A) 1017 rad/sec
B) 1/2p ´ 1012 rad/sec
C) 2 p ´ 1012 rad/sec
D) 4 p ´ 1012 rad/sec
Correct Answer: A
Solution :
Magnetic field due to revolution of electron \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi i}{r}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi .\left( \frac{e\omega }{2\pi } \right)}{r}={{10}^{-7}}\times \frac{e\omega }{r}\] \[\Rightarrow 16={{10}^{-7}}\times \frac{1.6\times {{10}^{-19}}\omega }{1\times {{10}^{-10}}}\Rightarrow \omega ={{10}^{17}}rad/\sec .\]
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