H Field in a Rectangular Loop Carrying Constant Current
\$\mathbf{\vec H \text{ at } (2,2,0)}\$:

\$\begin{align}\\
\vec H&amp;=\vec H_{AB}+\vec H_{BC}+\vec H_{CD}+\vec H_{DA}\\
&amp;{\begin{aligned}\\
=2\vec H_{AB}+\vec H_{BC}+\vec H_{DA} [\because \vec H_{AB}&amp;=\vec H_{CD}]\\
\end{aligned}\\}\\
\end{align}\\\$
Now, \$\tan\theta_1=\frac 22=1\implies\theta_1=45°\$
\$\therefore \alpha_1=180°-45°=135°\$ and
\$\cos\alpha_2=\frac{6}{\sqrt{2^2+6^2}}=0.949\$
According to Biot-savart law,
\$\vec H=\frac{I}{4\pi\rho}(\cos \alpha_2-\cos\alpha_1)\hat\phi\$
where \$\rho\$ denotes perpendicular distance from the current element.
\$\begin{align}\\
\therefore\color{blue}{\vec H_{AB}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times 2}(0.949-\cos135°)(\hat x&amp;\times\hat y)\\
\end{aligned}\\}\\
&amp;=0.659\hat z\\
\end{align}\\\$
\$\alpha_2=\tan^{-1}{\frac 26}=18.43°\\
\therefore\alpha_3=90°+18.43°=108.43°\\
\alpha_4=\tan^{-1}{\frac 62}=71.57°\$
So, \$\begin{align}\\
\color{blue}{\vec H_{BC}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times6}(\cos 71.56°-\cos 108.4&amp;°)\hat z\\
\end{aligned}\\}\\
&amp;=0.0838\hat z\\
\end{align}\\\$
\$\theta_3=\tan^{-1}{\frac 22}=45°\\
\alpha_5=180°-45°=135°\\
\alpha_6=90°-45°=45°\$
\$\begin{align}\\
\therefore\color{blue}{\vec H_{DA}}&amp;=\frac{10}{4\pi\times2}(\cos45°-\cos135°)\hat z\\
&amp;=0.5626\hat z\\
\end{align}\$
\$\begin{align}\\
\therefore\color{purple}{\vec H}&amp;=(2\times 0.659+0.0838+0.5626)\hat z\\
&amp;=1.964\hat z\\
\end{align}\\\$
\$\mathbf{\vec H \text{ at } (4,2,0)}\$:

\$\begin{align}\\
\vec H&amp;=\vec H_{AB}+\vec H_{BC}+\vec H_{CD}+\vec H_{DA}\\
&amp;{\begin{aligned}\\
=2\vec H_{AB}+2\vec H_{BC}[\because\vec H_{AB}&amp;=\vec H_{CD}\text{ and }\vec H_{BC}=\vec H_{DA}]\\
\end{aligned}\\}\\
\end{align}\\\$
Now, \$\theta_1=\tan^{-1}{\frac 24}=26.57°\$
\$\begin{aligned}\\
\therefore\alpha_1=180°-26.57°=153.43&amp;°\text{ and }\\
\end{aligned}\\\$
\$\alpha_2=\tan^{-1}{\frac 24}=26.57°\$
\$\begin{align}\\
\color{blue}{\vec H_{AB}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times2}(\cos 26.57°-\cos 153.43&amp;°)\hat z\\
\end{aligned}\\}\\
&amp;=0.712\hat z\\
\end{align}\\\$
\$\alpha_3=90°+26.57°=116.57°,\\
\alpha_4=\tan^{-1}{\frac 42}=63.43°\$
\$\begin{align}\\
\color{blue}{\vec H_{BC}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times4}(\cos 63.43°-\cos 116.57&amp;°)\hat z\\
\end{aligned}\\}\\
&amp;=0.178\hat z\\
\end{align}\\\$
Now, \$\begin{align}\\
\color{purple}{\vec H}&amp;=(2\times 0.712+2\times 0.178)\hat z\\
&amp;=1.78\hat z\\
\end{align}\\\$
\$\mathbf{\vec H \text{ at } (4,8,0)}\$:

\$\begin{align}\\
\vec H&amp;=\vec H_{AB}+\vec H_{BC}+\vec H_{CD}+\vec H_{DA}\\
&amp;{\begin{aligned}\\
=\vec H_{AB}+2\vec H_{BC}+\vec H_{CD}[\because\vec H_{BC}&amp;=\vec H_{DA}]\\
\end{aligned}\\}\\
\end{align}\\\$
\$\theta_1=\tan^{-1}{\frac 84}=63.43°\\
\alpha_1=180°-63.43°=116.57°\\
\alpha_2=63.43°\$
\$\begin{align}\\
\color{blue}{\vec H_{AB}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times8}(\cos 63.43°-\cos 116.57&amp;°)(\hat x\times\hat y)\\
\end{aligned}\\}\\
&amp;=0.089\hat z\\
\end{align}\\\$
\$\alpha_3=90°+63.43°=153.43°\\
\theta_2=\tan^{-1}{\frac 44}=45°\\
\alpha_4=90°+45°=135°\$
\$\begin{align}\\
\color{blue}{\vec H_{BC}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times 4}(\cos 135°-\cos 153.43&amp;°)(\hat y\times -\hat x)\\
\end{aligned}\\}\\
&amp;=0.037\hat z\\
\end{align}\\\$
\$\alpha_5=180°-45°=135°\\
\alpha_6=\tan^{-1}{\frac 44}=45°\$
\$\begin{align}\\
\color{blue}{\vec H_{CD}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\times4}(\cos45°-\cos135°)(-\hat x&amp;\times\hat y)\\
\end{aligned}\\}\\
&amp;=-0.281\hat z\\
\end{align}\\\$
\$\begin{align}\\
\color{purple}{\vec H}&amp;=(0.089+2\times 0.037-0.281)\hat z\\
&amp;=-0.118\hat z\\
\end{align}\\\$
\$\mathbf{\vec H \text{ at } (0,0,2)}\$:

\$\vec H=\vec H_{AB}+\vec H_{BC}+\vec H_{CD}+\vec H_{DA}\$

\$\alpha_1=90°, \cos\alpha_2=\frac{8}{\sqrt{68}}\$
\$\begin{align}\\
\color{blue}{\vec H_{AB}}&amp;=\frac{10}{4\pi\times2}(\frac{8}{\sqrt{68}}-\cos90°)(\hat x\times\hat z)\\
&amp;=-0.386\hat y\\
\end{align}\\\$

\$\alpha_3=90°, \cos\alpha_4=\frac{4}{\sqrt{84}}\$
We know, \$\hat\phi=\hat l\times\hat\rho\$
Now, \$\hat\rho=\frac{(0-8)\hat x+(2-0)\hat z}{\sqrt{68}}=\frac{-8\hat x+2\hat z}{\sqrt{68}}\\
\hat\phi=\hat y\times\frac{-8\hat x+2\hat z}{\sqrt{68}}=\frac{8\hat z+2\hat x}{\sqrt{68}}\$
\$\begin{align}\\
\color{blue}{\vec H_{BC}}&amp;=\frac{I}{4\pi\rho}(\cos\alpha_4-\cos\alpha_3)\hat\phi\\
&amp;=\frac{10}{4\pi\times\sqrt{68}}\left(\frac{4}{\sqrt{84}}\right)\frac{8\hat z+2\hat x}{\sqrt{68}}\\
&amp;=\frac{10}{68\pi\sqrt{84}}(8\hat z+2\hat x)\\
&amp;=0.04\hat z+0.012\hat x\\
\end{align}\\\$

\$\theta_1=\tan^{-1}{\frac{\sqrt{20}}{8}}=29.2°\\
\therefore \alpha_6=180°-29.2°=150.8°\\
\alpha_5=90°\\
\hat\phi=-\hat x\times\frac{-4\hat y+2\hat z}{\sqrt{20}}=\frac{4\hat z+2\hat y}{\sqrt{20}}\$
\$\begin{align}\\
\color{blue}{\vec H_{CD}}&amp;{\begin{aligned}\\
=\frac{10}{4\pi\sqrt{20}}(0-\cos&amp;150.8)\left(\frac{4\hat z+2\hat y}{\sqrt{20}}\right)\\
\end{aligned}\\}\\
&amp;=0.1389\hat z+0.0694 \hat y\\
\end{align}\\\$

\$\theta_2=\tan^{-1}{\frac 24}=26.56°\\
\alpha_7=180°-26.56°=153.44°\\
\alpha_8=90°\\
\hat\phi=-\hat y\times\hat z=-\hat x\$
\$\begin{align}\\
\color{blue}{\vec H_{DA}}&amp;=\frac{10}{4\pi\times2}(0-\cos 153.44°)(-\hat x)\\
&amp;=-0.3557\hat x\\
\end{align}\\\$
\$\begin{align}\\
\color{purple}{\vec H}&amp;{\begin{aligned}\\
=(0.012-0.3557)\hat x&amp;+(0.386+0.0694)\hat y&amp;+(0.04+0.1389)\hat z\\
\end{aligned}\\}\\
&amp;=-0.3437\hat x-0.3166\hat y+0.1789\hat z\\
\end{align}\\\$