Magnetic Field at Centre of Helium Nucleus Orbit – Rankers Physics

Magnetic Field Due to Circular Current Carrying Wire: Practice Problem & Solution

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 Explained:

To solve this problem, we apply the core principles of Magnetic Field Due to Circular Current Carrying Wire. Understanding the underlying formula is key to arriving at the correct answer below:

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

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): A paramagnetic sample shows greater degree of magnetisation when cooled. Reason (R): Magnetisation of paramagnetic material is inversely proportional to temperature.  
2. The magnetic field of given length of wire for single turn coil at its centre is 'B' then its value for two turns coil for the same wire is: (2002)
3. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has magnitude: (2015)
4. 19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020) ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)
5. Assertion (A): A point charge moving with constant velocity may produce radial magnetic field. Reason (R): Rest point charge produces radial electric field.