Field Cancellation

The formula for the magnetic field due to an infinite wire is:

$\vec{B} = \frac{\mu_o I}{2\pi \lvert \vec{a} \rvert} \left(d\vec{l} \times \hat{a}\right)$

Where $\vec{a}$ is the vector joining the wire and the test point and directed towards the test point and $d\vec{l}$ is in the direction of current flow in the wire.

Let the coordinates of the test point be $(x,0)$ .

Applying this to the scenario '0' gives a net field in the negative Y direction and the magnitude of the field is:

$B_{To} = \frac{\mu_o I_{Ao}}{2\pi \left(x-1\right)} + \frac{\mu_o I_{Bo}\left(x + 0.5\right)}{2\pi \left(\left(x+0.5\right)^2 + \left(-\sqrt{3}/2\right)^2\right)} + \frac{\mu_o I_{Co}\left(x + 0.5\right)}{2\pi \left(\left(x+0.5\right)^2 + \left(\sqrt{3}/2\right)^2\right)}$

Doing the same for scenario '1' gives the net field again along the Y direction. The field due to wire A is along negative Y while that due to B and C is along the positive Y direction:

$B_{T1} = \frac{\mu_o I_{A1}}{2\pi \left(x-1\right)} + \frac{\mu_o I_{B1}\left(x + 0.5\right)}{2\pi \left(\left(x+0.5\right)^2 + \left(-\sqrt{3}/2\right)^2\right)} + \frac{\mu_o I_{C1}\left(x + 0.5\right)}{2\pi \left(\left(x+0.5\right)^2 + \left(\sqrt{3}/2\right)^2\right)}$

The fields along X in both scenarios cancel out. Having obtained the fields, the next step is to evaluate the ratio:

$\frac{B_{T1} }{B_{To}} =R= \frac{x+1}{2\,x^2}$

The simplification carried out to obtain the above expression is left out. From this point, the integral is trivial to evaluate:

$\boxed{I = \int_{1.1}^{30} R \ dx = \dfrac{\ln\left(30\right)-\ln\left(\frac{11}{10}\right)+\frac{289}{330}}{2} \approx 2.0908}$

It can be seen that as $x$ increases, the ratio reduces, which indicates that the field in scenario 1 reduces in magnitude relative to that in scenario 0. The field in scenario 1 tends towards cancellation as $x$ increases.

Total induction of magnetic field at $(x_T,0)$ due to the three currents is $B_{tot}=\dfrac{\mu_0}{2π}\times (\dfrac{I_A}{x_T-1}+\dfrac{I_B\times (2x_T+1)}{2x_T^2+2x_T+2}+\dfrac{I_C\times (2x_T+1)}{2x_T^2+2x_T+2})$ . In the first case, $I_{A0}=I_{B0}=I_{C0}=1$ and so $B_0(x_T)=\dfrac{\mu_0}{2π}\times \dfrac{3x_T^2}{x_T^3-1}$ , while in the second case, $I_{A1}=1, I_{B1}=I_{C1}=-\dfrac{1}{2}$ . So, $B_1(x_T)=\dfrac{\mu_0}{2π}\times \dfrac{3x_T+3}{2x_T^3-2}$ . Therefore $\dfrac{B_1(x_T)}{B_0(x_T)}=\dfrac{1}{2x_T}+\dfrac{1}{2x_T^2}$ , and the value of the given integral is $\dfrac{\ln {\dfrac{30}{1.1}}}{2}+\dfrac{1}{2}(\dfrac{1}{1.1}-\dfrac{1}{30})=\boxed {2.0908...}$ .