جواب تمرین 3-1 صفحه 90 درس 3 فیزیک یازدهم ریاضی (مغناطیس)

\(\begin{array}{l}\left\{ \begin{array}{l}q = 1/6 \times {10^{ - 19}}\\\\\theta = {30^ \circ }\\\\B = 320G = 320 \times {10^{ - 4}}\,T\\\\F = 5/12 \times {10^{ - 14}}\,N\end{array} \right.\\\\F = \left| q \right|vB\sin \theta \Rightarrow v = \frac{F}{{\left| q \right|B\sin \theta }}\\\\ \Rightarrow v = \frac{{5/12 \times {{10}^{ - 14}}}}{{1/6 \times {{10}^{ - 19}} \times 320 \times {{10}^{ - 4}} \times \sin {{30}^ \circ }}} = \\\\ \Rightarrow v = 2 \times {10^7}\frac{m}{s} = 20,000\frac{{km}}{s}\end{array}\)

نیروی وارد بر بار (1):

\(\begin{array}{l}\left\{ \begin{array}{l}q = 6/15\,nC = 6/15 \times {10^{ - 9}}\,C\\\\\theta = {150^ \circ }\\\\B = 0/165\,T\\\\v = 46\,\frac{m}{s}\end{array} \right.\\\\F = \left| q \right|vB\sin \theta = \\\\6/15 \times {10^{ - 9}} \times 46 \times 0/165 \times \sin {150^ \circ }\\\\ \Rightarrow F = 2/33 \times {10^{ - 8}}\,N\end{array}\)

نیروی وارد بر بار (2):

\(\begin{array}{l}\left\{ \begin{array}{l}q = 6/15\,nC = 6/15 \times {10^{ - 9}}\,C\\\\\theta = {90^ \circ }\\\\B = 0/165\,T\\\\v = 46\,\frac{m}{s}\end{array} \right.\\\\F = \left| q \right|vB\sin \theta = \\\\6/15 \times {10^{ - 9}} \times 46 \times 0/165 \times \sin {90^ \circ }\\\\ \Rightarrow F = 4/67 \times {10^{ - 8}}\,N\end{array}\)

نیروی وارد بر بار (3):

\(\begin{array}{l}\left\{ \begin{array}{l}q = 6/15\,nC = 6/15 \times {10^{ - 9}}\,C\\\\\theta = {30^ \circ }\\\\B = 0/165\,T\\\\v = 46\,\frac{m}{s}\end{array} \right.\\\\F = \left| q \right|vB\sin \theta = \\\\6/15 \times {10^{ - 9}} \times 46 \times 0/165 \times \sin {30^ \circ }\\\\ \Rightarrow F = 2/33 \times {10^{ - 8}}\,N\end{array}\)