A) 1.12 mA
B) 3 mA
C) 0.75 mA
D) 2.25 mA
Correct Answer: A
Solution :
Frequency of revolution of electron, \[f=\frac{v}{2\pi r}=\frac{2.2\times {{10}^{6}}}{2\pi (5\times {{10}^{-11}})}=7.0\times {{10}^{15}}Hz\] Current associated, \[i=q\,f\] \[=(1.6\times {{10}^{-19}})(7.0\times {{10}^{15}})\] \[=11.2\times {{10}^{-4}}A=1.12\,mA\]You need to login to perform this action.
You will be redirected in
3 sec