{"id":4857,"date":"2020-01-09T14:14:56","date_gmt":"2020-01-09T14:14:56","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4857"},"modified":"2024-09-04T14:04:34","modified_gmt":"2024-09-04T14:04:34","slug":"resolucao-comentada-vestibulares-recentes-eletromagnetismo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletromagnetismo-2018-2019\/resolucao-comentada-vestibulares-recentes-eletromagnetismo\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes \u2013 Eletromagnetismo"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes \u2013 Eletromagnetismo<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span><\/p>\n

Inseparabilidade dos polos magn\u00e9ticos<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Se voc\u00ea\u00a0<\/b><\/span><\/span>quebrar um im\u00e3 (corte transversal)\u00a0<\/b><\/span><\/span>e em seguida\u00a0continuar\u00a0<\/b><\/span>dividindo os im\u00e3s<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

resultantes,\u00a0<\/b><\/span><\/span>voc\u00ea observar\u00e1 que\u00a0<\/b><\/span><\/span>cada peda\u00e7o partido continuar\u00e1 sendo um novo im\u00e3 com dois polos, Norte e Sul de maneira que cada peda\u00e7o atraia o outro.\u00a0N\u00e3o existem polos isolados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span>partir um im\u00e3 conforme a figura abaixo (corte longitudinal),<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>for\u00e7a entre eles ser\u00e1 de repuls\u00e3o.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

02-<\/span><\/span><\/span><\/p>\n

I- Verdadeira<\/b><\/span><\/span><\/span><\/span><\/p>\n

Aurora polar<\/b><\/span><\/span>\u00a0\u2013 Trata-se de um\u00a0<\/b><\/span><\/span>fen\u00f4meno luminoso gerado nas camadas mais elevadas da atmosfera (400 a 800 quil\u00f4metros de altura) e observado com maior freq\u00fc\u00eancia nas regi\u00f5es pr\u00f3ximas aos polos do planeta. No Polo Norte, chama-se aurora boreal; no Sul, austral.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Essas auroras ocorrem quando part\u00edculas el\u00e9tricas (f\u00f3tons e el\u00e9trons) provenientes do Sol chegam \u00e0s vizinhan\u00e7as da Terra e s\u00e3o atra\u00eddas por seu campo magn\u00e9tico.\u00a0<\/b><\/span><\/span>Ao\u00a0<\/b><\/span><\/span>alcan\u00e7arem a atmosfera, essas part\u00edculas se chocam com os \u00e1tomos de oxig\u00eanio e nitrog\u00eanio \u2013 num processo semelhante \u00e0 ioniza\u00e7\u00e3o (eletrifica\u00e7\u00e3o) de gases que faz acender o tubo de uma l\u00e2mpada fluorescente<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

II- Verdadeira<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>Terra se comporta como um grande im\u00e3\u00a0<\/b><\/span><\/span>onde o\u00a0<\/b><\/span><\/span>polo Sul magn\u00e9tico est\u00e1 aproximadamente localizado no polo Norte geogr\u00e1fico e vice versa.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span>pendurar um im\u00e3 em forma de barra pelo seu centro\u00a0<\/b><\/span><\/span>ou\u00a0<\/b><\/span><\/span>observar a agulha magn\u00e9tica<\/b><\/span><\/span>\u00a0de uma\u00a0<\/b><\/span><\/span>b\u00fassola voc\u00ea ver\u00e1 que seus polos ficam sempre alinhados na dire\u00e7\u00e3o norte-sul.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

O\u00a0<\/b><\/span><\/span>polo que indicar o polo norte geogr\u00e1fico\u00a0<\/b><\/span><\/span>recebe o nome de\u00a0<\/b><\/span><\/span>polo norte e estar\u00e1 indicando\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>polo sul magn\u00e9tico da Terra.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>polo que indicar o polo sul geogr\u00e1fico\u00a0<\/b><\/span><\/span>recebe o nome de\u00a0<\/b><\/span><\/span>polo sul\u00a0<\/b><\/span><\/span>e estar\u00e1\u00a0<\/b><\/span><\/span>indicando o polo norte magn\u00e9tico da Terra.\u00a0<\/b><\/span><\/span>Tudo isso ocorre\u00a0<\/b><\/span><\/span>porque polos de nomes opostos se atraem<\/b><\/span><\/span><\/span><\/span><\/p>\n

III- Falsa<\/b><\/span><\/span><\/span><\/span><\/p>\n

Analise\u00a0<\/b><\/span><\/span>na figura abaixo as\u00a0<\/b><\/span><\/span>linhas de indu\u00e7\u00e3o do campo magn\u00e9tico terrestre\u00a0<\/b><\/span><\/span>e observe que a\u00a0<\/b><\/span><\/span>agulha magn\u00e9tica de uma b\u00fassola (indica\u00e7\u00e3o do campo magn\u00e9tico terrestre) tem a propriedade de se alinhar de acordo com as linhas do campo geomagn\u00e9tico.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe na figura acima que\u00a0<\/b><\/span><\/span>o campo magn\u00e9tico terrestre, em cada ponto \u00e9 tangente \u00e0s linhas de indu\u00e7\u00e3o, variando em dire\u00e7\u00e3o e sentido e aumentando em m\u00f3dulo \u00e0 medida que se aproxima dos polos.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

03-<\/span><\/span><\/span><\/p>\n

As linhas de campo magn\u00e9tico comportam-se como esbo\u00e7ado na figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

As\u00a0<\/b><\/span><\/span>linhas de campo magn\u00e9tico partem do polo norte em dire\u00e7\u00e3o ao polo sul, exceto dentro do \u00edm\u00e3, em que elas partem do polo sul para o norte, como \u00e9 poss\u00edvel observar pela figura<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observando a figura do enunciado<\/b><\/span><\/span>, temos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Em que os vetores B representam o campo de magn\u00e9tico, e suas orienta\u00e7\u00f5es dependem da posi\u00e7\u00e3o do \u00edm\u00e3<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sabendo que se o polo mais pr\u00f3ximo for o norte, a linha de campo deve estar saindo, e se o sul for o polo mais pr\u00f3ximo, a linha de campo de deve estar entrando no seu respectivo \u00edm\u00e3.<\/b><\/span><\/span>\u00a0Para encontrar a resultante, devemos fazer a\u00a0<\/b><\/span><\/span>soma vetorial<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Na dire\u00e7\u00e3o de\u00a0<\/b><\/span><\/span>, logo a alternativa correta \u00e9 D.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

04-<\/span><\/span><\/span><\/p>\n

Veja abaixo um resumo da teoria:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas do Vetor Campo El\u00e9trico<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

For\u00e7a sobre q devido ao campo el\u00e9trico\u00a0<\/b><\/span><\/span>\u00a0E = F\/q\u00a0\u00a0<\/b><\/span><\/span>F = q.E (I)<\/b><\/span><\/span><\/span><\/span><\/p>\n

For\u00e7a sobre q devido ao campo magn\u00e9tico\u00a0\u00a0F = q.V.B (II)<\/b><\/span><\/span><\/span><\/span><\/p>\n

(I) = (II)\u00a0<\/b><\/span><\/span>\u00a0q.E = q.V.B\u00a0\u00a0<\/b><\/span><\/span>V = E\/B.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

05-<\/span><\/span><\/span><\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

S\u00e3o dados: q\u00a0<\/b><\/span><\/span>= 3,2\u00b5C\u00a0<\/b><\/span><\/span>= 3,2.10-6\u00a0C<\/b><\/span><\/span>;\u00a0<\/b><\/span><\/span>V = 2.104\u00a0m\/s<\/b><\/span><\/span>;\u00a0<\/b><\/span><\/span>F = 1,6.102\u00a0N\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>B = ?<\/b><\/span><\/span><\/span><\/span><\/p>\n

Fm\u00a0= q.V.B\u00a0<\/b><\/span><\/span>\u00a01,6.102\u00a0= 3,2.10-6.2.104.B\u00a0\u00a01,6.102\u00a0= 6,4.10-2.B\u00a0\u00a0B = 1,6.102\/6,4.10-2\u00a0\u00a0<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span>= 0,25.104\u00a0=\u00a0<\/b><\/span><\/span>2,5.103\u00a0T.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

06-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe que,\u00a0<\/b><\/span><\/span>neste caso<\/b><\/span><\/span>\u00a0o \u00e2ngulo entre\u00a0\u00a0e\u00a0\u00a0\u00e9\u00a0<\/b><\/span><\/span>90o\u00a0(s\u00e3o perpendiculares) e que sen90o\u00a0= 1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

O<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>per\u00edodo T<\/b><\/span><\/span>\u00a0(tempo que a carga q demora para efetuar uma volta completa) \u00e9\u00a0<\/b><\/span><\/span>fornecido por<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>V = \u0394S \/ \u0394t<\/b><\/span><\/span>\u00a0\u00a0\u00a0 numa\u00a0<\/b><\/span><\/span>volta completa<\/b><\/span><\/span>\u00a0\u00a0\u00a0 \u0394S = 2\u03c0R e \u0394t = T\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>V = 2\u03c0R \/ T<\/b><\/span><\/span>\u00a0(II).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Substituindo<\/b><\/span><\/span>\u00a0II\u00a0<\/b><\/span><\/span>em<\/b><\/span><\/span>\u00a0I\u00a0\u00a0\u00a0 R = m. (2\u03c0R\/T)\/q.B\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>T = 2\u03c0m\/q.B\u00a0\u00a0(per\u00edodo do MCU).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe que o\u00a0<\/b><\/span><\/span>per\u00edodo (T) do movimento circular<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>n\u00e3o depende da velocidade<\/b><\/span><\/span>\u00a0com que a\u00a0<\/b><\/span><\/span>part\u00edcula q penetra no campo magn\u00e9tico e nem do raio da circunfer\u00eancia.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

01.Falso<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013 s\u00e3o\u00a0<\/b><\/span><\/span>perpendiculares com \u00e2ngulo sempre de 90o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(veja teoria acima).<\/b><\/span><\/span><\/span><\/span><\/p>\n

02 \u2013 Correto<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2013\u00a0<\/b><\/span><\/span>Fm\u00a0= q.v.B\u00a0<\/b><\/span><\/span>= 4.10-6.2.102.7.10-2\u00a0= 56.10-6<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 5,6.10-5\u00a0N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

04. Correto \u2013\u00a0veja teoria acima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

08- Correto \u2013<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>pelo\u00a0<\/b><\/span><\/span>enuncia a massa m e a e o m\u00f3dulo da velocidade v<\/b><\/span><\/span>\u00a0do feixe de part\u00edculas \u00e9\u00a0<\/b><\/span><\/span>constante\u00a0<\/b><\/span><\/span>e pelas\u00a0<\/b><\/span><\/span>express\u00f5es<\/b><\/span><\/span>\u00a0da\u00a0<\/b><\/span><\/span>energia cin\u00e9tica Ec\u00a0= m.v2\/2 e da quantidade de movimento Q = m.v voc\u00ea verifica que Ec\u00a0e Q tamb\u00e9m s\u00e3o constantes.<\/b><\/span><\/span><\/span><\/span><\/p>\n

16. Correto<\/b><\/span><\/span>\u00a0\u2013 veja na express\u00e3o\u00a0<\/b><\/span><\/span>R = q.B\/m.v que o raio R \u00e9 fun\u00e7\u00e3o da massa m, sendo B e v constantes.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 ( 02 + 04 + 08<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>+ 16) = 30<\/b><\/span><\/span><\/span><\/span><\/p>\n

07-<\/span><\/span><\/span><\/p>\n

Breve resumo da teoria<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Regra da m\u00e3o esquerda<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Aplicando a\u00a0<\/b><\/span><\/span>regra da m\u00e3o esquerda no ponto onde o el\u00e9tron\u00a0<\/b><\/span><\/span>(carga negativa)\u00a0<\/b><\/span><\/span>est\u00e1 penetrando na regi\u00e3o de campo magn\u00e9tico<\/b><\/span><\/span>\u00a0voc\u00ea verificar\u00e1 que o\u00a0<\/b><\/span><\/span>campo magn\u00e9tico B est\u00e1 saindo da folha<\/b><\/span><\/span>, mas como\u00a0<\/b><\/span><\/span>\u00e9 carga negativa estar\u00e1 entrando.\u00a0<\/b><\/span><\/span>(lembre-se de que em cada ponto Vo\u00a0\u00e9 tangente \u00e0 trajet\u00f3ria e F \u00e9 dirigida para o centro da circunfer\u00eancia). \u201cveja figura abaixo\u201d.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

08-<\/span><\/span><\/span><\/p>\n

For\u00e7a magn\u00e9tica sobre carga el\u00e9trica<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

09-<\/span><\/span><\/span><\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe que,\u00a0<\/b><\/span><\/span>neste caso<\/b><\/span><\/span>\u00a0o \u00e2ngulo entre\u00a0\u00a0e\u00a0\u00a0\u00e9\u00a0<\/b><\/span><\/span>90o\u00a0(s\u00e3o perpendiculares) e que sen90o\u00a0= 1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

10-<\/span><\/span><\/span><\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/span><\/p>\n

11-<\/span><\/span><\/span><\/p>\n

For\u00e7a magn\u00e9tica sobre uma carga m\u00f3vel imersa num campo magn\u00e9tico<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

12-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Veja abaixo as duas condi\u00e7\u00f5es para que a for\u00e7a magn\u00e9tica sobre a carga seja nula.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

13-<\/span><\/span><\/span><\/p>\n

Nesse caso precisamos lembrar de uma f\u00f3rmula importante:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Isolando a velocidade:<\/b><\/span><\/span><\/span><\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>a part\u00edcula se movimenta em linha reta\u00a0<\/b><\/span><\/span><\/span>quer dizer que\u00a0<\/b><\/span><\/span><\/span>as for\u00e7as el\u00e9tricas e magn\u00e9ticas se anulam,<\/b><\/span><\/span><\/span>\u00a0ou seja, elas\u00a0<\/b><\/span><\/span><\/span>s\u00e3o iguais em intensidade, por\u00e9m possuem sentidos contr\u00e1rios<\/b><\/span><\/span><\/span>. Essa informa\u00e7\u00e3o \u00e9 fundamental, pois:<\/b><\/span><\/span><\/span><\/p>\n

Fm\u00a0= Fe<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

04 \u2013\u00a0Bom, sabemos que\u00a0<\/b><\/span><\/span><\/span>a cada 30 anos o C\u00e9sio perde metade de sua massa inicial.\u00a0<\/b><\/span><\/span><\/span>2077 \u00e9\u00a0<\/b><\/span><\/span><\/span>60 anos depois de 2017,\u00a0<\/b><\/span><\/span><\/span>sendo que\u00a0<\/b><\/span><\/span><\/span>em 2017 j\u00e1 se passaram 30 anos,\u00a0<\/b><\/span><\/span><\/span>ent\u00e3o a quantidade total de anos\u00a0<\/b><\/span><\/span><\/span>\u00e9 30 + 60 = 90 anos.\u00a0<\/b><\/span><\/span><\/span>Se a cada 30 anos se vai uma meia vida de C\u00e9sio, ent\u00e3o\u00a0<\/b><\/span><\/span><\/span>em 90 anos se v\u00e3o 3 meias vidas.<\/b><\/span><\/span><\/span>\u00a0Isso equivale \u00e0\u00a0<\/b><\/span><\/span><\/span>metade da metade da metade da massa inicial,<\/b><\/span><\/span><\/span>\u00a0ou:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O que equivale \u00e0 1\/8 da massa inicial, alternativa C.<\/b><\/span><\/span><\/span><\/p>\n

14-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

15-<\/span><\/span><\/span><\/p>\n

Campo magn\u00e9tico originado por um condutor retil\u00edneo extenso percorrido por corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Dire\u00e7\u00e3o e sentido do vetor campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Utilizando a regra da m\u00e3o direita em cada fio voc\u00ea verificar\u00e1 que o campo magn\u00e9tico resultante estar\u00e1 perpendicular ao plano determinado pelos fios, entrando nele.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

16-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a de intera\u00e7\u00e3o entre dois fios condutores paralelos percorridos por corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

Considere\u00a0<\/b><\/span><\/span><\/span>dois condutores retil\u00edneos 1 e 2\u00a0<\/b><\/span><\/span><\/span>percorridos,\u00a0<\/b><\/span><\/span><\/span>respectivamente por correntes el\u00e9tricas i1\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>i2,\u00a0e separados por uma dist\u00e2ncia\u00a0d.\u00a0<\/b><\/span><\/span><\/span>Tem-se\u00a0<\/b><\/span><\/span><\/span>duas situa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>As correntes el\u00e9tricas tem mesmo sentido<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Utilizando a\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita, voc\u00ea determina\u00a0B1\u00a0<\/b><\/span><\/span><\/span>que \u00e9 o\u00a0<\/b><\/span><\/span><\/span>vetor indu\u00e7\u00e3o magn\u00e9tica que o condutor 1 produz onde est\u00e1 o condutor 2.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>condutor 2, imerso no campo magn\u00e9tico de\u00a0B1\u00a0<\/b><\/span><\/span><\/span>ficar\u00e1 sujeito a uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica\u00a0Fm2, fornecida pela regra da m\u00e3o esquerda.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Assim, o\u00a0<\/b><\/span><\/span><\/span>condutor 2\u00a0<\/b><\/span><\/span><\/span>fica sujeito a uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica\u00a0Fm2\u00a0vertical e para cima.<\/b><\/span><\/span><\/span><\/p>\n

Analogamente<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>condutor 2 origina, onde est\u00e1 o condutor 1, um campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span>B2 \u00a0que, fornecido pela\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita estar\u00e1 saindo da folha de papel.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o surgir\u00e1 sobre o\u00a0<\/b><\/span><\/span><\/span>condutor 1 uma for\u00e7a vertical e para baixo, fornecida pela regra da m\u00e3o esquerda.<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o final:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

\u00a0\u00a0Se as\u00a0<\/b><\/span><\/span><\/span>correntes el\u00e9tricas tiverem sentidos contr\u00e1rios,\u00a0<\/b><\/span><\/span><\/span>procedendo da\u00a0<\/b><\/span><\/span><\/span>mesma forma que no item anterior, voc\u00ea observar\u00e1 que os dois fios sofrem for\u00e7a de repuls\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>caso do exerc\u00edcio<\/b><\/span><\/span><\/span>, como as\u00a0<\/b><\/span><\/span><\/span>correntes possuem mesmo sentido\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica entre eles ser\u00e1 de<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

atra\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

17-<\/span><\/span><\/span><\/p>\n

Campo magn\u00e9tico originado por um condutor retil\u00edneo extenso percorrido por corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Dire\u00e7\u00e3o e sentido do vetor campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

\u00a0Intensidade do campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

18-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

19-<\/span><\/span><\/span><\/p>\n

For\u00e7a de intera\u00e7\u00e3o entre dois fios condutores paralelos percorridos por corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

Considere\u00a0<\/b><\/span><\/span><\/span>dois condutores retil\u00edneos 1 e 2\u00a0<\/b><\/span><\/span><\/span>percorridos,\u00a0<\/b><\/span><\/span><\/span>respectivamente por correntes el\u00e9tricas i1\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>i2,\u00a0e separados por uma dist\u00e2ncia\u00a0d.\u00a0<\/b><\/span><\/span><\/span>Tem-se\u00a0<\/b><\/span><\/span><\/span>duas situa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>As correntes el\u00e9tricas tem mesmo sentido<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Utilizando a\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita, voc\u00ea determina\u00a0B1\u00a0<\/b><\/span><\/span><\/span>que \u00e9 o\u00a0<\/b><\/span><\/span><\/span>vetor indu\u00e7\u00e3o magn\u00e9tica que o condutor 1 produz onde est\u00e1 o condutor 2.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>condutor 2, imerso no campo magn\u00e9tico de\u00a0B1\u00a0<\/b><\/span><\/span><\/span>ficar\u00e1 sujeito a uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica\u00a0Fm2, fornecida pela regra da m\u00e3o esquerda.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Assim, o\u00a0<\/b><\/span><\/span><\/span>condutor 2\u00a0<\/b><\/span><\/span><\/span>fica sujeito a uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica\u00a0Fm2\u00a0vertical e para cima.<\/b><\/span><\/span><\/span><\/p>\n

Analogamente<\/b><\/span><\/span><\/span>, o\u00a0<\/b><\/span><\/span><\/span>condutor 2 origina, onde est\u00e1 o condutor 1, um campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span>B2\u00a0\u00a0que, fornecido pela\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita estar\u00e1 saindo da folha de papel.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o surgir\u00e1 sobre o\u00a0<\/b><\/span><\/span><\/span>condutor 1 uma for\u00e7a vertical e para baixo, fornecida pela regra da m\u00e3o esquerda.<\/b><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o final:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

\u00a0\u00a0Se as\u00a0<\/b><\/span><\/span><\/span>correntes el\u00e9tricas tiverem sentidos contr\u00e1rios,\u00a0<\/b><\/span><\/span><\/span>procedendo da\u00a0<\/b><\/span><\/span><\/span>mesma forma que no item anterior, voc\u00ea observar\u00e1 que os dois fios sofrem for\u00e7a de repuls\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>no exerc\u00edcio\u00a0<\/b><\/span><\/span><\/span>as\u00a0<\/b><\/span><\/span><\/span>correntes el\u00e9tricas\u00a0<\/b><\/span><\/span><\/span>possuem\u00a0<\/b><\/span><\/span><\/span>sentidos opostos\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a entre os fios\u00a0<\/b><\/span><\/span><\/span>ser\u00e1 de\u00a0<\/b><\/span><\/span><\/span>repuls\u00e3o (intera\u00e7\u00e3o repulsiva).<\/b><\/span><\/span><\/span><\/p>\n

Intensidade dessa for\u00e7a, de atra\u00e7\u00e3o ou de repuls\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o:\u00a0<\/b><\/span><\/span><\/span>Essas\u00a0<\/b><\/span><\/span><\/span>for\u00e7as<\/b><\/span><\/span><\/span>\u00a0possuem a\u00a0<\/b><\/span><\/span><\/span>mesma intensidade<\/b><\/span><\/span><\/span>\u00a0(m\u00f3dulo), pois\u00a0<\/b><\/span><\/span><\/span>constituem par a\u00e7\u00e3o e<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>rea\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>(mesma intensidade, mesma dire\u00e7\u00e3o mas sentidos contr\u00e1rios).<\/b><\/span><\/span><\/span><\/p>\n

20-<\/span><\/span><\/span><\/p>\n

Embaixo est\u00e1 um resumo.\u00a0Se estiver interessado, a teoria completa encontra-se em fisicaevestibular.com.br \u2013 Eletricidade \u2013 Eletromagnetismo \u2013\u00a0Campo magn\u00e9tico originado por um condutor retil\u00edneo extenso percorrido por corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

21-<\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>leitor da etiqueta emite um campo de radiofrequ\u00eancia<\/b><\/span><\/span>\u00a0(por meio de ondas magn\u00e9ticas de r\u00e1dio)\u00a0<\/b><\/span><\/span>\u00e0 etiqueta eletr\u00f4nica identificadora (tags) que retornam esses sinais de radiofrequ\u00eancia para os leitores.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Os leitores<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>captam essas informa\u00e7\u00f5es quando o fluxo magn\u00e9tico vari\u00e1vel produzido no retorno dessas ondas faz surgir na bobina (espira) do leitor uma corrente el\u00e9trica induzida.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Pela\u00a0<\/b><\/span><\/span>Lei de Faraday essa corrente el\u00e9trica origina na bobina um campo magn\u00e9tico tamb\u00e9m vari\u00e1vel que \u00e9 detectado pelo leitor que registra essas informa\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

22-<\/span><\/span><\/span><\/p>\n

Primeiro usamos a f\u00f3rmula para calcular o\u00a0<\/b><\/span><\/span>fluxo \u03a6B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de um\u00a0<\/b><\/span><\/span>campo magn\u00e9tico uniforme B<\/b><\/span><\/span>\u00a0passando por uma\u00a0<\/b><\/span><\/span>\u00e1rea A<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

A \u00e1rea mencionada acima \u00e9 a\u00a0<\/b><\/span><\/span>\u00e1rea do ret\u00e2ngulo por onde passa o campo magn\u00e9tico<\/b><\/span><\/span>, como apenas\u00a0<\/b><\/span><\/span>uma parte\u00a0<\/b><\/span><\/span>da espira est\u00e1 sobre o campo, dizemos que a \u00e1rea \u00e9 ent\u00e3o o produto do\u00a0<\/b><\/span><\/span>lado b<\/b><\/span><\/span>\u00a0por um\u00a0<\/b><\/span><\/span>comprimento x que depende do tempo, pois conforme a espira se move para dentro do campo, esse x aumenta<\/b><\/span><\/span>, assim a rela\u00e7\u00e3o acima se torna:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

 <\/p>\n

23-<\/span><\/span><\/span><\/p>\n

Observando a imagem:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

De acordo com a\u00a0<\/b><\/span><\/span>Lei de Faraday\u00a0<\/b><\/span><\/span>temos que:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Portanto a for\u00e7a eletromotriz induzida ir\u00e1\u00a0<\/b><\/span><\/span>depender da varia\u00e7\u00e3o do fluxo magn\u00e9tico<\/b><\/span><\/span>, precisamos encontra-lo. O fluxo magn\u00e9tico j\u00e1 depende de:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

No caso de\u00a0<\/b><\/span><\/span>um solenoide\u00a0<\/b><\/span><\/span>o seu campo magn\u00e9tico \u00e9:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como a corrente varia com o tempo pela equa\u00e7\u00e3o i = Kt. Observe tamb\u00e9m que o enunciado informa que a espira tem N voltas por unidade de comprimento, ou seja, n = N\/L voltas\/unidade de comprimento:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como sabemos a fem induzida, s\u00f3 nos resta saber a intensidade da corrente,\u00a0<\/b><\/span><\/span>podemos usar a lei de Ohm<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Alterativa correta \u00e9 a B.<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

24-<\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz induzida pelo movimento<\/b><\/span><\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz gerada por um condutor m\u00f3vel num campo el\u00e9trico uniforme<\/b><\/span><\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz\u00a0\u03b5<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>gerada por um\u00a0<\/b><\/span><\/span>condutor m\u00f3vel imerso num campo el\u00e9trico uniforme\u00a0\u00a0<\/b><\/span><\/span>\u00a0 esse\u00a0<\/b><\/span><\/span>condutor tem o comportamento de um gerador mec\u00e2nico de eletricidade de fem\u00a0\u03b5.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

O sentido da corrente el\u00e9trica induzida<\/b><\/span><\/span>\u00a0\u00e9 fornecido pela\u00a0<\/b><\/span><\/span>lei de Lenz\u00a0\u201ca for\u00e7a eletromotriz induzida e a corrente induzida geram um fluxo magn\u00e9tico que se op\u00f5e \u00e0 varia\u00e7\u00e3o do fluxo causador da indu\u00e7\u00e3o\u201d<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

I. Verdadeira \u2013<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Ao\u00a0<\/b><\/span><\/span>deslocarmos o circuito e o \u00edm\u00e3 juntos em sentidos contr\u00e1rios,<\/b><\/span><\/span>\u00a0ao\u00a0<\/b><\/span><\/span>longo da dire\u00e7\u00e3o y surge corrente no circ<\/b><\/span><\/span>uito porque est\u00e1\u00a0<\/b><\/span><\/span>ocorrendo a varia\u00e7\u00e3o (no caso diminui\u00e7\u00e3o) da \u00e1rea da espira.<\/b><\/span><\/span><\/span><\/span><\/p>\n

II. Verdadeira \u2013\u00a0est\u00e1 tamb\u00e9m variando a \u00e1rea da espira.<\/b><\/span><\/span><\/span><\/span><\/p>\n

III. Verdadeira \u2013\u00a0veja teoria acima.<\/b><\/span><\/span><\/span><\/span><\/p>\n

IV. Falsa \u2013\u00a0n\u00e3o ocorre varia\u00e7\u00e3o da \u00e1rea.<\/b><\/span><\/span><\/span><\/span><\/p>\n

V. Fasa \u2013\u00a0veja IV.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

25-<\/span><\/span><\/span><\/p>\n

Corrente el\u00e9trica induzida<\/b><\/span><\/span><\/span><\/span><\/p>\n

Para que voc\u00ea entenda o\u00a0<\/b><\/span><\/span>fen\u00f4meno da indu\u00e7\u00e3o eletromagn\u00e9tica<\/b><\/span><\/span>, considere uma\u00a0<\/b><\/span><\/span>\u00fanica espira\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

um\u00a0<\/b><\/span><\/span>im\u00e3 permanente.\u00a0<\/b><\/span><\/span>Quando voc\u00ea\u00a0<\/b><\/span><\/span>aproxima o im\u00e3 da espira,<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>n\u00famero de linhas de indu\u00e7\u00e3o do im\u00e3 que penetram na espira (fluxo magn\u00e9tico) aumenta<\/b><\/span><\/span>, fazendo\u00a0<\/b><\/span><\/span>surgir na espira uma corrente el\u00e9trica induzida num determinado sentido.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Quando voc\u00ea\u00a0<\/b><\/span><\/span>afasta o im\u00e3 da espira,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>n\u00famero de linhas de indu\u00e7\u00e3o do im\u00e3 (fluxo magn\u00e9tico)\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>penetram na espira diminui,\u00a0<\/b><\/span><\/span>fazendo\u00a0<\/b><\/span><\/span>surgir na espira uma corrente el\u00e9trica induzida com sentido oposto ao anterior.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

\u00a0Com o\u00a0<\/b><\/span><\/span>im\u00e3 im\u00f3vel,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span>n\u00famero de linhas de indu\u00e7\u00e3o que penetram na espira (fluxo magn\u00e9tico) n\u00e3o muda, n\u00e3o existindo corrente el\u00e9trica induzida.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Voc\u00ea obteria o\u00a0<\/b><\/span><\/span>mesmo resultado acima, mantendo o im\u00e3 fixo e movendo a espira\u00a0<\/b><\/span><\/span>ou\u00a0<\/b><\/span><\/span>movendo os dois,\u00a0<\/b><\/span><\/span>pois\u00a0<\/b><\/span><\/span>n\u00e3o importa quem se movimente,<\/b><\/span><\/span>\u00a0mas que\u00a0<\/b><\/span><\/span>haja varia\u00e7\u00e3o de fluxo magn\u00e9tico atrav\u00e9s da espira<\/b><\/span><\/span>, o que\u00a0<\/b><\/span><\/span>s\u00f3 ocorre se a velocidade relativa entre eles for diferente de zero.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>mesmo ser\u00e1 v\u00e1lido\u00a0<\/b><\/span><\/span>se,\u00a0<\/b><\/span><\/span>em vez\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>uma espira voc\u00ea tivesse um solenoide ou bobina\u00a0<\/b><\/span><\/span>(v\u00e1rias espiras circulares de mesmo raio,\u00a0<\/b><\/span><\/span>conc\u00eantricas e interligadas<\/b><\/span><\/span>)<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

R- E<\/b><\/span><\/span><\/span><\/span><\/p>\n

26-<\/span><\/span><\/span><\/p>\n

Se, ao\u00a0<\/b><\/span><\/span><\/span>fazer<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>gerador funcionar,<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>estudante obteve uma tens\u00e3o m\u00e1xima V\u00a0<\/b><\/span><\/span><\/span>e uma\u00a0<\/b><\/span><\/span><\/span>corrente de curto-circuito i, ent\u00e3o a tens\u00e3o m\u00e1xima U \u00e9 a for\u00e7a eletromotriz induzida\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>do gerador, ou seja, V =\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

Para\u00a0<\/b><\/span><\/span><\/span>n espiras a for\u00e7a eletromotriz induzida\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecida por\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>n.\u00a0= n.\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

Para\u00a0<\/b><\/span><\/span><\/span>variar o fluxo magn\u00e9tico no interior da associa\u00e7\u00e3o de espiras, podemos mudar\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>valor do campo magn\u00e9tico (B), da \u00e1rea (S), do \u00e2ngulo de giro () ou o n\u00famero de espiras (n).<\/b><\/span><\/span><\/span><\/p>\n

Mas o\u00a0<\/b><\/span><\/span><\/span>enunciado pede tamb\u00e9m que, al\u00e9m de dobrar\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>deve-se\u00a0<\/b><\/span><\/span><\/span>manter constante\u00a0a corrente de\u00a0curto circuito\u00a0icc<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>que \u00e9 fornecida por\u00a0<\/b><\/span><\/span><\/span>icc\u00a0=\u00a0\u00a0,\u00a0<\/b><\/span><\/span><\/span>onde\u00a0<\/b><\/span><\/span><\/span>r \u00e9 a resist\u00eancia interna devida \u00e0s n espiras<\/b><\/span><\/span><\/span>, o que\u00a0<\/b><\/span><\/span><\/span>s\u00f3 pode acontecer se voc\u00ea dobrar r.<\/b><\/span><\/span><\/span><\/p>\n

Mas, pela\u00a0<\/b><\/span><\/span><\/span>segunda lei de Ohm a resist\u00eancia r\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecida por:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como o\u00a0<\/b><\/span><\/span><\/span>exerc\u00edcio<\/b><\/span><\/span><\/span>\u00a0pede\u00a0<\/b><\/span><\/span><\/span>o dobro da d.d.p. induzida, sem mudar a corrente de curto-circuito icc,\u00a0<\/b><\/span><\/span><\/span>al\u00e9m de\u00a0<\/b><\/span><\/span><\/span>dobrarmos o fluxo\u00a0,\u00a0<\/b><\/span><\/span><\/span>devemos\u00a0<\/b><\/span><\/span><\/span>dobrar tamb\u00e9m a resist\u00eancia r do material.\u00a0<\/b><\/span><\/span><\/span>Isso pode acontecer\u00a0<\/b><\/span><\/span><\/span>atrav\u00e9s da mudan\u00e7a do material\u00a0, do comprimento L ou da \u00e1rea S de se\u00e7\u00e3o do fio.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>\u00fanica op\u00e7\u00e3o que satisfaz\u00a0<\/b><\/span><\/span><\/span>as\u00a0<\/b><\/span><\/span><\/span>duas condi\u00e7\u00f5es ao mesmo tempo \u00e9 dobrar o n\u00famero n de espiras,\u00a0<\/b><\/span><\/span><\/span>o que\u00a0<\/b><\/span><\/span><\/span>dobra a d.d.p. e a resist\u00eancia do material ao mesmo tempo,\u00a0<\/b><\/span><\/span><\/span>e que mant\u00e9m\u00a0<\/b><\/span><\/span><\/span>a corrente de curto inalterada.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

27-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria pode encontra-la em fisicaevestibular.com.br \u2013 Eletricidade \u2013 Eletromagnetismo -for\u00e7a eletromotriz induzida.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Com\u00a0<\/b><\/span><\/span><\/span>x variando entre 0 e L (x < 0 < L),<\/b><\/span><\/span><\/span>\u00a0espira\u00a0<\/b><\/span><\/span><\/span>entrando na regi\u00e3o do campo\u00a0<\/b><\/span><\/span><\/span>at\u00e9 ficar totalmente imersa, ent\u00e3o a varia\u00e7\u00e3o de\u00a0<\/b><\/span><\/span><\/span>fluxo magn\u00e9tico no interior da espira est\u00e1 aumentando.<\/b><\/span><\/span><\/span><\/p>\n

Nesse caso,\u00a0<\/b><\/span><\/span><\/span>pela lei de Lenz,\u00a0<\/b><\/span><\/span><\/span>deve surgir um\u00a0<\/b><\/span><\/span><\/span>fluxo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>se oponha a esse aumento de fluxo,\u00a0<\/b><\/span><\/span><\/span>ou seja,\u00a0<\/b><\/span><\/span><\/span>deve surgir um fluxo magn\u00e9tico que esteja\u00a0saindo\u00a0da folha.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Com\u00a0<\/b><\/span><\/span><\/span>saindo da folha,\u00a0<\/b><\/span><\/span><\/span>usando a\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita, o sentido da corrente el\u00e9trica induzida est\u00e1 indicado nas figuras acima (sentido anti hor\u00e1rio).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Enquanto a\u00a0<\/b><\/span><\/span><\/span>espira estiver totalmente imersa<\/b><\/span><\/span><\/span>\u00a0no interior da regi\u00e3o do\u00a0<\/b><\/span><\/span><\/span>campo magn\u00e9tico (x < L< 3L)<\/b><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

a\u00a0<\/b><\/span><\/span><\/span>corrente el\u00e9trica induzida \u00e9 nula<\/b><\/span><\/span><\/span>\u00a0(n\u00e3o ocorre varia\u00e7\u00e3o de fluxo magn\u00e9tico no interior da espira).<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Com\u00a0<\/b><\/span><\/span><\/span>x variando entre 0 e L (3L < x < 4L),<\/b><\/span><\/span><\/span>\u00a0espira\u00a0<\/b><\/span><\/span><\/span>saindo regi\u00e3o do campo\u00a0<\/b><\/span><\/span><\/span>at\u00e9 ficar totalmente fora dessa regi\u00e3o, ent\u00e3o a varia\u00e7\u00e3o de\u00a0<\/b><\/span><\/span><\/span>fluxo magn\u00e9tico no interior da espira est\u00e1 diminuindo.<\/b><\/span><\/span><\/span><\/p>\n

Nesse caso, pela\u00a0<\/b><\/span><\/span><\/span>lei de Lenz,\u00a0<\/b><\/span><\/span><\/span>deve surgir um\u00a0<\/b><\/span><\/span><\/span>fluxo magn\u00e9tico que se oponha a essa diminui\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>de\u00a0<\/b><\/span><\/span><\/span>fluxo<\/b><\/span><\/span><\/span>, ou seja, deve\u00a0<\/b><\/span><\/span><\/span>surgir um fluxo magn\u00e9tico que esteja\u00a0entrando\u00a0na folha.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Com\u00a0\u00a0<\/b><\/span><\/span><\/span>entrando da folha<\/b><\/span><\/span><\/span>, usando a\u00a0<\/b><\/span><\/span><\/span>regra da m\u00e3o direita, o sentido da corrente el\u00e9trica induzida est\u00e1 indicado nas figuras acima (sentido hor\u00e1rio).<\/b><\/span><\/span><\/span><\/p>\n

R- C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

28-<\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz induzida pelo movimento<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz gerada por um condutor m\u00f3vel num campo el\u00e9trico uniforme<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a eletromotriz\u00a0\u03b5<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>gerada por um\u00a0<\/b><\/span><\/span><\/span>condutor m\u00f3vel imerso num campo el\u00e9trico uniforme\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0 esse\u00a0<\/b><\/span><\/span><\/span>condutor tem o comportamento de um gerador mec\u00e2nico de eletricidade de fem\u00a0\u03b5.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

S\u00e3o fornecidos:\u00a0<\/b><\/span><\/span><\/span>m\u00f3dulo do campo magn\u00e9tico\u00a0<\/b><\/span><\/span><\/span>B = 6,0 T<\/b><\/span><\/span><\/span>; comprimento da barra condutora\u00a0<\/b><\/span><\/span><\/span>L = 1,0 m\u00a0<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>velocidade<\/b><\/span><\/span><\/span>\u00a0de deslocamento da barra condutora\u00a0<\/b><\/span><\/span><\/span>v = 2 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>for\u00e7a eletromotriz induzida ()\u00a0<\/b><\/span><\/span><\/span>na barra condutora pela express\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00a0= B.L.v\u00a0<\/b><\/span><\/span><\/span>= 6.1.2\u00a0<\/b><\/span><\/span><\/span><\/p>\n

= 12 V.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Estando a\u00a0<\/b><\/span><\/span><\/span>barra descendo com velocidade constante<\/b><\/span><\/span><\/span>\u00a0(equil\u00edbrio din\u00e2mico \u201cfor\u00e7a resultante nula\u201d) ent\u00e3o\u00a0<\/b><\/span><\/span><\/span>deve surgir sobre ela uma for\u00e7a magn\u00e9tica de intensidade Fm<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(vertical e para cima) que deve\u00a0<\/b><\/span><\/span><\/span>anular a for\u00e7a externa de intensidade F\u00a0<\/b><\/span><\/span><\/span>(vertical e para baixo).<\/b><\/span><\/span><\/span><\/p>\n

Essa\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica tem as seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>caso do exerc\u00edcio\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica tem dire\u00e7\u00e3o vertical e sentido para cima\u00a0<\/b><\/span><\/span><\/span>(figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

\u00a0<\/span>C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>intensidade da for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span>\u00a0pela express\u00e3o\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0= B.i.L.sen<\/b><\/span><\/span><\/span>sendo dados:\u00a0<\/b><\/span><\/span><\/span>B =<\/b><\/span><\/span><\/span>\u00a06 T;\u00a0<\/b><\/span><\/span><\/span>i =<\/b><\/span><\/span><\/span>\u00a012 A;\u00a0<\/b><\/span><\/span><\/span>L =<\/b><\/span><\/span><\/span>\u00a01 m e\u00a0<\/b><\/span><\/span><\/span>\u00a0= 90o:<\/b><\/span><\/span><\/span><\/p>\n

Fm\u00a0=<\/b><\/span><\/span><\/span>\u00a06.12.1.1\u00a0\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0= 72 N.<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 E<\/b><\/span><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes \u2013 Eletromagnetismo 01- Inseparabilidade dos polos magn\u00e9ticos \u00a0Se voc\u00ea\u00a0quebrar um im\u00e3 (corte transversal)\u00a0e em seguida\u00a0continuar\u00a0dividindo os im\u00e3s resultantes,\u00a0voc\u00ea observar\u00e1 que\u00a0cada peda\u00e7o partido continuar\u00e1 sendo um novo im\u00e3 com dois polos, Norte e Sul de maneira que cada peda\u00e7o atraia o outro.\u00a0N\u00e3o existem polos isolados. Se voc\u00ea\u00a0partir um im\u00e3 conforme a figura abaixo (corte longitudinal),\u00a0a\u00a0for\u00e7a entre<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4854,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-4857","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4857","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=4857"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4857\/revisions"}],"predecessor-version":[{"id":4858,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4857\/revisions\/4858"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4854"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4857"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}