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

The formula for the magnetic field due to a moving charge is given by: <br/><br/> $\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2}$ <br/><br/> 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. <br/><br/> 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: <br/><br/> $\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2} = \frac{\mu_0}{4 \pi} \frac{e v}{r^2}$ <br/><br/> 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 $0.52 \times 10^{-10} \mathrm{m}$. <br/><br/> Substituting the values, we get: <br/><br/> $\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{e v}{r^2} =\frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{0.52 \times 0.52 \times 10^{-20}} = 40 ~\mathrm{T}$ <br/><br/> 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.