Self Inductance of Triangular Wire Loop

Consider an equilateral triangle wire loop with a side length S = 3 S = \sqrt{3} and a wire radius R = 0.01 R = 0.01 . 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 \vec{P}_1 , P 2 \vec{P}_2 , and P 3 \vec{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 ) = ( 1 2 , 3 2 ) P 3 = ( P 3 x , P 3 y ) = ( 1 2 , 3 2 ) \vec{P}_1 = (P_{1x}, P_{1y}) = (1,0) \\ \vec{P}_2 = (P_{2x}, P_{2y}) = \Big(-\frac{1}{2},\frac{\sqrt{3}}{2} \Big) \\ \vec{P}_3 = (P_{3x}, P_{3y}) = \Big(-\frac{1}{2}, -\frac{\sqrt{3}}{2} \Big)

2) Calculate the resulting magnetic flux through the interior surface area bounded by a triangle with vertices P 4 \vec{P}_4 , P 5 \vec{P}_5 , and P 6 \vec{P}_6 . The triangles from steps 1 1 and 2 2 are both at z = 0 z = 0 .

P 4 = 0.98 P 1 P 5 = 0.98 P 2 P 6 = 0.98 P 3 \vec{P}_4 = 0.98 \, \vec{P}_1 \\ \vec{P}_5 = 0.98 \, \vec{P}_2 \\ \vec{P}_6 = 0.98 \, \vec{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 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 .

L f = 3 μ 0 S 2 π [ n ( S R ) 1.405 ] L_f = \frac{3 \mu_0 S}{2 \pi} \, \Big[ \ell n\Big(\frac{S }{R} \Big) - 1.405 \Big]

In this particular problem, S = 3 S = \sqrt{3} , R = 0.01 R = 0.01 , and μ 0 = 1 \mu_0 = 1 . In the equation, n \ell n denotes the natural logarithm.

What is 1000 L a L f \Big \lfloor 1000 \frac{L_a}{L_f} \Big \rfloor ? Note that \lfloor \cdot \rfloor denotes the floor function.


The answer is 989.

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1 solution

Karan Chatrath
Jan 23, 2021

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 P_i and P j P_j , and directed towards p j p_j , where i j i \ne j is:

r i j = P i + t i j ( P j P i ) \vec{r}_{ij} = \vec{P}_i + t_{ij} (\vec{P}_j - \vec{P}_i) t i j [ 0 , 1 ] t_{ij} \in [0,1] d r i j = d t i j ( P j P i ) \implies d\vec{r}_{ij} = dt_{ij} (\vec{P}_j - \vec{P}_i)

Now, the magnetic vector potential due to the unit current carrying element along the line joining points P 1 P_1 and P 2 P_2 at a point along the line joining the points P 4 P_4 and P 5 P_5 is:

d A 12 45 = μ o 4 π ( d r 12 r 45 r 12 ) d\vec{A}_{12-45} = \frac{\mu_o}{4\pi}\left( \frac{d\vec{r}_{12} }{\vec{r}_{45}-\vec{r}_{12}}\right)

The line integral of this magnetic vector potential along the line joining the points P 4 P_4 and P 5 P_5 is:

L 12 45 = 4 5 1 2 d A 12 45 d r 45 L_{12-45} = \int_{4 \to 5}\int_{1 \to 2}d\vec{A}_{12-45} \cdot d\vec{r}_{45}

Repeating the same procedure and computing the line integral due to the magnetic vector potential due to the current-carrying segment joining P 1 P_1 and P 2 P_2 and directed towards P 2 P_2 , along the line joining P 5 P_5 and P 6 P_6 is:

L 12 56 = 5 6 1 2 d A 12 56 d r 56 L_{12-56} = \int_{5 \to 6}\int_{1 \to 2}d\vec{A}_{12-56} \cdot d\vec{r}_{56}

Repeating the same procedure and computing the line integral due to the magnetic vector potential due to the current-carrying segment joining P 1 P_1 and P 2 P_2 and directed towards P 2 P_2 , along the line joining P 6 P_6 and P 4 P_4 is:

L 12 64 = 6 4 1 2 d A 12 64 d r 64 L_{12-64} = \int_{6 \to 4}\int_{1 \to 2}d\vec{A}_{12-64} \cdot d\vec{r}_{64}

Now, by Stokes' theorem, the total flux through the inner triangle due to the current-carrying segment joining P 1 P_1 and P 2 P_2 is:

Φ = L 12 45 + L 12 56 + L 12 64 \Phi = L_{12-45}+L_{12-56}+L_{12-64}

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 L_a = 3\Phi Q = L a L f Q = \frac{L_a}{L_f} A N S W E R = 1000 Q = 989 \mathrm{ANSWER}= \lfloor1000Q\rfloor = 989

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.

You can also use "Biot-Savard's Law" to calculate B ( r ) \vec{B}(\vec{r}) for the bottom side of the triangle analytically. The flux integral Φ 1 = A B 1 ( r ) d A \Phi_1=\int_A\vec{B}_1(\vec{r})\:d\vec{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 \Phi=3\Phi_1=L_aI .

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 (:

Carsten Meyer - 1 month, 1 week ago

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