A proton is revolving on a circular path of radius $2 \mathrm{mm}$ with frequency $10 \mathrm{Hz}$. Magnetic dipole moment associated with proton is

Given, charge on proton, $q=e$
$=1.6\times {10}^{-19} \mathrm{C}$

Radius, $r=2 \mathrm{mm}=2\times {10}^{-3} \mathrm{cm}$

Frequency, $f=10 \mathrm{Hz}$

Current associated by proton is given by

$I=\frac{q}{t}$                       $[T\to$ time period $]$

$=qf \left[\because T=\frac{1}{f}\right]$

$=1.6\times {10}^{-19}\times 10=1.6\times {10}^{-18} \mathrm{A}$

$\therefore$ Magnetic dipole moment, $M=IA$

$=I·\pi r^2=1.6\times {10}^{-18}\times 3.14\times {\left(2\times {10}^{-3}\right)}^2$
$=2\times {10}^{-24} \mathrm{A}-{\mathrm{m}}^2$