A) \[2.5\times {{10}^{-10}}\,N\]
B) \[7.6\times {{10}^{-11}}\,N\]
C) \[2.5\times {{10}^{-11}}\,N\]
D) \[7.6\times {{10}^{-12}}\,N\]
Correct Answer: D
Solution :
\[F=qvB=1.6\times {{10}^{-19}}\times \left[ \sqrt{\frac{2E}{m}} \right]\ 2.5\] \[=4\times {{10}^{-19}}\sqrt{\frac{2\times 2\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{1.66\times {{10}^{-27}}}}\]\[=7.6\times {{10}^{-12}}N\]
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