A) \[2\times {{10}^{-24}}A{{m}^{2}}\]
B) \[4\times {{10}^{-24}}A{{m}^{2}}\]
C) \[8\times {{10}^{-24}}A{{m}^{2}}\]
D) \[16\times {{10}^{-24}}A{{m}^{2}}\]
Correct Answer: C
Solution :
The time period of electron motion in circular orbit is \[T=\frac{2\pi a}{v}=\frac{2\pi \times 5.0\times {{10}^{-11}}}{2.0\times {{10}^{6}}}=5\pi \times {{10}^{-17}}\sec .\] Therefore, the equivalent current is \[I=\frac{\text{Charge}}{\text{Time}}=\frac{e}{T}=\frac{1.6\times {{10}^{-19}}}{5\pi \times {{10}^{-17}}}=\frac{1.6}{5\pi }\times {{10}^{-2}}A\] Equivalent magnetic dipole moment = current\[\times \]area of circular orbit. \[=I\times (\pi {{a}^{2}})=\frac{1.6\times {{10}^{-2}}}{5\pi }\times \pi {{(5\times {{10}^{-11}})}^{2}}=8\times {{10}^{-24}}\,A{{m}^{2}}.\] Hence, the correction option is [c].
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