PhysicsEasy49×since 2003Q7626An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 ×\times× 10^−-−4 Wbm^−-−2. The frequency of revolution of the electron will be : (Take mass of electron = 9.0 ×\times× 10^−-−31 kg)A1.6×105 Hz1.6 \times 10^{5} \mathrm{~Hz}1.6×105 HzB5.6×105 Hz5.6 \times 10^{5} \mathrm{~Hz}5.6×105 HzC2.8×106 Hz2.8 \times 10^{6} \mathrm{~Hz}2.8×106 HzD1.8×106 Hz1.8 \times 10^{6} \mathrm{~Hz}1.8×106 HzCheck answerSkip