InterviewSolution
Saved Bookmarks
| 1. |
An electron is moving in a circular path under the influence fo a transerve magnetic field of `3.57xx10^(-2)T`. If the value of `e//m` is `1.76xx10^(141) C//kg`. The frequency of revolution of the electron isA. 1 GHzB. 100 MHzC. 62.8 MHzD. 6.28 MHz |
|
Answer» Correct Answer - A `f=(qB)/(2pim)` `=((q)/(m))*(B)/(2pi)` `=(1.76xx10^(11)xx3.56xx10^(-2))/(2xx3.14)` `1xx10^(9)Hz` `=1GHz` |
|