Consider an equilateral triangle wire loop with a side length S = 3 and a wire radius R = 0 . 0 1 . Calculate the self inductance of the wire loop by making the following assumptions:
1) One unit of current flows around the perimeter of a triangle with vertices P 1 , P 2 , and P 3 . There is no cross-sectional area associated with the current loop. This loop corresponds to the central axis of the wire.
P 1 = ( P 1 x , P 1 y ) = ( 1 , 0 ) P 2 = ( P 2 x , P 2 y ) = ( − 2 1 , 2 3 ) P 3 = ( P 3 x , P 3 y ) = ( − 2 1 , − 2 3 )
2) Calculate the resulting magnetic flux through the interior surface area bounded by a triangle with vertices P 4 , P 5 , and P 6 . The triangles from steps 1 and 2 are both at z = 0 .
P 4 = 0 . 9 8 P 1 P 5 = 0 . 9 8 P 2 P 6 = 0 . 9 8 P 3
Because the magnetic source (current loop) is separated from the target surface, this simplified problem is mathematically well-behaved. Let the approximate self inductance derived from the above calculations be L a .
I found a formula online for approximating the self inductance of an equilateral triangle wire loop. Let this value be L f .
L f = 2 π 3 μ 0 S [ ℓ n ( R S ) − 1 . 4 0 5 ]
In this particular problem, S = 3 , R = 0 . 0 1 , and μ 0 = 1 . In the equation, ℓ n denotes the natural logarithm.
What is ⌊ 1 0 0 0 L f L a ⌋ ? Note that ⌊ ⋅ ⌋ denotes the floor function.
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You can also use "Biot-Savard's Law" to calculate B ( r ) for the bottom side of the triangle analytically. The flux integral Φ 1 = ∫ A B 1 ( r ) d A can also be solved analytically. By symmetry, the flux from all three sides is the same, so Φ = 3 Φ 1 = L a I .
The term that approaches the singularity and the constant term from the analytic solution combined result in the approximation given. Convergence is tricky and the flux integral is a bit nasty - I made a mistake in one coefficient so I only reached the correct result in attempt no. 4 (:
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Another good one. Before diving into the problem, I am introducing the notation I used. The position vector of a point lying on the line joining P i and P j , and directed towards p j , where i = j is:
r i j = P i + t i j ( P j − P i ) t i j ∈ [ 0 , 1 ] ⟹ d r i j = d t i j ( P j − P i )
Now, the magnetic vector potential due to the unit current carrying element along the line joining points P 1 and P 2 at a point along the line joining the points P 4 and P 5 is:
d A 1 2 − 4 5 = 4 π μ o ( r 4 5 − r 1 2 d r 1 2 )
The line integral of this magnetic vector potential along the line joining the points P 4 and P 5 is:
L 1 2 − 4 5 = ∫ 4 → 5 ∫ 1 → 2 d A 1 2 − 4 5 ⋅ d r 4 5
Repeating the same procedure and computing the line integral due to the magnetic vector potential due to the current-carrying segment joining P 1 and P 2 and directed towards P 2 , along the line joining P 5 and P 6 is:
L 1 2 − 5 6 = ∫ 5 → 6 ∫ 1 → 2 d A 1 2 − 5 6 ⋅ d r 5 6
Repeating the same procedure and computing the line integral due to the magnetic vector potential due to the current-carrying segment joining P 1 and P 2 and directed towards P 2 , along the line joining P 6 and P 4 is:
L 1 2 − 6 4 = ∫ 6 → 4 ∫ 1 → 2 d A 1 2 − 6 4 ⋅ d r 6 4
Now, by Stokes' theorem, the total flux through the inner triangle due to the current-carrying segment joining P 1 and P 2 is:
Φ = L 1 2 − 4 5 + L 1 2 − 5 6 + L 1 2 − 6 4
By symmetry of the problem, the flux through the inner triangle due to the other two segments of the current-carrying triangle will be the same in magnitude and sign. Therefore, the total flux through the inner triangle due to the outer current-carrying triangle is:
L a = 3 Φ Q = L f L a A N S W E R = ⌊ 1 0 0 0 Q ⌋ = 9 8 9
The resulting integrands can be found by simplifying the expressions and the double integrals can be solved numerically using a script or outsourced to Wolfram-Alpha. I did the former and have left out those steps. This solution focuses on providing the method to find the solution.