1.

An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57xx10^(-2)T. If the value of e/m is 1.76xx10^(11)C/(kg), the frequency of revolution of the electron is

Answer»

62.8 MHz
6.28 MHz
1 GHz
100 MHz

Solution :For CIRCULAR orbit, `(MV^(2))/r=Bev`
`thereforev=(BER)/m`
`thereforeromega=(Ber)/m""[becausev=romega]`
`thereforeomega=(Be)/m`
`therefore2pif=(Be)/m""[becauseomega=2pif]`
`thereforef=B/(2pi)xxe/m=(3.57xx10^(-2)xx1.76xx10^(11))/(2xx3.14)`
`thereforef=10^(9)Hz=1GHz`


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