A) \[8\times {{10}^{-20}}N\]
B) \[3.2\times {{10}^{-19}}N\]
C) \[8\times {{10}^{-18}}N\]
D) \[1.6\times {{10}^{-19}}N\]
E) zero
Correct Answer: C
Solution :
\[F=Bev\] and \[B=\frac{{{\mu }_{0}}i}{2\pi r}\] \[\therefore \] \[F=\frac{{{\mu }_{0}}i}{2\pi r}\times ev\] \[=\frac{4\pi \times {{10}^{-7}}\times 5}{2\pi \times 0.1}\times 1.6\times {{10}^{-19}}\times 5\times {{10}^{6}}\] \[=8\times {{10}^{-18}}N\]
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