// Numbas version: exam_results_page_options {"name": "FIS5047_S8_Numbas", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "Aplicacion Faraday", ""], "variable_overrides": [[], [], [], []], "questions": [{"name": "Faraday", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Diana Amaro", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4562/"}, {"name": "Ivan Munoz", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18334/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Conectar un galvanómetro $G$ (dispositivo que indica el paso de corriente eléctrica) a una bobina $B$ (hilo conductor enrollado en espiral) e introducir a lo largo de esta bobina se coloca una barra magnetizada $M$. Inmediatamente se mueve la aguja del galvanómetro, comprobándose así que el alambre es atravesado por una corriente eléctrica, aunque en el conjunto no hay batería ni generador de ningún tipo. El simple movimiento de la barra imantada da origen a la corriente eléctrica. Solo hay corriente eléctrica en el alambre cuando la barra se mueve. Si la barra se detiene, la aguja del galvanómetro vuelve inmediatamente a cero.

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De la experiencia descrita en el texto, se concluye que;

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Un campo eléctrico siempre crea un campo magnético.
", "Un campo magnético siempre genera una corriente eléctrica.", "Una corriente eléctrica puede crear un campo magnético", "Una barra magnetizada en movimiento puede generar una corriente eléctrica."], "matrix": [0, 0, 0, "5"], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Circuito 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ivan Munoz", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18334/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Un componente tiene 3 A de corriente que fluye a través de él durante 20 segundos.

", "advice": "

a) La cantidad total de corriente que pasa a través del componente es $Q=I\\cdto t$. Por lo tanto, $Q=3×20=60 C.$

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b) La diferencia de potencial a través del componente es, $V=WQ$. Por lo tanto, $V=\\frac{120}{60}=2 V$.

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c) La resistencia del componente es, $R=\\frac{V}{I}=\\frac{2}{3}=0.67\\Omega$.

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¿Cuál es la cantidad total de carga, en Coulombs(C), que pasa a través del componente durante este tiempo?

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[[0]]C

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El trabajo total realizado al mover todas las cargas que habían pasado en 20 segundos es de 120 J. ¿Cuál es la diferencia de potencial entre los componentes en voltios (V)?

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[[0]]J

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¿Cuál es la resistencia del componente en ohmios Ω? Da tu respuesta con 2 decimales.

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[[0]]$\\Omega$

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Considere el riel de la figura, formado por dos alambres conductores.
La barra de masa $m=${m} kg tiene resistencia $R=${R} $\\Omega$ y desliza libremente.
El campo magnético uniforme $\\vec{B}=${B}T actúa vertical hacia arriba. $L=${L} es la distancia entre los alambres y $\\varepsilon=$ {ep} $V$ es la fem del circuito


Asuma que la barra se mueve con velocidad $v=${v} m/s

", "advice": "

La fem externa producirá una corriente que pasará por la barra que posee corriente $R$.

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Como hay corriente en la barra, el campo magnético externo afectará a la barra. La barra al moverse el flujo magnético a través del loop variara, generando una fem inducida. El flujo a través del loop formado por los rieles y la barra es,
$\\Phi_B=\\int_S \\vec{B}\\cdot d\\vec{S}=BS=BLx.$

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La variación del flujo será
$\\dfrac{d\\Phi_B}{dt}=BL\\dot{x}=BLv(t)$

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donde $v(t)$ es la velocidad de la barra.

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Por ley de Lenz, la fem inducida,
$\\varepsilon_{\\text{ind}}=-BLv$.

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Usando leyes de Kirchhoff al circuito visto desde arriba, escogeremos la corriente en sentido anti-horario, por lo que
$\\varepsilon+\\varepsilon_{\\text{ind}}-IR=0\\Rightarrow \\varepsilon-BLv=IR$

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Por lo tanto,
$I=\\dfrac{\\varepsilon}{R}-\\dfrac{BLv}{R}.$

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Obtenga la corriente que fluye por el circuito

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\n

$I=$[[0]]

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Considere un alambre muy largo. Considere que por el alambre pasa una corriente $I=${II}A

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Consideremos un cable recto por el que pasa una corriente constante $I$. Entonces, de la ley de Ampere, escogeremos como línea Amperiana un circulo de radio $r$ al rededor del cable.

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Entonces, de la ley de Ampere tendremos,

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$\\int B dl=B(r) (2 \\pi r)=\\mu_0 I_{enc}$

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Donde usamos que estamos cerrando la línea amperiana en un círculo por lo que tendremos el perimetro de un círculo ($2 \\pi r$).

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Por lo tanto $B(r)=\\frac{\\mu_0}{2\\pi r} I_{enc}$

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Como la corriente es constante, la corriente encerrada sera $I_{enc}=I$.

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\n

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Obtenga el campo magnético a una distancia $r=${d} m del cable.

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$B=$[[0]] T

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