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9.4: Solenoide Largo

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    131665
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    Coloquemos un solenoide infinitamente largo de\(n\) giros por unidad de longitud para que su eje coincida con el\(z\) -eje de coordenadas, y la corriente\(I\) fluya en el sentido de aumentar\(\phi\). En ese caso, ya sabemos que el campo dentro del solenoide es uniforme y está\(\mu\, n\, I\, \hat{\textbf{z}}\) dentro del solenoide y cero afuera. Dado que el campo solo tiene un\(z\) componente, el potencial vectorial\(\textbf{A}\) puede tener solo un componente\(\phi\) -.

    Supondremos que el radio del solenoide es\(a\). Ahora considere un círculo de radio\(r\) (menor que\(a\)) perpendicular al eje del solenoide (y por lo tanto al campo\(\textbf{B}\)). El flujo magnético a través de este círculo (es decir, la integral de la superficie\(\textbf{B}\) a través del círculo) es\(\pi r^2B = \pi r^2 nI\). Ahora, como todos saben, la integral de superficie de un campo vectorial a través de una curva cerrada es igual a la integral de línea de su rizo alrededor de la curva, y esto es igual a\(2\pi r A_\phi\). Así, dentro del solenoide el potencial del vector es

    \[\textbf{A}=\frac{1}{2}\mu n r I \hat{\boldsymbol{\phi}}.\label{9.4.1}\]

    Se deja al lector argumentar que, fuera del solenoide\((r > a)\), el potencial del vector magnético es

    \[\textbf{A}=\frac{\mu na^2 I}{2r}\hat{\boldsymbol{\phi}}.\]


    This page titled 9.4: Solenoide Largo is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.