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Q.

An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57×10-2  T. If the value of e/m is 1.76×1011 C kg-1, the frequency of revolution of the electron is

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a

1 GHz

b

100 MHz

c

62.8 MHz

d

6.28 MHz

answer is A.

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Detailed Solution

Here,B=3.57×10-2 T,em=1.76×1011Ckg-1Frequency of revolution of the electron, v=1T=v2πr……..(1) Also, mv2r=evB⇒vr=eBm……….(2)From eqns. (i) and (ii)v=12π×eBm=12×3.14×1.76×1011×3.57×10-2=109 Hz=1GHz
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