A) 1 mA
B) 1.5 mA
C) 2.5 mA
D) \[1.5\times {{10}^{-2}}mA\]
Correct Answer: A
Solution :
Time period of revolution of electron \[T=\frac{2\pi }{\omega }=\frac{2\pi r}{v}\] Hence corresponding electric current \[i=\frac{e}{T}=\frac{ev}{2\pi r}\] \[\Rightarrow \,\,i=\frac{1.6\times {{10}^{-19}}\times 2\times {{10}^{6}}}{2\times 3.14\times 0.5\times {{10}^{-10}}}=1\,mA.\]
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