Magnetic Effects Of Current Question 194
In hydrogen atom, the electron is making $ 6.6\times 10^{15},rev/\sec $ around the nucleus in an orbit of radius 0.528 Å. The magnetic moment $ (A\cdot m^{2}) $ will be [MP PET 1999]
Options:
A) $ 1\times {{10}^{-15}} $
B) $ 1\times {{10}^{-10}} $
C) $ 1\times {{10}^{-23}} $
D) $ 1\times {{10}^{-27}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ i=6.6\times 10^{15}\times 1.6\times {{10}^{-19}}=1.05\times {{10}^{-3}}amp $ $ A=\pi R^{2}=3.142\times {{( 0.528 )}^{2}}\times {{10}^{-20}}m^{2} $
Þ $ M=iA=10.5\times {{10}^{-4}}\times 3.142\times {{( 0.528 )}^{2}}\times {{10}^{-20}} $ $ =10\times {{10}^{-23}}units=1\times {{10}^{-22}}units $
BETA
