An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76 × 10⁶ ms⁻¹ in an orbit of radius 0.52 A°. The magnetic field produced at the nucleus of the hydrogen atom is _________ T.

Solution

Physics sir · 4 steps

The revolving electron is a tiny current loop. Find its current, then use the field at the centre of a loop.

1. Electron as current. The electron goes round once in time T = (2π r)/v. So the charge e passes any point once every T, which is a current I = e/T = ev/(2π r).
2. Field at centre. The nucleus sits at the centre of this circular loop, where Biot-Savart gives B = (μ₀ I)/2r. Substitute the current.

   B = μ₀/2r·ev/(2π r) = μ₀/4π·ev/r²

3. Put in numbers. r = 0.52 × 10⁻¹⁰ m, so r² = 2.704 × 10⁻²¹ m². Also ev = 1.6 × 10⁻¹⁹ × 6.76 × 10⁶ = 1.0816 × 10⁻¹². Then B = 10⁻⁷ × (1.0816 × 10⁻¹²)/(2.704 × 10⁻²¹) = 40 T.
4. So. The magnetic field at the nucleus is 40 T.

The idea to keep. Any charge moving in a circle is a current loop with I = qv/(2π r). At the centre, B = μ₀/4πqv/r², so remember to square the radius.

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