An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76 × 10⁶ m s⁻¹ in an orbit of radius 0.52 A. The magnetic field produced at the nucleus of the hydrogen atom is _________ T. Numerical answer type. Work it out, then open the answer.

Answer: 40. I = (ev)/(2π r), so B = (μ₀ I)/(2r) = (μ₀)/(4π)ev/r². B = 10⁻⁷ × 1.6 × 10⁻¹⁹ × 6.76 × 10⁶(0.52 × 10⁻¹⁰)² B = 1.0816 × 10⁻¹⁹2.704 × 10⁻²¹ = 40 T So the answer is 40 T.

An electron in a hydrogen atom revolves around its nucleus with a… | JEE Main 2023 (15 Apr, Shift 1)

Q58JEE Main 2023 · 15 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

An electron in a hydrogen atom revolves around its nucleus with a speed of $6.76 \times 10^{6}\text{ m s}^{-1}$ in an orbit of radius $0.52\ \mathring{\text{A}}$ . The magnetic field produced at the nucleus of the hydrogen atom is _________ T.

Numerical answer type. Work it out, then open the answer.

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Answer: 40

$I = \frac{ev}{2\pi r}$ , so $B = \frac{\mu_0 I}{2r} = \frac{\mu_0}{4\pi}\frac{ev}{r^2}$ .
$B = 10^{-7} \times \frac{1.6 \times 10^{-19} \times 6.76 \times 10^{6}}{(0.52 \times 10^{-10})^2}$
$B = \frac{1.0816 \times 10^{-19}}{2.704 \times 10^{-21}} = 40\text{ T}$
So the answer is $40\text{ T}$ .

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