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

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Homework Equations


Biot-Savart law

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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

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

r2 = R2 + s2

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Solving the integral

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after a few steps

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

Thanks in advance
 

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kuruman said:
What about the magnetic field from segments 5-6 and 6-1?

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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)
 
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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

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so
|B| ≅ |-2 · 10-5| T ≅ 2 · 10-5 T
The minus sign suggests an entering direction for the magnetic field
 

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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.
SquareLoop.png
 

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