An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 × 10⁻⁴ Wbm⁻². The frequency of revolution of the electron will be :
(Take mass of electron = 9.0 × 10⁻³¹ kg)

Solution

Physics sir · 4 steps

The energy is a decoy here: cyclotron frequency depends only on q, B and m, never on speed.

1. Force balance. The magnetic force qvB supplies the centripetal force mv²/r, so the radius is r = mv/qB. A faster electron moves in a bigger circle.
2. Speed cancels. The time for one turn is T = (2π r)/v = (2π m)/qB. Here v cancels out, so the 0.1 keV energy is not needed.

   f = qB/(2π m)

3. Put in numbers. f = (1.6 × 10⁻¹⁹ × 1 × 10⁻⁴)/(2π × 9.0 × 10⁻³¹) = (1.6 × 10⁻²³)/(5.65 × 10⁻³⁰) ≈ 2.83 × 10⁶ Hz.
4. So. The frequency of revolution is about 2.8 × 10⁶ Hz, option (c).

The idea to keep. For a charge moving perpendicular to a uniform field, f = qB/(2π m). Energy or speed changes only the radius, never the frequency, so do not spend time computing v.

Answer from the NTA final answer key. Solution by PrepWiser.