question_answer

A charge particle having charge\[1\times {{10}^{-19}}C\]revolves in an orbit of radius\[1\overset{o}{\mathop{A}}\,\]such that the frequency of revolution is\[{{10}^{16}}Hz\]. The resulting magnetic moment in SI units will be

A) \[1.57\times {{10}^{-21}}\]        

B) \[3.14\times {{10}^{-21}}\]

C)  \[1.57\times {{10}^{-23}}\]       

D) \[3.14\times {{10}^{-23}}\]

Correct Answer: D

Solution :

 Current associated with the revolving charged particle \[I=\frac{q\omega }{2\pi }\] Magnetic moment \[M=IA=\frac{q\omega }{2\pi }\times \pi {{r}^{2}}\] \[=\frac{q\omega {{r}^{2}}}{2}=\frac{1\times {{10}^{-19}}\times 2\pi \times {{10}^{16}}\times {{10}^{-20}}}{2}\] \[=\pi \times {{10}^{-39}}\times {{10}^{16}}=3.14\times {{10}^{-23}}\]