Đáp án:
\(\begin{array}{l}
11.\\
{r_1} = 5cm\\
12.\\
a.\\
B = 7,{2.10^{ - 6}}\\
b.\\
B = {10.10^{ - 6}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
11.\\
\frac{B}{{{B_1}}} = \frac{{{r_1}}}{r}\\
\frac{B}{{2B}} = \frac{{{r_1}}}{{10}}\\
\Rightarrow {r_1} = 5cm\\
12.\\
a.\\
{B_1} = {2.10^{ - 7}}\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{{2,4}}{{0,2}} = 2,{4.10^{ - 6}}\\
{B_2} = {2.10^{ - 7}}\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{{2,4}}{{0,1}} = 4,{8.10^{ - 6}}\\
B = {B_1} + {B_2} = 2,{4.10^{ - 6}} + 4,{8.10^{ - 6}} = 7,{2.10^{ - 6}}\\
b.\\
{B_1} = {2.10^{ - 7}}\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{{2,4}}{{0,08}} = {6.10^{ - 6}}\\
{B_2} = {2.10^{ - 7}}\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{{2,4}}{{0,06}} = {8.10^{ - 6}}\\
B = \sqrt {B_1^2 + B_2^2} = \sqrt {{{({{6.10}^{ - 6}})}^2} + {{({{8.10}^{ - 6}})}^2}} = {10.10^{ - 6}}
\end{array}\)