #3
    For the top line of the rectangle:

    dB = (μ0I / 4π) ⋅ dl x r / r2

    ∫ dB = (μ0 I / 4π) ⋅ ∫ dl sinθ / r2

    B = (μ0 I / 4π) ∫ (dl / R2 + x2) ⋅ (R / √R2+x2)

    B = (μ0 I R / 4π) ∫ 1 / (R2+x2)3/2 dl

    bounds of integration for the top line of the rectangle are from R=(0,4)

    B = μ0I/4πR (4 / √R2 + 42)

    ---

    For the bottom line of the rectangle everything is the same except R is now 3R because it is further away.

    B = μ0I/4π3R (4 / √3R2 + 42)


    I've essentially followed this YouTube video to calculate B for the top and bottom line:
     
  5. Aug 7, 2017 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Pretty close. A couple of things: (1) dl should be written as dx in the integral since you are letting x represent position along the wire. (2) The limits of integration are x = 0 to x = 4R.

    This isn't quite right because of the mistake in the limits mentioned above.

    Note: You listed a relevant equation that represents the result of doing the integration. So, if you understand that equation, you don't need to do the integration. (Or, you can use the equation to check your result for the integration.)

    In the denominator of you answer, it looks like only the R2 is under the square root. Use parentheses if you want to indicate that all of R2 + 4 is inside the square root.

    Again, close but not quite right because of a mistake in the limits of integration.

    What about the vertical sides? Does either one of those contribute to the answer?
     
  6. Aug 7, 2017 #5
    Ok, so I used the equation that I mentioned in my OP.

    dBz = μ0I / 4πz ⋅ (sinθ2-sinθ1)

    For the top and bottom lines of the rectangle sinθ1 = 0 so dBz reduces to:

    dBz = μ0I / 4πz ⋅ (sinθ2)

    For the top line of the rectangle:

    z = √(R2 + (4R)2)
    sinθ2 = R / z

    dBz = μ0 I / 4πz ⋅ (R/z)

    dBz = μ0 I R / 4πz2
     
  7. Aug 7, 2017 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Last edited: Aug 7, 2017
  8. Aug 7, 2017 #7
    I was using z as the hypotenuse, which I think is where the confusion has come from. I've renamed the hypotenuse to h for simplicity

    dBz = μ0 I / 4πz ⋅ (sinθ2-sinθ1)

    z = R (for the top line of the rectangle)

    For the top line of the triangle dBz reduces to:

    dBz = μ0 I / 4πR ⋅ (sinθ2)

    sinθ2 = R/h

    Therefore:

    dBz = (μ0 I / 4πR) ⋅ (R/h)

    dBz = (μ0 I / 4πh)

    Expressing the answer in terms of R:

    dBz = (μ0 I / 4π√(R2 +(4R)2)
     
  9. Aug 7, 2017 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note that θ2 is the angle shown:
    upload_2017-8-7_23-40-28.png

    So, sinθ2 ≠ R/h. Otherwise, I think your work is OK.
     
  10. Aug 8, 2017 #9
    ohhh in my OP the YouTube I linked it uses the θ from the other corner.

    sinθ2 = 4R / h

    dBz = μ0 I / 4πR ⋅ (4R / h)

    Therefore:

    dBz = (μ0 I 4R / 4πRh)

    dBz = (μ0 I / πh)

    Expressing the answer in terms of R:

    dBz = (μ0 I / πh)

    Bz = μ0 I / π √(R2+(4R)2)

    That's the magnetic field at the point P due to the top line of the rectangle only
     
  11. Aug 8, 2017 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes

    OK. I would simplify the square root expression.
     
  12. Aug 8, 2017 #11
    simplified:
    Bz = μ0 I / π [R2+(4R)2]3/2
     
  13. Aug 8, 2017 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I don't see that. I just meant that you can simplify √(R2 + (4R)2) to R√17.
     
  14. Aug 8, 2017 #13
    For the line 2R on the right hand side of the rectangle:

    sinθ2 = 4R / h2

    sinθ1 = 4R / h1

    dBz = μ0 I / 4π3R ⋅ [4R / h2 - 4R / h1 ]

    h2 = 4R / √((4R)2+(3R)2) = 4R / √ 25R2

    h1 = 4R / √((4R)2+(R)2) = 4R / √ 17R2

    ∴ dBz = μ0 I / 4π3R ⋅ [ 4R / √ 25R2 - 4R / √ 17R2 ]
     
  15. Aug 8, 2017 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    For the right hand side of length 2R, the meaning of Z, θ1, and θ2 are as shown
    upload_2017-8-8_0-54-53.png
     
  16. Aug 8, 2017 #15
    sinθ2 = 3R / √((4R)2 + (3R)2)

    sinθ1 = R / √((4R)2 + (R)2)

    dBz = μ0 I / 4π4R ⋅ [sinθ2 - sinθ1]

    dBz = μ0 I / 16πR ⋅ [3R / √((4R)2 + (3R)2) - R / √((4R)2 + (R)2)]
     
  17. Aug 8, 2017 #16
    I'm not sure what to do for the left hand side of the rectangle. If both sinθ cancel, do i integrate 'R' from 3R to 0?
     
  18. Aug 8, 2017 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think that's correct.
     
  19. Aug 8, 2017 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What are the values of θ1 and θ2 for this side? What does the formula for B give?

    Or, apply the Biot-Savart law to an infinitesimal length dl of this side of the rectangle.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Magnetic Field at the point P
Loading...