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 (a) 1 GHz (b) 100 MHz (c) 62.8 MHz (d) 6.28 MHz

Answer: (a). Frequency of revolution f = (qB)/(2π m) = 1/2π(e/m)B f = 1.76×10¹¹× 3.57×10⁻²2π = 6.28×10⁹6.28 f = 1×10⁹ Hz So the answer is (a) 1 GHz.

An electron is moving in a circular path under the influence of a… | NEET 2016 (Phase 2)

Q97NEET 2016 · PHASE 2Easy

Moving Charge in a Magnetic Field

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

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Answer: (a)

Frequency of revolution $f = \dfrac{qB}{2\pi m} = \dfrac{1}{2\pi}\left(\dfrac{e}{m}\right)B$
$f = \dfrac{1.76\times10^{11}\times 3.57\times10^{-2}}{2\pi} = \dfrac{6.28\times10^{9}}{6.28}$
$f = 1\times10^{9}\text{ Hz}$
So the answer is (a) 1 GHz.

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