Đáp án:

 $\begin{align}
  & a)B={{5.10}^{-5}}T \\ 
 & b)B=9,{{33.10}^{-5}}T \\ 
 & c)B=1,{{4.10}^{-4}}T \\ 
\end{align}$

Giải thích các bước giải:

 ${{I}_{1}}=10A;{{I}_{2}}=20A;R=8cm$

a) M cách đều 2 dây

$\begin{align}
  & {{B}_{1}}={{2.10}^{-7}}.\dfrac{{{I}_{1}}}{{{r}_{1}}}={{2.10}^{-7}}.\dfrac{10}{0,04}={{5.10}^{-5}}T \\ 
 & {{B}_{2}}={{2.10}^{-7}}.\dfrac{{{I}_{2}}}{{{r}_{2}}}={{2.10}^{-7}}.\dfrac{20}{0,04}={{10}^{-4}}T \\ 
\end{align}$

2 dây cùng chiều:

$B=\left| {{B}_{1}}-{{B}_{2}} \right|={{5.10}^{-5}}T$

b) ta có:

$\begin{array}{*{35}{l}}
   \text{ }\!\!~\!\!\text{ } & {{B}_{1}}={{2.10}^{-7}}.\dfrac{{{I}_{1}}}{{{r}_{1}}}={{2.10}^{-7}}.\dfrac{10}{0,02}={{10}^{-4}}T  \\
   \text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ } & {{B}_{2}}={{2.10}^{-7}}.\dfrac{{{I}_{2}}}{{{r}_{2}}}={{2.10}^{-7}}.\frac{20}{0,04}=6,{{67.10}^{-6}}T  \\
   \text{ }\!\!~\!\!\text{ } & {}  \\
\end{array}$

2 dây cùng chiều, M nằm trong 

$B=\left| {{B}_{1}}-{{B}_{2}} \right|=9,{{33.10}^{-5}}T$

c) $\begin{align}
  & {{B}_{1}}={{2.10}^{-7}}.\dfrac{{{I}_{1}}}{{{r}_{1}}}={{2.10}^{-7}}.\dfrac{10}{0,02}={{10}^{-4}}T \\ 
 & {{B}_{2}}={{2.10}^{-7}}.\dfrac{{{I}_{2}}}{{{r}_{2}}}={{2.10}^{-7}}.\dfrac{20}{0,1}={{4.10}^{-5}}T \\ 
\end{align}$

M nằm ngoài

$B={{B}_{1}}+{{B}_{2}}=1,{{4.10}^{-4}}T$