Magnetic Field at Centre of Helium Nucleus Orbit – Rankers Physics
Topic: Magnetic Effects of Current
Subtopic: Magnetic Field Due to Circular Current Carrying Wire

Magnetic Field at Centre of Helium Nucleus Orbit

A helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:
\(\mu_0 \times 10^{-19}\text{ T}\)
\(\frac{10^{-19}}{\mu_0}\text{ T}\)
\(2 \times 10^{-19}\text{ T}\)
\(\frac{2 \times 10^{-19}}{\mu_0}\text{ T}\)

Solution:

Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).

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