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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 25 July 2022, Shift 2, Physics Q41

Question 41 of 90 in this shift, Physics question 11 of 30, Section A.

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1×10−41 \times 10^{-4} Wbm−2^{-2}. The frequency of revolution of the electron will be (Take mass of electron = 9.0×10−319.0 \times 10^{-31} kg)
  1. (1)1.6×1051.6 \times 10^{5} Hz
  2. (2)5.6×1055.6 \times 10^{5} Hz
  3. (3)2.8×1062.8 \times 10^{6} HzOfficial answer
  4. (4)1.8×1061.8 \times 10^{6} Hz

Official answer

Option 3

NTA final key.