question_answer

A charge particle having charge \[1\times {{10}^{-19}}\,C\] revolves in an orbit of radius 1 A such that the frequency of revolution is\[1016\text{ }Hz\]. 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 \[M=IA=\frac{q\omega }{2\pi \times \pi {{r}^{2}}}\] \[=\frac{q\omega {{r}^{2}}}{2}=\frac{1\times {{10}^{-19}}\times 2\pi \times {{10}^{16}}\times {{10}^{-20}}}{2}\] \[=\pi \times {{10}^{-39}}\times {{10}^{16}}=3.14\times {{10}^{-23}}\]