Ta có: \(\left\{ \begin{array}{l}{B_1} = {2.10^{ - 7}}\dfrac{{{I_1}}}{{{r_1}}}\\{B_2} = {2.10^{ - 7}}\dfrac{{{I_2}}}{{{r_2}}}\end{array} \right.\)

a)

\(\overrightarrow {{B_A}}  = \overrightarrow {{B_1}}  + \overrightarrow {{B_2}} \)

\(\left\{ \begin{array}{l}{r_1} = {r_2} = 0,05m\\{B_1} = {4.10^{ - 5}}T\\{B_2} = {8.10^{ - 5}}T\end{array} \right.\) và \(\overrightarrow {{B_1}}  \uparrow  \uparrow \overrightarrow {{B_2}} \)

\( \Rightarrow {B_A} = {B_1} + {B_2} = {12.10^{ - 5}}T\)

b)

\(\overrightarrow {{B_B}}  = \overrightarrow {{B_1}}  + \overrightarrow {{B_2}} \)

\(\left\{ \begin{array}{l}{r_1} = 0,04m\\{r_2} = 0,14m\\{B_1} = {5.10^{ - 5}}T\\{B_2} = \dfrac{{20}}{7}{.10^{ - 5}}T\end{array} \right.\) và \(\overrightarrow {{B_1}}  \uparrow  \downarrow \overrightarrow {{B_2}} \)

\( \Rightarrow {B_B} = {B_1} - {B_2} = \dfrac{{15}}{7}{.10^{ - 5}}T\)

c)

\(\overrightarrow {{B_C}}  = \overrightarrow {{B_1}}  + \overrightarrow {{B_2}} \)

\(\left\{ \begin{array}{l}{r_1} = {r_2} = 0,1m\\{B_1} = {2.10^{ - 5}}T\\{B_2} = {4.10^{ - 5}}T\end{array} \right.\) và \(\widehat {\left( {\overrightarrow {{B_1}} ,\overrightarrow {{B_2}} } \right)} = {120^0}\)

\( \Rightarrow {B_C} = \sqrt {B_1^2 + B_2^2 + 2{B_1}{B_2}cos{{120}^0}}  = 2\sqrt 3 {.10^{ - 5}}T\)

d)

\(\overrightarrow {{B_D}}  = \overrightarrow {{B_1}}  + \overrightarrow {{B_2}} \)

\(\left\{ \begin{array}{l}{r_1} = 0,8m\\{r_2} = 0,6m\\{B_1} = 2,{5.10^{ - 5}}T\\{B_2} = \dfrac{{20}}{3}{.10^{ - 5}}T\end{array} \right.\) và \(\overrightarrow {{B_1}}  \bot \overrightarrow {{B_2}} \)

\( \Rightarrow {B_D} = \sqrt {B_1^2 + B_2^2}  = 7,{12.10^{ - 5}}T\)