Đáp án:
\(\begin{array}{l}
a.\\
{B_M} = {4.10^{ - 5}}\\
b.\\
{B_N} = 1,{8667.10^{ - 5}}\\
c.\\
{B_M} = {2.10^{ - 5}}\\
d.\\
{B_M} = 1,{744.10^{ - 5}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{B_1} = {2.10^{ - 7}}.\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{4}{{0,05}} = 1,{6.10^{ - 5}}\\
{B_2} = {2.10^{ - 7}}.\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{6}{{0,05}} = 2,{4.10^{ - 5}}\\
{B_M} = {B_1} + {B_2} = 1,{6.10^{ - 5}} + 2,{4.10^{ - 5}} = {4.10^{ - 5}}\\
b.\\
{B_1} = {2.10^{ - 7}}.\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{4}{{0,15}} = 0,{533.10^{ - 5}}\\
{B_2} = {2.10^{ - 7}}.\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{6}{{0,05}} = 2,{4.10^{ - 5}}\\
{B_N} = |{B_1} - {B_2}| = |0,{533.10^{ - 5}} - 2,{4.10^{ - 5}}| = 1,{8667.10^{ - 5}}\\
c.\\
{B_1} = {2.10^{ - 7}}.\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{4}{{0,06}} = 1,{333.10^{ - 5}}\\
{B_2} = {2.10^{ - 7}}.\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{6}{{0,08}} = 1,{5.10^{ - 5}}\\
{B_M} = \sqrt {{B_1}^2 + {B_2}^2} = \sqrt {{{(1,{{333.10}^{ - 5}})}^2} + {{(1,{{5.10}^{ - 5}})}^2}} = {2.10^{ - 5}}\\
d.\\
{B_1} = {2.10^{ - 7}}.\frac{{{I_1}}}{{{r_1}}} = {2.10^{ - 7}}.\frac{4}{{0,1}} = 0,{8.10^{ - 5}}\\
{B_2} = {2.10^{ - 7}}.\frac{{{I_2}}}{{{r_2}}} = {2.10^{ - 7}}.\frac{6}{{0,1}} = 1,{2.10^{ - 5}}\\
{B_M} = \sqrt {{B_1}^2 + 2{B_1}{B_2}\cos 60 + {B_2}^2} = \sqrt {{{(0,{{8.10}^{ - 5}})}^2} + 2.0,{{8.10}^{ - 5}}.1,{{2.10}^{ - 5}}\cos 60 + {{(1,{{2.10}^{ - 5}})}^2}} = 1,{744.10^{ - 5}}
\end{array}\)
