Consider an electron revolving in a circular orbit of a hydrogen...

The electrostatic force provides the necessary centripetal force for the electron:

$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}$

Rearranging for $r$, we get:

$r = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{mv^2}$

This shows that $r \propto \dfrac{1}{v^2}$.

The revolving electron constitutes an electric current $I$:

$I = \dfrac{e}{T} = \dfrac{ev}{2\pi r}$

The magnetic field $B$ at the center of the circular orbit is:

$B = \dfrac{\mu_0 I}{2r} = \dfrac{\mu_0 e v}{4\pi r^2}$

Thus, $B \propto \dfrac{v}{r^2}$.

Substituting $r \propto \dfrac{1}{v^2}$ into the expression for $B$:

$B \propto \dfrac{v}{\left(\dfrac{1}{v^2}\right)^2} = v \times v^4 = v^5$

Comparing this with $B \propto v^x$, we get $x = 5$.

Answer: $5$