B-Field from Equilateral

Great problem as always. I start with a general situation. Consider a current carrying wire segment with start and end points $\vec{R}_A=(x_A,y_A,z_A)$ and $\vec{R}_B=(x_B,y_B,z_B)$ respectively. Current flows from point A to B and has a magnitude of $I_{AB}$ .

Any general point lying on the wire can be parameterised as follows:

$\vec{R}_P= R_A + t(\vec{R}_B - \vec{R}_A)$ $0 \le t \le 1$

We define:

$\vec{L} = \vec{R}_P -\vec{R}_A$ $\implies d\vec{L} = dt \ (\vec{R}_B - \vec{R}_A)$

The vector $d\vec{L}$ is a current carrying element on the wire. The magnetic field at a general point $\vec{R}_T$ die to the current carrying element is:

$d\vec{B}_{AB} = \frac{\mu_o}{4 \pi} \left(\frac{I_{AB} \ d\vec{L} \times (\vec{R}_T - \vec{R}_P)}{\lvert \vec{R}_T - \vec{R}_P \rvert^3}\right)$

Applying this general recipe to compute the magnetic field at $(1,1,1)$ leads to:

$d\vec{B}_{31} = \frac{\mu_o}{4 \pi} \left(\frac{I_{31} \ d\vec{L} \times (\vec{R}_T - \vec{R}_P)}{\lvert \vec{R}_T - \vec{R}_P \rvert^3}\right)$ $\vec{R}_P= \vec{P}_3 + t(\vec{P}_1 - \vec{P}_3)$ $d\vec{L} = dt \ (\vec{P}_1 - \vec{P}_3)$ $d\vec{B}_{31} = \left(f_{x,31}(t) \ \hat{i} + f_{y,31}(t) \ \hat{j} + f_{z,31}(t) \ \hat{k} \right) \ dt$ $\vec{B}_{31} = \left(\int_{0}^{1} f_{x,31}(t) \ dt \right) \ \hat{i} + \left(\int_{0}^{1} f_{y,31}(t) \ dt \right) \ \hat{j} + \left(\int_{0}^{1} f_{z,31}(t) \ dt \right) \ \hat{k}$

$d\vec{B}_{12} = \frac{\mu_o}{4 \pi} \left(\frac{I_{12} \ d\vec{L} \times (\vec{R}_T - \vec{R}_P)}{\lvert \vec{R}_T - \vec{R}_P \rvert^3}\right)$ $\vec{R}_P= \vec{P}_1 + t(\vec{P}_2 - \vec{P}_1)$ $d\vec{L} = dt \ (\vec{P}_2 - \vec{P}_1)$ $d\vec{B}_{12} = \left(f_{x,12}(t) \ \hat{i} + f_{y,12}(t) \ \hat{j} + f_{z,12}(t) \ \hat{k} \right) \ dt$ $\vec{B}_{12} = \left(\int_{0}^{1} f_{x,12}(t) \ dt \right) \ \hat{i} + \left(\int_{0}^{1} f_{y,12}(t) \right) \ dt \ \hat{j} + \left(\int_{0}^{1} f_{z,12}(t) \ dt \right) \ \hat{k}$

$d\vec{B}_{32} = \frac{\mu_o}{4 \pi} \left(\frac{I_{32} \ d\vec{L} \times (\vec{R}_T - \vec{R}_P)}{\lvert \vec{R}_T - \vec{R}_P \rvert^3}\right)$ $\vec{R}_P= \vec{P}_3 + t(\vec{P}_2 - \vec{P}_3)$ $d\vec{L} = dt \ (\vec{P}_2 - \vec{P}_3)$ $d\vec{B}_{32} = \left(f_{x,32}(t) \ \hat{i} + f_{y,32}(t) \ \hat{j} + f_{z,32}(t) \ \hat{k} \right) \ dt$ $\vec{B}_{32} = \left(\int_{0}^{1} f_{x,32}(t) \ dt \right) \ \hat{i} + \left(\int_{0}^{1} f_{y,32}(t) \ dt \right) \ \hat{j} + \left(\int_{0}^{1} f_{z,32}(t) \ dt \right) \ \hat{k}$

Finally, the net magnetic field magnitude is:

$\vec{B}_T = \lvert \vec{B}_{31} + \vec{B}_{12} + \vec{B}_{32} \rvert \approx 0.488$ $dt = 10^{-6}$

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clear all
clc

% Branch currents (From Ohm's law):
I31    = 10/2;
I12    = 10/2;
I32    = 10;

% Points on equilateral triangle:
P1     = [1;0;0];
P2     = [-1/2;-sqrt(3)/2;0];
P3     = [-1/2;sqrt(3)/2;0];

% Point at which B field is to be computed:
RT     = [1;1;1];

dt     = 1e-6;

% Calcluating field due to segment 3-1:
B31    = 0;

for t = 0:dt:1

    RP         = P3 + t*(P1 - P3);
    dL         = dt*(P1 - P3);

    % Biot Savart:
    dB31       = (1/(4*pi))*cross(I31*dL,RT-RP)/norm(RT - RP)^3;

    % Numerical integration:
    B31        = B31 + dB31;
end

% Calcluating field due to segment 1-2:
B12    = 0;

for t = 0:dt:1

    RP         = P1 + t*(P2 - P1);
    dL         = dt*(P2 - P1);

    % Biot Savart:
    dB12       = (1/(4*pi))*cross(I12*dL,RT-RP)/norm(RT - RP)^3;

    % Numerical integration:
    B12        = B12 + dB12;
end

% Calcluating field due to segment 3-2:
B32    = 0;

for t = 0:dt:1

    RP         = R3 + t*(P2 - P3);
    dL         = dt*(P2 - P3);

    % Biot Savart:
    dB32       = (1/(4*pi))*cross(I32*dL,RT-RP)/norm(RT - RP)^3;

    % Numerical integration:
    B32        = B32 + dB32;
end

% Resultant magnitude:
B     = norm(B31 + B12 + B32)
% B   = 0.488