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z4(~vKR*?_d$M!PoO|d0_;2V32aDw^>@MKde(-I7&!Q!K7_dWQXOgD2*(3gMEBh@N$ z9(Z`w?C83)!1=%{_ie`6F@w6czd)1e?ZG1ryr61Uw5R4g)7k_80-CnOb3ypT8_Vn9gtGu`ARd2OH|0+k{N#t^1;K$L=jcew3L_6pjvkEhN#cM$k8N3$gNqsx_u4 z>0+)T$qef7IJ4^^eRj}`>V|pbYS~QG?euY<#oy0G2a}M=Y!KrHH_Rldh)d3IIDKutkQS@uu$@o6uHYU|8lS_xbrM%Yf)&dQYA{MAVx861j?Ub6TivieW~Q&UebC z;*Cq^K!oilQdO(#{9aw6wSEuWROTz=DtJv+e zfCC6#jgBFF31@h9RjVWC(PFb!hy35^l4uB?tF9s3-l)0zLn>-d%Wci@BI@$+;bay5 zxY`Hg&gB`L8%&0\)) in un mezzo conduttivo in presenza di un'onda elettromagnetica di questo tipo: +\begin{align} + \vec{E}(z,t)&= E_0 \cos (kz- \omega t) \vec{\hat{i}} & + \vec{B}(z,t)&=\frac{E_0}{v} \cos (kz - \omega t) \vec{\hat{j}} +\end{align} +\begin{figure}[H] + \begin{minipage}[t]{0.3\textwidth} + \centering + \includegraphics[width=\linewidth]{screenshots/2024-04-18-09-25-40.png} + \end{minipage}% + \hfill% + \begin{minipage}[t]{0.3\textwidth} + \centering + \includegraphics[width=\linewidth]{screenshots/2024-04-18-09-34-09.png} + \end{minipage} +\end{figure} +Il campo elettrico mette in moto la carica e, essendo in un mezzo conduttivo dove vale \(\vec{E}=\rho _R \vec{J}=\rho _R n q \vec{v}\), otteniamo che \(\vec{v} \propto \vec{E}\). Avendo una velocità verso le x positive, la carica subisce una forza di Lorentz diretta lungo l'asse z positivo per tutta la fase dell'onda in cui la componente del campo elettrico è positiva. Quando il campo elettrico si ribalta (quindi anche la velocità della carica) e va verso il basso, anche il campo magnetico inverte il suo verso e quindi il risultato netto è che la forza di Lorentz è diretta anche in questo caso nel verso di propagazione. Nel caso di una carica negativa non cambia nulla! Complessivamente la forza di Lorentz dipende da \(q ^{2} \), quindi ha sempre lo stesso verso indipendentemente dal segno della carica. + +Consideriamo ora una piastra metallica con superficie \(\Sigma \) e \(N\) cariche libere sulla superficie. Su ogni carica agisce una forza data da \(\vec{F}=q \vec{E} + q \vec{v} \cp \vec{B}\), quindi sull'intera piastra si avrà una forza complessiva pari a \(\vec{F}_{tot} = Nq (\vec{E}+ \vec{v} \cp \vec{B})\). La forza elettrica oscilla e quindi mediamente è nulla, ma quella di natura magnetica ha sempre la stessa direzione. La potenza media fornita dall'onda alla piastra è data da +\begin{equation} + \langle \mathcal{P} \rangle = \langle \vec{F}_{tot} \cdot \vec{v} \rangle = \langle N q \vec{E} \cdot \vec{v} \rangle +\end{equation} +che ci dà la potenza spesa per unità di area: +\begin{equation} + \left\langle \frac{\mathcal{P} }{\Sigma } \right\rangle = + \left\langle \frac{Nq \vec{E}\cdot \vec{v}}{\Sigma } \right\rangle + = \sigma \langle \vec{E}\cdot \vec{v} \rangle +\end{equation} +Se l'onda elettromagnetica viene completamente assorbita dalla piastra, si ha che la potenza spesa per unità di area è uguale all'intensità dell'onda iniziale: +\begin{equation} + I = \left\langle \frac{\mathcal{P} }{\Sigma } \right\rangle + = \sigma \langle \vec{E}\cdot \vec{v} \rangle +\end{equation} + +Abbiamo visto che la forza di Lorentz è diretta sempre nella direzione di propagazione dell'onda. Per onde elettromagnetiche che si incontrano quotidianamente le oscillazioni sono molto rapide, il campo magnetico ha modulo \(\vert \vec{B} \vert = \quotient{\vert \vec{E} \vert }{c} \) e le velocità massime degli elettroni sono significativamente inferiori a quelle della luce. Quindi la forza elettrica è sempre dominante rispetto a quella di Lorentz e si può approssimare la direzione di \(\vec{v}\) con quella di \(\vec{E}\), ottenendo quindi: +\begin{equation} + \langle \vec{F}_{tot} = N \langle \vert q\vec{v} \cp \vec{B} \vert \rangle \rangle + = \frac{Nq \langle \vec{v} \cdot \vec{E} \rangle }{c} +\end{equation} +Da questo si trova che la pressione è +\begin{equation} + p = \left\langle \frac{\vec{F}_{tot} }{\Sigma } \right\rangle = + \frac{\sigma }{c} \langle \vec{E} \cdot \vec{v} \rangle = + \frac{I}{c}= + \frac{\langle \vert \vec{S} \vert \rangle }{c} +\end{equation} + +\paragraph{Nota storica} Misurare la pressione di radiazione è molto difficile perché è necessaria un'intensità molto alta! I primi radiometri dell''800 funzionavano al contrario, perché l'effetto dell'onda era di scaldare una delle due facce della piastra metallica e quindi le molecole di gas spingevano leggermente la piastra avendo velocità leggermente diverse da un lato della piastra e dall'altro. Oggi la pressione di radiazione è utilizzata per i viaggi spaziali con le vele solari. \ No newline at end of file