Đáp án:
\(\begin{array}{l}
2.\\
MN = \sqrt {A{M^2} + A{N^2}} = \sqrt {{8^2} + {6^2}} = 10cm\\
a.\\
{F_{AM}} = 1,{2.10^{ - 3}}\\
{F_{AN}} = {9.10^{ - 4}}\\
{F_{MN}} = 1,{5.10^{ - 3}}\\
b.\\
{F_{AM}} = 1,{2.10^{ - 3}}\\
{F_{AN}} = 0\\
{F_{MN}} = 1,{2.10^{ - 3}}c.\\
c.\\
{F_{AM}} = 0,{72.10^{ - 3}}\\
{F_{AN}} = 0,{72.10^{ - 3}}\\
{F_{MN}} = 0\\
4.\\
\alpha = 45
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2.\\
MN = \sqrt {A{M^2} + A{N^2}} = \sqrt {{8^2} + {6^2}} = 10cm\\
a.\\
{F_{AM}} = BI.AM.\sin 90 = {3.10^{ - 3}}.5.0,08 = 1,{2.10^{ - 3}}\\
{F_{AN}} = BI.AN\sin 90 = {3.10^{ - 3}}.5.0,06 = {9.10^{ - 4}}\\
{F_{MN}} = BI.MNsin90 = {3.10^{ - 3}}.5.0,1 = 1,{5.10^{ - 3}}\\
b.\\
{F_{AM}} = BI.AM.\sin 90 = {3.10^{ - 3}}.5.0,08 = 1,{2.10^{ - 3}}\\
{F_{AN}} = BI.AN\sin 0 = {3.10^{ - 3}}.5.0,06.0 = 0\\
{F_{MN}} = BI.MNsin\alpha = {3.10^{ - 3}}.5.0,1.\frac{{0,08}}{{0,1}} = 1,{2.10^{ - 3}}c.\\
c.\\
{F_{AM}} = BI.AM.\sin \beta = {3.10^{ - 3}}.5.0,08.\frac{{0,06}}{{0,1}} = 0,{72.10^{ - 3}}\\
{F_{AN}} = BI.AN\sin \alpha = {3.10^{ - 3}}.5.0,06\frac{{0,08}}{{0,1}} = 0,{72.10^{ - 3}}\\
{F_{MN}} = BI.MNsin0 = {3.10^{ - 3}}.5.0,1.0 = 0\\
4.\\
\tan \alpha = \frac{F}{p} = \frac{{BIl\sin 90}}{{mg}} = \frac{{0,5.2.0,05\sin 90}}{{0,005.10}} = 1\\
\alpha = 45
\end{array}\)