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. `62.8 MHz`B. `6.28 MHz`C. `1 GHz`D. `100 MHz`

Answer» Correct Answer - C
`f = (eB)/(2pi m)`
`f = (1.76xx10^(11)xx3.57xx10^(-2))/(2xx3.14)`
`f = 10^(9) Hz` or `1 GHz`


Discussion

No Comment Found

Related InterviewSolutions