{"id":1927,"date":"2016-02-11T23:21:41","date_gmt":"2016-02-11T23:21:41","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1927"},"modified":"2024-08-19T17:39:16","modified_gmt":"2024-08-19T17:39:16","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-magnetico-gerado-por-uma-espira-circular-ou-por-um-solenoide","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletromagnetismo\/campo-magnetico-gerado-por-uma-espira-circular-ou-por-um-solenoide\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-magnetico-gerado-por-uma-espira-circular-ou-por-um-solenoide\/","title":{"rendered":"Campo Magn\u00e9tico – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Campo Magn\u00e9tico gerado por uma espira circular ou por um solenoide<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A\u00a0<\/b><\/span><\/span><\/span>— veja teoria<\/b><\/span><\/span><\/span><\/p>\n

02<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a001- Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

02. Falsa —\u00a0veja 1<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa — tem sentidos opostos, fornecidos pela regra da m\u00e3o direita<\/b><\/span><\/span><\/span><\/p>\n

08- Falsa —\u00a0veja na express\u00e3o B=\u03bc.i\/2R que i \u00e9 diretamente proporcional a B\u00a0 —\u00a0 se i dobra, B tamb\u00e9m dobra<\/b><\/span><\/span><\/span><\/p>\n

16- Falsa — veja na express\u00e3o B=\u03bc.i\/2R que R \u00e9 inversamente proporcional a B\u00a0 —\u00a0 se R dobra, B cai pela metade<\/b><\/span><\/span><\/span><\/p>\n

32- Falsa — a dire\u00e7\u00e3o n\u00e3o se altera, mas o sentido sim<\/b><\/span><\/span><\/span><\/p>\n

R- 1<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Usando a regra da m\u00e3o direita\u00a0 —\u00a0 o campo magn\u00e9tico\"\"\u00a0estar\u00e1 saindo da folha de papel e \u00e9 nesta dire\u00e7\u00e3o que estar\u00e1 a agulha da b\u00fassola, com o p\u00f3lo norte no sentido de A para B\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os vetores indu\u00e7\u00e3o magn\u00e9tica devem se anular e, para isso, as correntes nas espiras circulares devem ter mesma intensidade e sentidos contr\u00e1rios\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A intensidade do campo magn\u00e9tico no centro da espira \u00e9 fornecido por Be=\u03bci\/2\u03c0R e o campo magn\u00e9tico no centro da esfera originado pelo fio retil\u00edneo \u00e9 fornecido por Bf=\u03bci<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2R\u00a0 —\u00a0 para que o campo resultante seja nulo\u00a0 —\u00a0 Be=Bf\u00a0 — \u03bci\/2\u03c0R= \u03bci<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2R\u00a0 —\u00a0 i=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2\u03c0\u00a0 —\u00a0 i\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\u00a0 —\u00a0 o campo magn\u00e9tico\"\"\u00a0no centro da espira originado pela corrente no fio \u00e9 perpendicular ao plano da folha e entrando nela\u00a0 —\u00a0 o campo magn\u00e9tico no centro da espira originado por ela deve estar saindo da folha de papel para que eles possam se anular e, para isso, a corrente na espira deve ser no sentido anti-hor\u00e1rio\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>A dire\u00e7\u00e3o e sentido do campo magn\u00e9tico originado no centro da espira pelo fio retil\u00edneo \u00e9 fornecida pela regra da m\u00e3o direita\u00a0 e tem dire\u00e7\u00e3o perpendicular ao papel e entrando nele (figura 1) e tem intensidade B<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=\u03bci<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/2\u03c0R\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=62,8\u03bc\/2\u03c0R —\u00a0 para que o campo magn\u00e9tico resultante no centro da espira seja nulo, o campo magn\u00e9tico a\u00ed deve ser perpendicular ao papel e saindo dele e, para que isso ocorra o sentido da corrente el\u00e9trica deve ser anti-hor\u00e1rio (fornecido pela regra da m\u00e3o direita \u2013 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura 2) e deve ter intensidade\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u03bc.i<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\/2R\u00a0 —\u00a0\u00a0 como o campo magn\u00e9tico resultante deve ser nulo\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=B<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u03bc.i<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\/2R=62,8\u03bc\/2\u03c0R\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=62,8\/\u03c0\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u224820 A e sentido anti-hor\u00e1rio<\/b><\/span><\/span><\/span><\/p>\n

07<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>a) Observe na figura abaixo que, se R<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=8\u2126\u00a0 —\u00a0 Ri<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8\/4=2\u2126 e Ri<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3.8\/4=6\u2126\u00a0 —\u00a0 observe tamb\u00e9m <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

que a diferen\u00e7a de potencial U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a mesma para os dois trechos\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=Ri<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=Ri<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=6.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=6i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(I)\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4 \u2013 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 4 \u2013 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1 A\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3 A<\/b><\/span><\/span><\/span>
\nb) Pela regra da m\u00e3o direita a corrente el\u00e9trica i1 originar\u00e1 no centro da espira um\u00a0\"\"\u00a0entrando na folha de papel e i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0originar\u00e1 um<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

saindo da folha de papel\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=(1\/4)\u03bci<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2R=\u03bc.3\/4.2R\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3\u03bc\/8R\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(3\/4)\u03bci<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/2R=3\u03bc\/4.2R\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3\u03bc\/8R\u00a0 —\u00a0 observe que B<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como eles tem sentidos opostos e mesma intensidade, eles se anulam\u00a0 —\u00a0 o campo magn\u00e9tico no centro da circunfer\u00eancia\u00a0<\/b><\/span><\/span><\/span>ser\u00e1 nulo<\/b><\/span><\/span><\/span>.
\n<\/b><\/span><\/span><\/span>08-<\/b><\/span><\/span><\/span> Os dois campos B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0devem se anular\u00a0 —\u00a0 B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03bci<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03bci<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/0,8R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03bci<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03bci<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/0,4R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03bci<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 I<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A espira produz em C um campo magn\u00e9tico B2 perpendicular \u00e0 folha e saindo dela (regra da m\u00e3o direita)\u00a0 —\u00a0 para que os campos se anulem a corrente no fio deve ser da direita para a esquerda, produzindo em C, um campo perpendicular \u00e0 folha e entrando nela (regra da m\u00e3o direita)<\/b><\/span><\/span><\/span><\/p>\n

b) espira —\u00a0 B1=\u03bci1\/2R\u00a0 —\u00a0 fio\u00a0 —\u00a0 B2= \u03bci2\/2\u03c0R —\u00a0 B1=B2\u00a0 —\u00a0 \u03bci1\/2R= \u03bci2\/2\u03c0R\u00a0 —<\/b><\/span><\/span><\/span>\u00a0i2\/i1=\u03c0=3<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esquematizando a situa\u00e7\u00e3o na figura abaixo, e, com a corrente no sentido anti-hor\u00e1rio vista pelo observador O, utilizando a regra<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

da m\u00e3o direita, a espira se tornar\u00e1 um pequeno im\u00e3 com p\u00f3lo norte e p\u00f3lo sul indicados:<\/b><\/span><\/span><\/span><\/p>\n

I. Correta —\u00a0elas se repelem (p\u00f3los de mesmo nome) e estas for\u00e7as constituem um par a\u00e7\u00e3o e rea\u00e7\u00e3o (mesma intensidade, mesma dire\u00e7\u00e3o mas sentidos opostos).<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa —\u00a0a for\u00e7a entre eles \u00e9 de repuls\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

III. Fa=Fi —\u00a0ma.aa=mi.ai\u00a0 —\u00a0 ma.aa=2ma.ai\u00a0 —\u00a0 aa=2.a\u00adi\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/span><\/p>\n

IV. Falsa<\/b><\/span><\/span><\/span><\/p>\n

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

11-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Usando a regra da m\u00e3o direita as configura\u00e7\u00f5es das linhas de indu\u00e7\u00e3o do campo magn\u00e9tico do <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

solenoide est\u00e3o indicadas na figura —\u00a0 a agulha magn\u00e9tica tem seu polo norte sempre no sentido das linhas de indu\u00e7\u00e3o\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Fechando a chave, colocando a corrente no sentido indicado (sai do p\u00f3lo sul do gerador) e aplicando a regra da m\u00e3o direita voc\u00ea verifica que as linhas de indu\u00e7\u00e3o do campo magn\u00e9tico gerado pelo eletro\u00edm\u00e3 saem do p\u00f3lo norte, \u00e0 esquerda e entram no p\u00f3lo sul,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e0 direita\u00a0 —\u00a0 o im\u00e3 ser\u00e1 atra\u00eddo\u00a0 —\u00a0 p\u00f3los opostos se atraem\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Se uma corrente el\u00e9trica circular pelas espiras, a barra de ferro se comportar\u00e1 como um magneto (im\u00e3)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

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

00. Verdadeira\u00a0 —\u00a0 a pilha fornece corrente cont\u00ednua.<\/b><\/span><\/span><\/span><\/p>\n

01. Falsa\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

02. Falsa\u00a0 —\u00a0 s\u00f3 surgir\u00e1 campo magn\u00e9tico se a corrente nas espiras for corrente cont\u00ednua.<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Quando voc\u00ea liga as extremidades do fio aos p\u00f3los da pilha, este passa a ser percorrido por uma corrente el\u00e9trica que, por sua vez, gera um campo magn\u00e9tico ao seu redor. Este campo magn\u00e9tico gerado pela corrente el\u00e9trica imantar\u00e1 a haste e ela, comportando-se como um \u00edm\u00e3, passar\u00e1 a atrair pequenos objetos de ferro ou a\u00e7o.
\nb) Pela regra da m\u00e3o direita, o p\u00f3lo norte \u00e9 a extremidade da direita, de onde saem as linhas de indu\u00e7\u00e3o do campo magn\u00e9tico.
\nc. A extremidade da direita passa a ser o p\u00f3lo sul e a da esquerda o p\u00f3lo norte, ou seja, quando se inverte o sentido da corrente el\u00e9trica, inverte-se tamb\u00e9m a polaridade.<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> <\/b>R- C<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

19<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0I. Observe na express\u00e3o B=\u03bc.i.(N\/L)\u00a0 —\u00a0 (N\/L) \u00e9 o n\u00famero de espiras por unidade de comprimento\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/span><\/p>\n

II. A intensidade do campo magn\u00e9tico aumenta\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 observe na express\u00e3o B=\u03bc.i.(N\/L) que i \u00e9 proporcional a B\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/span><\/p>\n

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

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela regra da m\u00e3o direita a extremidade esquerda do solen\u00f3ide \u00e9 um p\u00f3lo sul\u00a0 —\u00a0 p\u00f3los de nomes opostos se atraem e o carrinho se aproximar\u00e1 do solen\u00f3ide\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I- Pela regra da m\u00e3o direita voc\u00ea verifica que est\u00e1 correta.<\/b><\/span><\/span><\/span><\/p>\n

II- Falsa — im\u00e3 atrai ferro independente da polaridade.<\/b><\/span><\/span><\/span><\/p>\n

III- Falsa — o vetor indu\u00e7\u00e3o magn\u00e9tica no interior do im\u00e3 \u00e9 proporcional a corrente el\u00e9trica i que ir\u00e1 aumentar.<\/b><\/span><\/span><\/span><\/p>\n

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

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Aplicando a fun\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 8=E \u2013 5.200.10-3\u00a0 —\u00a0 8=E \u2013 1\u00a0 —\u00a0 E=9,0V<\/b><\/span><\/span><\/span><\/p>\n

b) B=4.\u03c0.10-7.(200\/0,2).200.10-3=4.\u03c0.10-7.103.2.10-1\u00a0 —\u00a0 B=8,0\u03c0.10-5T<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) B=4\u03c0.10-7.(2.000\/0,5).3\u00a0 —\u00a0 B=48.\u03c0.10-4\u00a0 —\u00a0 B=4,8.\u03c0.10-3T<\/b><\/span><\/span><\/span><\/p>\n

b) Como ele \u00e9 lan\u00e7ado paralelamente ao eixo, o \u00e2ngulo entre\"\"\u00a0e\"\"\u00a0\u00e9 nulo\u00a0 —\u00a0 Fm=q.V.B.sen0o=q.V.B.0\u00a0 —\u00a0 Fm=0\u00a0 —\u00a0 se Fm=0, a acelera\u00e7\u00e3o do el\u00e9tron \u00e9 nula e ele se desloca em MRU.<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esquematizando a situa\u00e7\u00e3o (figura)\u00a0 —\u00a0 a barra AB s\u00f3 pode estar imantada pois N atrai A (que \u00e9 polo sul) j\u00e1 que p\u00f3los opostos se atraem e, consequentemente B ser\u00e1 um polo norte\u00a0\u00a0 —\u00a0 como S e B s\u00e3o repelidas pela bobina, esta ao ser percorrida por corrente i<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

originar\u00e1 em sua parte superior um p\u00f3lo S e na inferior um polo norte\u00a0 —\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span> <\/b>Quando se aumenta a resist\u00eancia R, a intensidade da corrente el\u00e9trica na bobina diminui e consequentemente diminui tamb\u00e9m o campo magn\u00e9tico gerado por ela —<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Com a chave A aberta e a C fechada, para que o sistema permane\u00e7a em equil\u00edbrio, segundo o <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

enunciado, uma pequena massa M<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,008kg\u00a0 \u00e9 colocada no ponto m\u00e9dio do segmento P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a corrente no circuito \u00e9 no sentido hor\u00e1rio e, aplicando a regra da m\u00e3o esquerda no segmento P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, a for\u00e7a magn\u00e9tica sobre esse trecho de fio ser\u00e1 vertical e para cima\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=M<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g=0,008.10\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,08N\u00a0 —\u00a0 no equil\u00edbrio, F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,08N<\/b><\/span><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>=iB<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>L — 0,08=2.B<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>.0,2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=0,2T<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) Com a chave A fechada e C aberta o circuito ficar\u00e1 como o esquema indicado e a corrente el\u00e9trica <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

circular\u00e1 no sentido hor\u00e1rio\u00a0 —\u00a0 utilizando a regra da m\u00e3o esquerda, a for\u00e7a magn\u00e9tica no trecho P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0ser\u00e1 vertical e para baixo e no trecho P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, vertical e para cima\u00a0 —\u00a0 nos trechos P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0n\u00e3o haver\u00e1 for\u00e7a magn\u00e9tica, pois neles, o campo magn\u00e9tico \u00e9 paralelo \u00e0 corrente el\u00e9trica\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

Essas duas for\u00e7as produzem um torque resultante, que tenta girar a espira em torno de OO\u2019 no sentido hor\u00e1rio\u00a0—\u00a0 para a espira ficar em equil\u00edbrio, voc\u00ea deve colocar uma massa M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>numa dada posi\u00e7\u00e3o de modo que ela produza um torque em sentido contr\u00e1rio (anti-hor\u00e1rio), produzindo um torque resultante nulo\u00a0 —\u00a0 para que isso ocorra voc\u00ea pode colocar M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0em qualquer posi\u00e7\u00e3o do<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

segmento P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, mas \u00e9 mais coerente coloc\u00e1-la no ponto m\u00e9dio do mesmo — as intensidades das for\u00e7as magn\u00e9ticas em cada fio s\u00e3o iguais (mesmo campo magn\u00e9tico, mesma corrente e mesmo comprimento-F=iBL)\u00a0 —\u00a0 para neutralizar o bin\u00e1rio a for\u00e7a peso da massa M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0dever\u00e1 ter o torque de mesmo m\u00f3dulo, mas sentido oposto, ou seja, a massa M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0dever\u00e1 ser o dobro de M<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e logo M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2.0,008\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 0,016 kg.<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(01) Errada\u00a0 —\u00a0 cargas el\u00e9tricas somente geram campo magn\u00e9tico quando est\u00e3o em movimento.<\/b><\/span><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 corrente el\u00e9trica \u00e9 formada pelo movimento de cargas el\u00e9tricas.<\/b><\/span><\/span><\/span><\/p>\n

(04) Errada\u00a0 —\u00a0 se a corrente \u00e9 alternada os p\u00f3los da bobina se alternam, diferente de um \u00edm\u00e3 que tem p\u00f3los fixos.<\/b><\/span><\/span><\/span><\/p>\n

(08) Correta.<\/b><\/span><\/span><\/span><\/p>\n

(16) Errada.<\/b><\/span><\/span><\/span><\/p>\n

R- (02 + 08) = 10<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que cada uma das faces do cubo percorrida por corrente comporta-se como uma espira quadrada —\u00a0 o campo magn\u00e9tico no centro do cubo gerado por uma face tem a mesma dire\u00e7\u00e3o do campo no centro dessa face\u00a0 —\u00a0 o sentido desse campo \u00e9 dado pela regra da m\u00e3o direita (m\u00e3o aberta, polegar no sentido da corrente, o sentido da corrente \u00e9 fornecido pela \u201cfechada\u201d da m\u00e3o,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

passando pelo interior da espira)\u00a0 —\u00a0 para o cubo da figura (a) onde a corrente circula s\u00f3 na face esquerda, a Fig. 1 ilustra o vetor indu\u00e7\u00e3o magn\u00e9tica citado no enunciado\u00a0 —\u00a0 para o cubo da figura (b) voc\u00ea deve analisar os campos gerados pelas faces opostas (pois t\u00eam a mesma dire\u00e7\u00e3o) e confront\u00e1-los quanto ao sentido\u00a0 —\u00a0 a contribui\u00e7\u00e3o para a intensidade do campo magn\u00e9tico de cada aresta percorrida por corrente \u00e9 B\/4\u00a0 — em cada um dos tr\u00eas casos abaixo, a intensidade do campo <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

magn\u00e9tico resultante vale B<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=B\/2 + B\/2\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=B\u00a0 —\u00a0 observe pelo sistema de coordenadas fornecido que B<\/b><\/span><\/span><\/span>x\u00a0<\/b><\/span><\/span><\/sub><\/span>= -B; B<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= B e B<\/b><\/span><\/span><\/span>z<\/b><\/span><\/span><\/sub><\/span>\u00a0= B\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dados\u00a0 —\u00a0 \u03bc<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1,3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0T.m\/A\u00a0 —\u00a0 N = 25.000 espiras\u00a0 —\u00a0 L = 0,65 m\u00a0 — \u00a0i = 80 A\u00a0 —\u00a0 B= \u03bco.n.i\u00a0 —\u00a0 n=n\u00famero de espiras\/unidade de comprimento=25.000\/0,65\u00a0 —\u00a0 B= \u03bco(N\/L).i=1,3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>x(25.000\/0.65)x80\u00a0 — \u00a0<\/b><\/span><\/span><\/span>B=4.0 T<\/b><\/span><\/span><\/span><\/p>\n

b) Dados\u00a0 — \u00a0m = 200 g = 0,2 kg\u00a0 —\u00a0 d = 2 mm = 2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0m\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 — \u00a0\u00a0como o <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

im\u00e3 sobe em movimento acelerado\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0> P\u00a0 —\u00a0 pric\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P = m.a<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=0,2.10 + 0,2.0,5\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=2,1N\u00a0 —\u00a0 c\u00e1lculo do trabalho\u00a0 —\u00a0 W=F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d=2,1.2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=4,2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 N = 300 espiras\/m\u00a0 —\u00a0 \u00a0\u03bc<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 1,26.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>T.m\/A\u00a0 —\u00a0 \u00a0i = 100 mA = 0,1 A; sen 62\u00ba = 0,88, cos 62\u00ba = 0,47 e tg 62\u00ba = 1,87\u00a0 —\u00a0 pela regra da m\u00e3o direita, o vetor indu\u00e7\u00e3o magn\u00e9tica no interior do solenoide \u00e9 horizontal para direita\u00a0 —\u00a0 o vetor indu\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

magn\u00e9tica terrestre \u00e9 orientado para o norte\u00a0 —\u00a0 a figura ilustra esses vetores\u00a0 —\u00a0 nessa figura\u00a0 —\u00a0 tg <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

62<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0= B<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/B<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 B<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=B<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>tg62<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> —\u00a0 B<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=N.\u03bc<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.i.tg62<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=300×1,26.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>x0,1×1,87\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=7,1.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0T<\/b><\/span><\/span><\/span><\/p>\n

31-(FMABC-SP)<\/b><\/span><\/span><\/span><\/p>\n

Quando um solenoide \u00e9 percorrido por corrente el\u00e9trica ele produz um campo magn\u00e9tico, campo este que cria um fluxo que atravessa as espiras\u00a0 —\u00a0 a capacidade de uma bobina (solenoide) em criar o fluxo magn\u00e9tico em seu interior em fun\u00e7\u00e3o da\u00a0 corrente que a percorre \u00e9 denominada Indut\u00e2ncia (s\u00edmbolo L) medida em “henry” cujo s\u00edmbolo \u00e9 H\u00a0 —\u00a0 em fun\u00e7\u00e3o dos dados do exerc\u00edcio, a indut\u00e2ncia \u00e9 fornecida pela express\u00e3o matem\u00e1tica\u00a0 —\u00a0 L= \u03bco.n2.S\/d, sendo\u00a0 —\u00a0 \u03bco\u00a0\u2013 permeabilidade magn\u00e9tica do meio no interior do solenoide, no caso \u00e9 o v\u00e1cuo – \u03bco=4\u03c0.10-7Tm\/A\u00a0 —\u00a0 n \u2013 n\u00famero de espiras \u2013 n=500 espiras\u00a0 —\u00a0 S \u2013 \u00e1rea da se\u00e7\u00e3o reta do solenoide (ou de cada espira) \u2013 S=\u03c0R2=3.(15.10-3)2\u00a0=3×225.10-6\u00a0 —\u00a0 S=675.10-6m2\u00a0 —\u00a0 d \u2013 comprimento do solenoide \u2013 d=25cm=25.10-2m\u00a0 —\u00a0 L= \u03bco.n2.S\/d,\u00a0 —\u00a0 L=4.3.10-7x(500)2×675.10-6\/25.10-2\u00a0 —\u00a0 L=81.10-5=0,81.10-3H\u00a0 —\u00a0 L=0,81mH\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Campo Magn\u00e9tico gerado por uma espira circular ou por um solenoide 01-\u00a0R- A\u00a0— veja teoria 02–\u00a001- Correta\u00a0 —\u00a0 veja teoria 02. Falsa —\u00a0veja 1 04. Falsa — tem sentidos opostos, fornecidos pela regra da m\u00e3o direita 08- Falsa —\u00a0veja na express\u00e3o B=\u03bc.i\/2R que i \u00e9 diretamente proporcional a B\u00a0 —\u00a0 se i dobra,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1923,"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-1927","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1927","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=1927"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1927\/revisions"}],"predecessor-version":[{"id":10567,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1927\/revisions\/10567"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1923"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1927"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}