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