A) \[8\times {{10}^{-20}}N\]
B) \[3.2\times {{10}^{-19}}N\]
C) \[8\times {{10}^{-18}}N\]
D) \[1.6\times {{10}^{-19}}N\]
Correct Answer: C
Solution :
Magnetic field produced by wire is perpendicular to the motion of electron and it is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2i}{a}={{10}^{-7}}\times \frac{2\times 5}{0.1}={{10}^{-5}}\,Wb/{{m}^{2}}\] Hence force on electron \[F=qvB=(1.6\times {{10}^{-19}})\times 5\times {{10}^{6}}\times {{10}^{-5}}=8\times {{10}^{-18}}\,N\]
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