A) \[1\times {{10}^{-11}}\Omega \]
B) \[1.1\times {{10}^{-7}}\Omega \]
C) \[5.1\times {{10}^{-7}}\Omega \]
D) \[2.1\times {{10}^{-7}}\Omega \]
Correct Answer: A
Solution :
Using the releation \[i=\frac{q}{t}=\frac{electronic\text{ }charge}{time\text{ }period}=\frac{e}{t}\] Hence, \[t=\frac{2\pi r}{v}\] So, \[i=\frac{e}{2\pi rv}\] \[=\frac{ev}{2\pi r}\] \[=\frac{(1.6\times {{10}^{-19}})(4\times {{10}^{6}})}{2\times 3.14\times 1.0\times {{10}^{-2}}}\] \[=1\times {{10}^{-11}}\Omega \]
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