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