Magnetic Effects Of Current Question 442
Question: The orbital speed of electron orbiting around a nucleus in a circular orbit of radius 50 pm is $ 2.2\times 10^{6},m{{s}^{-1}} $ . Then the magnetic dipole moment of an electron is
Options:
A) $ 1.6\times {{10}^{-19}}Am^{2} $
B) $ 5.3\times {{10}^{-21}}Am^{2} $
C) $ 8.8\times {{10}^{-26}}Am^{2} $
D) $ 8.8\times {{10}^{-25}}Am^{2} $
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Answer:
Correct Answer: C
Solution:
[c] Magnetic dipole moment $ m=iA=\frac{e}{T}\times \pi r^{2}=\frac{e}{(2\pi r/v)}\times \pi r^{2}=\frac{erv}{2}. $ $ =\frac{1.6\times {{10}^{-19}}\times 50\times {{10}^{-12}}\times 2.2\times 10^{6}}{2} $ $ =8.8\times {{10}^{-25}},Am^{2} $ .
BETA
