Dissertativas 1 - 2ª Fase Dia 3 - ITA 2025

a) Pode-se escrever o fluxo magnético como: $\mathrm{\varphi }=\mathrm{\alpha }·{\mathrm{h}}^{\mathrm{x}}·{\mathrm{c}}^{\mathrm{y}}·{\mathrm{e}}^{\mathrm{z}}$

Sendo $\alpha$ uma constante adimensional.

Tem-se as dimensionais:

$$\begin{gathered} \left[\mathrm{\varphi }\right] =\left[\mathrm{B}\right]·\left[\mathrm{A}\right]=\frac{\left[\mathrm{F}\right]}{\left[\mathrm{i}\right]·\left[\mathrm{L}\right]}·\left[\mathrm{A}\right]=\frac{\mathrm{M}·\mathrm{L}·{\mathrm{T}}^{-2}}{\mathrm{I}·\mathrm{L}}·{\mathrm{L}}^2=\mathrm{M}·{\mathrm{L}}^2·{\mathrm{T}}^{-2}·{\mathrm{I}}^{-1} \\ \left[\mathrm{h}\right] =\frac{\left[\mathrm{E}\right]}{\left[\mathrm{f}\right]}=\frac{\mathrm{M}·{\mathrm{L}}^2·{\mathrm{T}}^{-2}}{{\mathrm{T}}^{-1}}=\mathrm{M}·{\mathrm{L}}^2·{\mathrm{T}}^{-1} \\ \left[\mathrm{c}\right]=\mathrm{L}·{\mathrm{T}}^{-1} \\ \left[\mathrm{e}\right]=\mathrm{T}·\mathrm{I} \end{gathered}$$

Pela homogeneidade dimensional:

$$\begin{gathered} \left[\mathrm{\varphi }\right]=\left[\mathrm{\alpha }\right]·{\left[\mathrm{h}\right]}^{\mathrm{x}}·{\left[\mathrm{c}\right]}^{\mathrm{y}}·{\left[\mathrm{e}\right]}^{\mathrm{z}} \\ \mathrm{M}·{\mathrm{L}}^2{\mathrm{T}}^{-2}·{\mathrm{I}}^{-1}=1·{\left(\mathrm{M}·{\mathrm{L}}^2·{\mathrm{T}}^{-1}\right)}^{\mathrm{x}}·{\left(\mathrm{L}·{\mathrm{T}}^{-1}\right)}^{\mathrm{y}}·{\left(\mathrm{T}·\mathrm{I}\right)}^{\mathrm{z}} \\ \mathrm{M}·{\mathrm{L}}^2·{\mathrm{T}}^{-2}·{\mathrm{I}}^{-1}={\mathrm{M}}^{\mathrm{x}}·{\mathrm{L}}^{2\mathrm{x}+\mathrm{y}}·{\mathrm{T}}^{-\mathrm{x}-\mathrm{y}+\mathrm{z}}·{\mathrm{I}}^{\mathrm{z}} \\ \left\{\begin{array}{l}\mathrm{x}=1\\ 2\mathrm{x}+\mathrm{y}=2\\ -\mathrm{x}-\mathrm{y}+\mathrm{z}=-2\\ \mathrm{z}=-1\end{array}\right. \\ (\mathrm{x},\mathrm{y},\mathrm{z}) = (1,0,-1) \end{gathered}$$

Logo: $\mathrm{\varphi }=\mathrm{\alpha }·\frac{\mathrm{h}}{\mathrm{e}}$

Considera-se $\mathrm{\alpha }=1\Rightarrow \mathrm{\varphi }=\frac{\mathrm{h}}{\mathrm{e}}$

Portanto, ${\mathrm{\varphi }}_0=\frac{\mathrm{\varphi }}{2}\Rightarrow \boxed{{\mathrm{\varphi }}_0==\frac{\mathrm{h}}{2\mathrm{e}}}$

b) O fluxo magnético dentro da órbita do elétron é: 

${\mathrm{\varphi }}_0=\mathrm{B}·{\mathrm{\pi R}}^2=\frac{\mathrm{h}}{2\mathrm{e}}\Rightarrow \mathrm{B}=\frac{\mathrm{h}}{2\mathrm{e}·{\mathrm{\pi R}}^2}$

A força magnética é a responsável por manter o elétron na órbita:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{m}}=\mathrm{e}·\mathrm{v}·\mathrm{B} =\frac{{\mathrm{m}}_{\mathrm{e}}·{\mathrm{V}}^2}{\mathrm{R}}\Rightarrow \mathrm{R}=\frac{{\mathrm{m}}_{\mathrm{e}}·\mathrm{V}}{\mathrm{e}·\mathrm{B}} \\ \mathrm{R}=\frac{{\mathrm{m}}_{\mathrm{e}}·\mathrm{\beta }·\mathrm{c}}{\mathrm{e}·{\displaystyle \frac{\mathrm{h}}{2\mathrm{e}·{\mathrm{\pi R}}^2}}}\Rightarrow \boxed{\mathrm{R}=\frac{\mathrm{h}}{2{\mathrm{\pi m}}_{\mathrm{e}}\mathrm{\beta c}}} \end{gathered}$$