91Ó°ÊÓ

Time period of electron is given by

$\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}\mathbf{r}}{\mathbf{v}}$

The current is given by

$\mathit{l}\mathbf{=}\frac{\mathbf{Q}}{t}$

The magnetic moment is given by

$\mathit{μ}\mathbf{=}\mathit{I}\mathit{A}\mathbf{=}\mathit{I}\mathit{π}{\mathit{r}}^{\mathbf{2}}$

The speed of proton is,$V=2.2×{10}^6\mathrm{m}/\mathrm{s}$ .

The radius of the orbit is,$r=5.3×{10}^{−11}\mathrm{m}$ .

(a)

Orbital period of an electron is given by

$T=\frac{2\mathrm{Ï€}r}{v}$

Substitute all the value in the above equation.

$\begin{array}{r}T=\frac{2\mathrm{π}\left(5.3×{10}^{−11}\mathrm{m}\right)}{2.2×{10}^6\mathrm{m}/\mathrm{s}}\\ =1.5×{10}^{−16}\mathrm{s}\end{array}$

Therefore, the orbital period of the electron is$\mathrm{T}=1.5×{10}^{-16}\mathrm{s}$

(b)

The current is given by

$\begin{array}{r}\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}\\ \mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}\\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{value}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}.\\ \mathrm{I}=\frac{1.6×{10}^{−19}\mathrm{C}}{1.5×{10}^{−16}\mathrm{s}}\\ =1.1×{10}^{−3}\mathrm{A}\end{array}$

Therefore, the current is$I=1.1×{10}^{-3}A$

(c)

The magnetic moment is given by

$\begin{array}{r}\mathrm{μ}=IA\\ \mathrm{μ}=I\mathrm{π}{r}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{μ}=\left(1.1×{10}^{−3}\mathrm{A}\right)\mathrm{π}{\left(5.3×{10}^{−11}\mathrm{m}\right)}^2\\ =9.71×{10}^{−24}{\mathrm{Am}}^2\end{array}$

Therefore, the magnetic moment is$\mathrm{μ}=9.71×{10}^{−24}{\mathrm{Am}}^2$