Question 7.1: The conducting triangular loop in Figure 7.6(a) carries a cu...

This example illustrates how eq. (7.12)

H=\frac{I}{4\pi\rho}(\cos\alpha_{2}-\cos\alpha_{1})a_{\phi}

is applied to any straight, thin, current carrying conductor.

The key point to keep in mind in applying eq. (7.12) is figuring out \alpha_{1},  \alpha_{2},  \rho, and a_{\phi }. To find H at (0,0,5) due to side 1 of the loop in Figure 7.6(a), consider Figure 7.6(b), where side 1 is treated as a straight conductor.

Notice that we join the point of interest (0,0,5) to the beginning and end of the line current. Observe that \alpha_{1},  \alpha_{2}, and \rho are assigned in the same manner as in Figure 7.5 on which eq. (7.12) is based:

\cos\alpha_{1}=\cos90^{\circ}=0,       \cos\alpha_{2}=\frac{2}{\sqrt{29}},      \rho=5

To determine a_{\phi} is often the hardest part of applying eq. (7.12). According to eq. (7.15):  a_{\phi }=a_{\ell}\times a_{\rho},

a_{\ell}=a_{x}  and  a_{\rho}=a_{z}, so

a_{\phi }=a_{x}\times a_{z}=-a_{y}

Hence

H_{1}=\frac{I}{4\pi\rho}(\cos\alpha_{2}-\cos\alpha_{1})a_{\phi}=\frac{10}{4\pi(5)}\left(\frac{2}{\sqrt{29}}-0\right)(-a_{y})=-59.1a_{y}mA/m