JEE Main 2022 · 25 July · morning · Single correct · +4 / −1

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by :

  1. (a) μ⃗_L=eL⃗/(2 m)
  2. (b) μ⃗_L=-eL⃗/(2 m)
  3. (c) μ⃗ₗ=-(e L⃗)/m
  4. (d) μ⃗ₗ=(2 eL⃗)/m
Answer and solution

Answer (b) μ⃗_L=-eL⃗/(2 m)

Solution

Physics sir · 4 steps

Treat the orbiting electron as a tiny current loop, then remember its negative charge flips the direction.

  1. Current of the orbit. An electron of charge magnitude e goes round a circle of radius r with speed v. It passes any point once every T = (2π r)/v, so the loop carries current I = e/T = ev/(2π r).
  2. Magnitude of moment. A current loop has magnetic moment μ = IA with A = π r². That gives μ = ev/(2π r) × π r² = evr/2. Since L = mvr, put vr = L/m.

    μ_L = e/2m L

  3. Direction. The electron is negative, so the conventional current flows opposite to its motion. By the right-hand rule, μ⃗_L then points opposite to L⃗, which puts a minus sign in the vector form.
  4. So. μ⃗_L = -eL⃗/2m, which is option (b).

The idea to keep. For any orbiting charge, μ/L = q/2m, the gyromagnetic ratio. For the electron, q is negative, so μ⃗ is always antiparallel to L⃗.

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

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Asked about 110 times in JEE Main since 2020. This question is from 2022.