91Ó°ÊÓ

Distance between electron and proton, $r=5.2×{10}^{-11}\text{m}$

The component of the proton’s magnetic dipole moment along z-axis,${\mathrm{μ}}_{s,z}=1.4×{10}^{−26}\raisebox{1ex}{$\text{ J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$

A moving charge produces both, an electric field as well as a magnetic field. The electric field is a region around a charge, in which any other charge experiences a force. Similarly, a magnetic field is a space around a magnet where any other magnet experiences a force.

The electric field is given due to a charge q at a separation r is given as-

$\mathrm{E}=\frac{}{4{\mathrm{Ï€Î\mu }}_0}×\left(\frac{\mathrm{q}}{{\mathrm{r}}^2}\right)$

The magnetic field is given as,

$\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{μ}}_{\mathrm{s},\mathrm{z}}}{2{\mathrm{Ï€°ù}}^3}$

The Bohr Magneton,

${\mathrm{μ}}_{\mathrm{b}}=\frac{\mathrm{eh}}{4{\mathrm{Ï€³¾}}_{\mathrm{e}}}$

The electric field using Coulomb’s law is,

$\begin{array}{rcl}E& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}×\left(\frac{q}{r^2}\right)\\ & =& 9×{10}^9\text{ }\raisebox{1ex}{${\text{Nm}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{${\text{C}}^{\text{2}}$}\right.×\left(\frac{1.6×{10}^{−19}\text{ C}}{{(5.2×{10}^{−11}\text{″¾})}^2}\right)\\ & =& 5.3×{10}^{11}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\end{array}$

Hence, the magnitude of the proton’s electric field will be $\begin{array}{rcl}& & 5.3×{10}^{11}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\end{array}$.

You can write the magnetic field as.

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_0{\mathrm{μ}}_{s,z}}{2\mathrm{Ï€}{r}^3}\\ & =& \frac{4\mathrm{Ï€}×{10}^{−7}{\displaystyle \raisebox{1ex}{$\text{ H}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.}×1.4×{10}^{−26}{\displaystyle \raisebox{1ex}{$\text{ J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}}{2\mathrm{Ï€}×{(5.2×{10}^{−11}\text{″¾})}^3}\\ & =& 2.0×{10}^{-2}\text{T}\end{array}$

Hence, the magnitude of the proton’s magnetic field will be $\begin{array}{rcl}& & 2.0×{10}^{-2}\text{T}\end{array}$.

You know that

${\mathrm{μ}}_b=\frac{eh}{4\mathrm{π}{m}_e}$

And you have

${\mathrm{μ}}_{s,z}=1.4×{10}^{-26}\text{ }\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$

Thus, you can take the ratio as $\frac{{\mathrm{μ}}_b}{{\mathrm{μ}}_{s,z}}$. You get

$\begin{array}{rcl}\frac{{\mathrm{μ}}_b}{{\mathrm{μ}}_{s,z}}& =& \frac{{\displaystyle \left(\frac{eh}{4\mathrm{π}{m}_e}\right)}}{1.4×{10}^{−26}{\displaystyle \raisebox{1ex}{$\text{ J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}}\\ & =& \frac{{\displaystyle 9.27×{10}^{−24}\raisebox{1ex}{${\displaystyle \text{ J}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{T}}$}\right.}}{{\displaystyle 1.4}{\displaystyle ×}{\displaystyle {{\displaystyle 10}}^{−26}}{\displaystyle \text{ }}\raisebox{1ex}{$\text{ J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}\\ & =& 6.6×{10}^2\end{array}$

Hence, the ratio of the spin magnetic dipole moment of the electron to that of proton is $6.6×{10}^2$.