Applicazione formula Laplace

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Ecco qui il problema completo :

Suddivido la spira in 4 lati $AB$ , $BD$ ,$DC$ e $CA$ . $B(O)=B_{AB}(O)+B_{BD}(O)+B_{DC}(O)+B_{CA}(O)$
Noto che $B_{AB}(O)=B_{DC}(O)$ e $B_{BD}(O)=B_{CA}(O)$ quindi basta calcolare $B_{AB}(O)$ e $B_{BD}(O)$

$\vec{B}_{AB}(O)=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{dx \times \vec{r}}{r^3}$ $\qquad$ dove $r=\sqrt(x^2 + (\frac{l}{2})^2)$
${B}_{AB}(O)=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{r\sin\psi dx}{r^3}=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{\sin\psi dx}{r^2}=\frac{\mu_0 i}{4\pi}\int_{\frac{-h}{2}}^{\frac{h}{2}} \frac{\frac{l}{2}dx}{((\frac{h}{2})^2 + x^2)^(\frac{3}{2}})$ $\qquad$ essendo $\sin\psi=\frac{l}{2r}$

$=\frac{\mu_0 il}{8\pi} [ \frac{8x}{l^2 \sqrt(l^2 +4x^2)}]_{\frac{-h}{2}}^{\frac{h}{2}}=\frac{\mu_0 ih}{\pi l \sqrt(l^2 +h^2} $


$\vec{B}_{BD}(O)=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{dy \times \vec{r}}{r^3}$ $\qquad$ dove $r=\sqrt(y^2 + (\frac{h}{2})^2)$
$B_{BD}(O)=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{r\sin\psi dy}{r^3}=\frac{\mu_0 i}{4\pi}\int_{\gamma} \frac{\sin\psi dy}{r^2}=\frac{\mu_0 i}{4\pi}\int_{\frac{-l}{2}}^{\frac{l}{2}} \frac{\frac{h}{2}dy}{((\frac{h}{2})^2 + x^2)^(\frac{3}{2}})$ $\qquad$ essendo $\sin\psi=\frac{h}{2r}$

$=\frac{\mu_0 ih}{8\pi} [ \frac{8y}{h^2 \sqrt(h^2 +4x^2)}]_{\frac{-h}{2}}^{\frac{h}{2}}=\frac{\mu_0 il}{\pi h \sqrt(l^2 +h^2} $

Sommando i vari contributi avrò :

$B(O)=2B_{AB}(O)+2B_{BD}(O)=\frac{2\mu_0 i \sqrt(l^2 + h^2)}{\pi l h}$

In modulo : $\vec{B}(O)=\frac{2\mu_0 i \sqrt(l^2 + h^2)}{\pi l h} \hat{z}$ $\qquad$ normale al piano che contiene la spira , uscente dal "foglio" .