Magnetic field in a closed path

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Faefnir
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Homework Statement


Find (a) the module and (b) the direction, entering or exiting the page plane, of the magnetic field at point P, knowing that a = 4.7 cm and i = 13 A

Diagram of a closed rectangular wire path with corners labeled 1 through 6, with side length 2a vertically and a horizontally, and a current arrow i on the left vertical segment; point P lies to the upper right, level with corner 1.


Homework Equations


Biot-Savart law

Biot-Savart law equation: dB vector equals mu-zero over 4 pi, times i times ds divided by r squared, times sin theta.


The Attempt at a Solution



For 1-2 and 4-5 segments, B = 0 because sin θ = 0 for all segment points.

Taking the 2-3 segment

R = distance from P to the wire (2-P line)
r = distance from P to the point 3. Assume the point 3 as the infinitesimal segment ds (3-P line)
s = wire length (2-3 line)

From the first rectangle triangles law

R = r sin (π-θ) = r sin (θ) ⇒ sin (θ) = R/r

Biot-Savart law equation: dB vector equals mu0 over 4 pi, times i times ds over r squared, times R over r.


r is triangle's hypotenuse with s and R as catetis

r2 = R2 + s2

Biot-Savart law term for a differential magnetic field dB: mu0/(4 pi) times i times R over (R squared plus s squared) to the power 3/2, times ds.


Equation for the magnetic field element dB = (mu0 / 4 pi) * i * R * 1 / (R^2 + s^2)^(3/2) * ds, typeset in a plain rendered font.


Equation for the magnetic field B equals mu-zero over 4 pi times i times R times the integral from 0 to 2a of 1 over (R squared plus s squared) to the three-halves power, ds.


Solving the integral

Equation for the magnetic field B equals mu-zero over 4 pi times i times R times 1 over R squared times the bracketed expression s over the square root of (R squared plus s squared), evaluated from 0 to 2a.


after a few steps

Formula: B = (μ0 / 2π) · (i / R) · (1 / √8), with B as a vector.


Regardless of the numerical result, the final formula I got is right, or have I made any mistakes?

Thanks in advance
 
Attachments
  • Diagram of a closed rectangular wire path with corners labeled 1 through 6, with side length 2a vertically and a horizontally, and a current arrow i on the left vertical segment; point P lies to the upper right, level with corner 1.
    dMyM7u8.png
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  • Biot-Savart law equation: dB vector equals mu-zero over 4 pi, times i times ds divided by r squared, times sin theta.
    gif.gif
    846 bytes · Views: 402
  • Biot-Savart law equation: dB vector equals mu0 over 4 pi, times i times ds over r squared, times R over r.
    gif.gif
    801 bytes · Views: 395
  • Biot-Savart law term for a differential magnetic field dB: mu0/(4 pi) times i times R over (R squared plus s squared) to the power 3/2, times ds.
    gif.gif
    1.1 KB · Views: 372
  • Equation for the magnetic field element dB = (mu0 / 4 pi) * i * R * 1 / (R^2 + s^2)^(3/2) * ds, typeset in a plain rendered font.
    gif.gif
    1.1 KB · Views: 409
  • Equation for the magnetic field B equals mu-zero over 4 pi times i times R times the integral from 0 to 2a of 1 over (R squared plus s squared) to the three-halves power, ds.
    gif.gif
    1.3 KB · Views: 416
  • Formula: B = (μ0 / 2π) · (i / R) · (1 / √8), with B as a vector.
    gif.gif
    724 bytes · Views: 398
  • Equation for the magnetic field B equals mu-zero over 4 pi times i times R times 1 over R squared times the bracketed expression s over the square root of (R squared plus s squared), evaluated from 0 to 2a.
    gif.gif
    1.5 KB · Views: 409
Last edited:
kuruman said:
What about the magnetic field from segments 5-6 and 6-1?

Equation for magnetic field B = (μ0/4π) · i · (1/R) · [s / sqrt(R² + s²)] evaluated from 0 to a.


Equation B = (μ0 / 4π) · (i / R) · (1 / √2), with B a vector, written as a typeset formula.
 
Attachments
  • Equation B = (μ0 / 4π) · (i / R) · (1 / √2), with B a vector, written as a typeset formula.
    gif.gif
    711 bytes · Views: 365
  • Equation for magnetic field B = (μ0/4π) · i · (1/R) · [s / sqrt(R² + s²)] evaluated from 0 to a.
    gif.gif
    1.3 KB · Views: 366
Is this both of them or just one? What about the total? What about the direction?
 
Both of 5-6 and 6-1.

About total and direction:
2-3 is entering
6-1 is exiting
Not sure about horizontal segments (3-4 and 5-6): Can you help me? (yes, I know sounds like ready-made food)

I supposed to rotate the image by 90° clockwise:
3-4 is entering
5-6 is exiting

So, the total is -1.86 · 10-5
The book reports 20 μA for a entering current (from chapter 29 exercises, Halliday-Resnick-Walker, Electromagnetism and Optics, 7th edition)
 
Last edited:
Faefnir said:
-1.86 · 10-5
Can you show me how you got this number? Please use the final equation with symbols that you used.
Faefnir said:
The book reports 20 μA for a entering current (from chapter 29 exercises, Halliday-Resnick-Walker, Electromagnetism and Optics, 7th edition)
Do you actually mean μA? If so, what current is this?
 
Solved. Taking ds verse agrees as the one of the current

Equation: B_tot = -B23 - B34 + B56 - B61 ≅ -2 · 10^-5 T, a sum of magnetic field contributions from four path segments.

so
|B| ≅ |-2 · 10-5| T ≅ 2 · 10-5 T
The minus sign suggests an entering direction for the magnetic field
 
Attachments
  • Equation: B_tot = -B23 - B34 + B56 - B61 ≅ -2 · 10^-5 T, a sum of magnetic field contributions from four path segments.
    gif.gif
    1.1 KB · Views: 348
I agree with your answer. Now that you have solved your problem, I will show you an easier way than calculating the separate contributions from each segment. Suppose you found an expression for the B-field at the corner of a regular square. The field for the loop in the problem is the superposition of two squares one of side ##2a## and counterclockwise current and one with side ##a## and clockwise current (see figure below). So once you find the expression for the square, you reuse it for the smaller square with a negative sign because the current is reversed.
Diagram showing an L-shaped closed loop with directional arrows equal to a large square loop plus a smaller square loop with reversed arrow direction.
 
Attachments
  • Diagram showing an L-shaped closed loop with directional arrows equal to a large square loop plus a smaller square loop with reversed arrow direction.
    SquareLoop.png
    1.5 KB · Views: 416
Last edited:
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