Magnetic Effects Of Current Question 124
Question: An electron is revolving round a proton, producing a magnetic field of 16 weber/m2 in a circular orbit of radius 1Å. It?s angular velocity will be [RPMT 2002]
Options:
A) 1017 rad/sec
B) 1/2p ´ 1012 rad/sec
C) 2 p ´ 1012 rad/sec
D) 4 p ´ 1012 rad/sec
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Answer:
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 .( \frac{e\omega }{2\pi } )}{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 . $
BETA
