A) \[1.18\times {{10}^{-13}}\]
B) \[1.28\times {{10}^{-13}}\]
C) \[1.6\times {{10}^{-13}}\]
D) \[1.72\times {{10}^{-13}}\]
Correct Answer: C
Solution :
\[\overrightarrow{v\,}=2\times {{10}^{5}}\hat{i}\]and \[\overrightarrow{B}=(\hat{i}+4\hat{j}-3\hat{k})\] \[\overrightarrow{F}=q\,(\overrightarrow{v\,}\times \overrightarrow{B})=-1.6\times {{10}^{-19}}[2\times {{10}^{5}}\hat{i}\times (i+4\hat{j}-3\hat{k})]\] \[=-1.6\times {{10}^{-19}}\times 2\times {{10}^{5}}[\hat{i}\times \hat{i}+4(\hat{i}\times \hat{j})-3(\hat{i}\times \hat{k})]\] \[=-3.2\times {{10}^{-14}}[0+4\hat{k}+3\hat{j}]=3.2\times {{10}^{-14}}(-4\hat{k}-3\hat{k})\] Þ \[|\overrightarrow{F}|\,=3.2\times {{10}^{-14}}\times 5=1.6\times {{10}^{-13}}N.\]
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