A) Zero
B) \[1.6\times {{10}^{-10}}N\]
C) \[3.2\times {{10}^{-8}}N\]
D) \[1.6\times {{10}^{-6}}N\]
Correct Answer: B
Solution :
\[F=qvB\] also Kinetic energy \[K=\frac{1}{2}m{{v}^{2}}\]\[\Rightarrow v=\sqrt{\frac{2K}{m}}\] \ \[F=q\sqrt{\frac{2K}{m}}B\] \[=1.6\times {{10}^{-19}}\sqrt{\frac{2\times 200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{1.67\times {{10}^{-27}}}}\times 5\] \[=1.6\times {{10}^{-10}}N\]
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