QJEEM23Ph1f4d: JEE Main 2023 Physics - Class 12 JEEM23Ph1f4d Solution: An electron in a hydrogen atom revolves around its nucleus with …

The formula for the magnetic field due to a moving charge is given by:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2}$

where $\mu_0$ is the permeability of free space, $q$ is the charge of the moving particle, $v$ is the speed of the particle, $\theta$ is the angle between the velocity vector and the position vector from the particle to the point where we want to calculate the magnetic field, and $r$ is the distance between the particle and the point where we want to calculate the magnetic field.

In this case, we're interested in the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom. Since the orbit is circular, the angle between the velocity vector and the position vector is 90 degrees, so $\sin \theta = 1$ . We can substitute the known values into the formula to find the magnetic field:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2} = \frac{\mu_0}{4 \pi} \frac{e v}{r^2}$

where $e$ is the charge of an electron. We know that the radius of the orbit is $0.52 \mathrm{~A}^{\circ}$ , which is equivalent to $0.52 \times 10^{-10} \mathrm{m}$ .

Substituting the values, we get:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{e v}{r^2} =\frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{0.52 \times 0.52 \times 10^{-20}} = 40 ~\mathrm{T}$

This means that the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom is 40 tesla, which is an incredibly strong magnetic field.