JEE Main 2022 · 25 Jul · Shift 2 · Physics Q41

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PhysicsSingle correct+4 / -1Question 41 of 90

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of $1\times10^{-4}$ Wbm $^{-2}$ . The frequency of revolution of the electron will be (Take mass of electron $=9.0\times10^{-31}$ kg)

1.6\times10^5

5.6\times10^5

2.8\times10^6

1.8\times10^6

Answer and worked solution

Official answer: C

Cyclotron frequency $\dfrac{eB}{2\pi m}=\dfrac{1.6\times10^{-19}\times10^{-4}}{2\pi\times9\times10^{-31}}\approx2.8\times10^6$ Hz (independent of energy).

(C)

Topics:Motion in magnetic and electric fields

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JEE Main 2022 · 25 Jul · Shift 2, Physics question 41