An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of $1 \times 10^{-4} \mathrm{Wbm}^{-2}$. The frequency of revolution of the electron will be (Take mass of electron $=9.0 \times 10^{-31} \mathrm{~kg}$ )

(C) $2.8 \times 10^{6} \mathrm{~Hz}$

(A) $1.6 \times 10^{5} \mathrm{~Hz}$

(B) $5.6 \times 10^{5} \mathrm{~Hz}$

(D) $1.8 \times 10^{6} \mathrm{~Hz}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Magnetic Effects of Current

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of

1 \times 1 0^{- 4} Wbm^{- 2}

. The frequency of revolution of the electron will be
(Take mass of electron

= 9.0 \times 1 0^{- 31} kg

)

Answer: (C)

The frequency of revolution of a charged particle in a magnetic field is given by:

f = \frac{e B}{2 π m}

where

e = 1.6 \times 1 0^{- 19}

B = 1 \times 1 0^{- 4}

T, and

m = 9.0 \times 1 0^{- 31}

kg.

Substituting:

f = \frac{1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 4}}{2 π \times 9 \times 1 0 ^{- 31}} = \frac{1.6 \times 1 0 ^{- 23}}{18 π \times 1 0 ^{- 31}} \approx 2.8 \times 1 0^{6} Hz

Note: The kinetic energy is not needed for this calculation as the frequency is independent of the particle's speed.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option