A) \[7.4\times {{10}^{12}}\,N\]
B) \[7.4\times {{10}^{-12}}\,N\]
C) \[7.4\times {{10}^{19}}\,N\]
D) \[7.4\times {{10}^{-19}}\,N\]
Correct Answer: B
Solution :
\[F=qvB\] and \[K=\frac{1}{2}m{{v}^{2}}\] Þ \[F=qB\sqrt{\frac{2k}{m}}\] \[=1.6\times {{10}^{-19}}\times 1.5\sqrt{\frac{2\times 5\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{1.7\times {{10}^{-27}}}}\] \[=7.344\times {{10}^{-12}}\,N\]
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