{"id":1918,"date":"2016-02-11T23:02:55","date_gmt":"2016-02-11T23:02:55","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1918"},"modified":"2024-08-19T17:44:12","modified_gmt":"2024-08-19T17:44:12","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-magnetico","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletromagnetismo\/campo-magnetico-originado-por-um-condutor-retilineo-extenso-percorrido-por-corrente-eletrica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-campo-magnetico\/","title":{"rendered":"Campo magn\u00e9tico – Resolu\u00e7\u00e3o comentada"},"content":{"rendered":"

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Usando a regra da m\u00e3o direita\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o B=\u03bc.i\/2\u03c0r que B \u00e9 inversamente proporcional a r\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>B=\u03bci\/2\u03c0r=4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>.5\/2\u03c00,2\u00a0 —\u00a0 B=5,0.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>T\u00a0 —\u00a0 regra da m\u00e3o direita\u00a0 —\u00a0 penetrando na folha de papel\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O polo norte da b\u00fassola est\u00e1 no sentido da orienta\u00e7\u00e3o do campo magn\u00e9tico, fornecida pela regra da m\u00e3o direita\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0—\u00a0veja teoria<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Regra da m\u00e3o direita —<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela regra da m\u00e3o direita o sentido de\"\"\u00a0\u00e9 para fora da p\u00e1gina (saindo dela) e pela express\u00e3o matem\u00e1tica B=\u03bc.i\/2\u03c0r, a intensidade de B \u00e9 inversamente proporciona \u00e0 dist\u00e2ncia r\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o B=\u03bc.i\/2\u03c0r que i \u00e9 diretamente proporcional a B\u00a0 —\u00a0 se i triplica, B tamb\u00e9m ser\u00e1 triplicado\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando n\u00e3o passa corrente el\u00e9trica pelo fio, a agulha da b\u00fassola indica a dire\u00e7\u00e3o norte sul magn\u00e9tica da Terra\u00a0—\u00a0 quando passa corrente el\u00e9trica no fio, surge outro campo magn\u00e9tico devido ao fio e a agulha da b\u00fassola ir\u00e1 indicar outra posi\u00e7\u00e3o que \u00e9 obtida pela soma vetorial dos dois campos magn\u00e9ticos, ou seja,\u00a0\"\"=\u00a0\"\"+\u00a0\"\"\u00a0\u00a0— \u00a0a esfera permanece na mesma posi\u00e7\u00e3o, pois est\u00e1 em repouso\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando n\u00e3o passa corrente el\u00e9trica pelo fio, a agulha da b\u00fassola indica a dire\u00e7\u00e3o norte sul magn\u00e9tica d Terra\u00a0 —\u00a0 quando passa corrente el\u00e9trica no fio, surge outro campo magn\u00e9tico devido ao fio e a agulha da b\u00fassola ir\u00e1 indicar outra posi\u00e7\u00e3o que\u00e9 obtida pela soma vetorial dos dois campos magn\u00e9ticos, ou seja,\u00a0\"\"=\u00a0\"\"+\u00a0\"\"\u00a0\u00a0—\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando, pela regra da m\u00e3o direita em cada regi\u00e3o, os campos originados por cada fio, as regi\u00f5es onde eles n\u00e3o podem se anular<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(soma vetorial), s\u00e3o as II e III\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o B=\u03bc.i\/2\u03c0r que, se i duplica, B duplica e que se r for reduzida \u00e0 metade B tamb\u00e9m duplica\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando, pela regra da m\u00e3o direita em cada regi\u00e3o, os campos originados por cada fio, as regi\u00f5es onde eles tem sentidos opostos<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0(um saindo da folha e o outro entrando na folha) s\u00e3o I e III\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O vetor campo magn\u00e9tico criado pela corrente el\u00e9trica \u00e9 ortogonal a ela e com sentido fornecido pela regra da m\u00e3o direita —\u00a0 observe que quando a b\u00fassola est\u00e1 sobre o condutor o campo magn\u00e9tico \u00e9 orientado para baixo (onde indica o p\u00f3lo norte da b\u00fassola) e quando a b\u00fassola est\u00e1 sob o condutor o campo magn\u00e9tico \u00e9 orientado para cima (onde indica o p\u00f3lo norte da b\u00fassola)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela regra da m\u00e3o direita o campo magn\u00e9tico criado pelo fio tem dire\u00e7\u00e3o vertical e sentido para baixo (para onde indica o polo norte da agulha) e o campo magn\u00e9tico da Terra tem dire\u00e7\u00e3o norte-sul, podendo estar para cima ou para baixo, mas como i \u00e9 muito alta,\u00a0\"\", o campo magn\u00e9tico resultante sempre ter\u00e1 dire\u00e7\u00e3o vertical e sentido para baixo\u00a0 —<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) O fio est\u00e1 no centro da figura e a corrente tem sentido “saindo do papel”.<\/b><\/span><\/span><\/p>\n

b) Observe a figura a seguir.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

17-\u00a0<\/b><\/span><\/span>Pela regra da m\u00e3o direita C<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0origina em P um campo magn\u00e9tico vertical e para cima e C<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>origina em P um campo magn\u00e9tico<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

vertical e para baixo\u00a0 —\u00a0 como a intensidade de B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00e9 maior que a de B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, pois B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0tem maior corrente que B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, o campo magn\u00e9tico resultante \u00e9 vertical e para baixo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0estiver saindo e i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0entrando na folha do papel\u00a0\"\"\u00a0e\u00a0\"\"\u00a0estar\u00e3o como na figura e devem se <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

anular, ou seja, 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>\/2\u03c0.2d= \u03bci<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2\u03c0.d\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2=i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Enquanto a corrente era nula, a agulha da b\u00fassola indicava a orienta\u00e7\u00e3o do campo magn\u00e9tico da Terra (\"\")\u00a0 —\u00a0 quando o fio \u00e9 ligado, surge nele corrente saindo do plano da p\u00e1gina, que pela regra da m\u00e3o direita, provocam linhas de indu\u00e7\u00e3o magn\u00e9tica em torno do fio, no sentido anti-hor\u00e1rio\u00a0 —\u00a0\u00a0 o vetor indu\u00e7\u00e3o magn\u00e9tica devido ao fio (\"\") \u00e9 tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto\u00a0 —\u00a0\u00a0 a agulha da b\u00fassola orienta-se, ent\u00e3o, no sentido do vetor indu\u00e7\u00e3o magn\u00e9tica resultante\u00a0 —\u00a0 \u00a0\"\"=\u00a0\"\"+\u00a0\"\"\u00a0\u00a0—\u00a0 <\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na express\u00e3o F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u03bc.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.L\/2.\u03c0.d que, duplicando a intensidade de cada corrente F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0ficar\u00e1 4 vezes maior e que duplicando a dist\u00e2ncia F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0ficar\u00e1 2 vezes menor —\u00a0 4F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\/2=2F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na experi\u00eancia I as correntes possuem mesmo sentido e a for\u00e7a entre os fios \u00e9 de atra\u00e7\u00e3o diminuindo a dist\u00e2ncia entre eles e na experi\u00eancia II, onde as corrente possuem sentidos contr\u00e1rios, a for\u00e7a entre eles \u00e9 de repuls\u00e3o, aumentando a dist\u00e2ncia entre eles\u00a0— <\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a \u00e9 de atra\u00e7\u00e3o e tem a mesma intensidade, pois obedecem ao princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja resolu\u00e7\u00e3o do exerc\u00edcio anterior\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo onde est\u00e3o as for\u00e7as que agem sobre cada fio:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as correntes tem mesmo sentido ir\u00e3o atrair-se\u00a0 —\u00a0 como a for\u00e7a \u00e9 inversamente proporcional \u00e0 dist\u00e2ncia, se dist\u00e2ncia dobrar, a for\u00e7a cair\u00e1 pela metade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a \u00e9 de repuls\u00e3o pois a correntes tem sentidos contr\u00e1rios e \u00e9 diretamente proporcional a i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u03bc.i.i.L\/2.\u03c0.d\u00a0 —\u00a0F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u03bc.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.L\/2.\u03c0.d\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelas configura\u00e7\u00f5es das linhas de indu\u00e7\u00e3o magn\u00e9tica voc\u00ea percebe que os fios se repelem e assim conduzem correntes de sentidos opostos. Essas for\u00e7as tem sempre a mesma intensidade (par a\u00e7\u00e3o e rea\u00e7\u00e3o) e a for\u00e7a que repele o fio da esquerda tem sentido da direita para a esquerda\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

29-\u00a0<\/b><\/span><\/span><\/span>Observe nas figuras abaixo a dire\u00e7\u00e3o e o sentido da for\u00e7a magn\u00e9tica sobre o fio, desviando-o para a direita\u00a0—\u00a0 a intensidade<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

dessa for\u00e7a \u00e9 fornecida pela express\u00e3o F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=B.i.l.sen\u03b8=B.i.l.sen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=B.i.l\u00a0 —\u00a0 essa express\u00e3o afirma que a for\u00e7a magn\u00e9tica \u00e9 diretamente proporcional \u00e0 intensidade da corrente el\u00e9trica i que percorre o fio\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

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

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

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

(16) Errada.\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

31-<\/b><\/span><\/span><\/span>Observe o esquema abaixo\u00a0 —\u00a0 no tri\u00e2ngulo OMP destacado \u00e0 direita na figura\u00a0 —\u00a0 tg\u03b8=(\u221a3.a\/2)\/(a\/2)\u00a0 —\u00a0 tg\u03b8=\u221a3\u00a0 —\u00a0 \u03b8=60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

— no mesmo tri\u00e2ngulo\u00a0 —\u00a0 sen \u03b8=(3.a\/2)\/2r\u00a0 —\u00a0 \u221a3\/2=\u221a(3.a)\/2r\u00a0 —\u00a0 r=a\u00a0 —\u00a0 \u00a0fios paralelos percorridos por correntes de mesmo sentido se atraem, e por correntes de sentidos opostos se repelem\u00a0 —\u00a0 para dois condutores de comprimento \u2113, distantes r entre si, percorridos por correntes I<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e I<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, num meio onde a permeabilidade magn\u00e9tica \u00e9 \u03bc, a intensidade dessa for\u00e7a \u00e9\u00a0 —\u00a0 F=\u03bcI<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.I<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>. \u2113\/2\u03c0r\u00a0 —\u00a0 no caso desse exerc\u00edcio, I<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=I<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=I e \u2113=r=a\u00a0 —\u00a0 assim, sobre os lados RM e NS, a intensidade da for\u00e7a magn\u00e9tica, de acordo com a express\u00e3o acima, \u00e9 dada por\u00a0 —\u00a0 F=\u03bcI<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>a\/2\u03c0a\u00a0 —\u00a0 F=\u03bcI<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u03c0 (I)\u00a0 —\u00a0 sendo os fios perpendiculares entre si, a for\u00e7a magn\u00e9tica \u00e9 nula\u00a0 — assim o fio n\u00e3o exerce for\u00e7a sobre os lados MN e RS da espira\u00a0 —\u00a0 baseado na figura abaixo que mostra a situa\u00e7\u00e3o vista de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cima\u00a0 —\u00a0 usando a regra da m\u00e3o direita nos lados NS e MR voc\u00ea encontra a dire\u00e7\u00e3o e o sentido dos campos magn\u00e9ticos\u00a0\"\"\u00a0e\u00a0\"\"\u00a0e, em seguida utilizando a regra da m\u00e3o esquerda voc\u00ea localiza as for\u00e7as\u00a0\"\"\u00a0e\u00a0\"\"\u00a0\u00a0—\u00a0 observe que quem tende a provocar a rota\u00e7\u00e3o no sentido hor\u00e1rio em torno do eixo z s\u00e3o as componentes Fcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0de cada uma das for\u00e7as\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/sup>\u00a0e\u00a0\"\"\u00a0\u00a0—\u00a0 cada uma dessas for\u00e7as tem a mesma intensidade fornecida pela equa\u00e7\u00e3o (I) acima\u00a0 —\u00a0 F=\u03bcI<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u03c0 (I)\u00a0 —\u00a0 a dist\u00e2ncia do ponto de aplica\u00e7\u00e3o das for\u00e7as at\u00e9 o eixo z vale \u221a3\/2.a\u00a0 —\u00a0 M<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>=2.Fd=2.(Fcos60<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>.\u221a3\/2.a)\u00a0 —\u00a0 M<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>=\u03bcI<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u03c0.0,5.\u221a3.a\u00a0 —<\/b><\/span><\/span>\u00a0M<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>=\u221a3\u03bcI<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.a\/4\u03c0<\/b><\/span><\/span><\/p>\n

32-\u00a0<\/b><\/span><\/span>A intensidade do campo magn\u00e9tico produzido por um fio retil\u00edneo \u00e9 dado pela express\u00e3o B=\u03bc<\/b><\/span><\/span>o\u00ad<\/b><\/span><\/span><\/sub>i\/2\u03c0r\u00a0 —\u00a0 observe que ela \u00e9 diretamente proporcional \u00e0 corrente el\u00e9trica. Sendo assim, se duplicarmos a corrente, duplicaremos tamb\u00e9m a intensidade do campo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo, onde, pela regra da m\u00e3o direita voc\u00ea obt\u00e9m os vetores\u00a0\"\"\u00a0\u00a0no fio da <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

aresta DA e\u00a0\"\"\u00a0no fio da aresta GH\u00a0 —\u00a0 como esses dois campos t\u00eam a mesma intensidade, pois B=\u03bci\/2r o campo magn\u00e9tico resultante,\u00a0\"\"\u00a0, tem o sentido CF, indicado na figura\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 as figuras abaixo mostram os campos magn\u00e9ticos originados, usando a regra da m\u00e3o direita, no fio 4, pelos fios 1<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\u00e3o direita, no fio 4, pelos fios 1\u00a0(\"\"), 2 (\"\") e 3 (\"\")\u00a0 —\u00a0 em seguida, somando-os vetorialmente foi obtido o vetor campo magn\u00e9tico resultante que aponta para o nordeste.<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 observe na figura abaixo a utiliza\u00e7\u00e3o da regra da m\u00e3o esquerda no fio 4 que fornece a for\u00e7a magn\u00e9tica\u00a0\"\"\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

indicando para o sudeste.<\/b><\/span><\/span><\/p>\n

04-\u00a0Falsa\u00a0 —\u00a0 observe que as correntes el\u00e9tricas nos fios 1 e 3 tem o mesmo sentido e que, dois fios condutores retil\u00edneos, paralelos e pr\u00f3ximos um do outro, sofrem for\u00e7as de atra\u00e7\u00e3o se as correntes que os percorrem tiverem mesmo sentido.<\/b><\/span><\/span><\/p>\n

08- Correta\u00a0 — observe na express\u00e3o\u00a0F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u03bc.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.L\/2.\u03c0.d que fornece a intensidade da for\u00e7a magn\u00e9tica entre dois fios longos e pr\u00f3ximos, separados pela dist\u00e2ncia d, que F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0\u00e9 inversamente proporcional \u00e0 d\u00a0 — como a dist\u00e2ncia entre 2 e 3 \u00e9 menor que a dist\u00e2ncia entre 1 e 3, a afirmativa est\u00e1 correta.<\/b><\/span><\/span><\/p>\n

16- Correta\u00a0 —\u00a0 observe atentamente a seq\u00fc\u00eancia das figuras abaixo\u00a0 —\u00a0 como B<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0> B<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>, o campo magn\u00e9tico resultante aponta para<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

o sudoeste.<\/b><\/span><\/span><\/p>\n

32- Falsa\u00a0 —\u00a0 aponta para o nordeste\u00a0 —\u00a0 veja a figura abaixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- (02 + 08 + 16)=26<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A figura abaixo mostra o campo magn\u00e9tico produzido pela corrente i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e a for\u00e7a de atra\u00e7\u00e3o que tal campo produz no fio onde<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

circula i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 campo\u00a0\"\"\u00a0produzido por i<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>no fio 2 (regra da m\u00e3o direita) —\u00a0 B<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03bco.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2\u03c0r (I)\u00a0 —\u00a0 for\u00e7a de atra\u00e7\u00e3o\u00a0\"\"\u00a0sobre o fio 2, fornecida pela regra da m\u00e3o esquerda\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=Bi\u2113sen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=Bi<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2113 (II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u2113\/2\u03c0r\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\/\u2113= \u03bc<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2\u03c0r\u00a0 —\u00a0 aplicando a lei de Ohm V=R.i\u00a0 —\u00a0 fio 1\u00a0 —\u00a0 1=1.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1 A\u00a0 —\u00a0 fio 2\u00a0 —\u00a0 1=2i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,5 A\u00a0 —\u00a0 substituindo os valores na equa\u00e7\u00e3o F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\/\u2113= \u03bc<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2\u03c0r\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\/\u2113=4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>.1.0,5\/2\u03c0.0,1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\/\u2113=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span><\/p>\n

b) Express\u00e3o do efeito fotoel\u00e9trico\u00a0 —\u00a0 W<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=h.f<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 h.f<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u2013 energia que um f\u00f3ton transporta\u00a0 —\u00a0 W<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 fun\u00e7\u00e3o trabalho do material\u00a0 —\u00a0 v=c=\u03bb<\/b><\/span><\/span>o.<\/b><\/span><\/span><\/sub>f<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 f<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=c\/\u03bb<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 W<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=h.c\/\u03bb<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3ev=6.10<\/b><\/span><\/span>-34<\/b><\/span><\/span><\/sup>.3.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\/\u03bb<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3×1,6.10<\/b><\/span><\/span>-19<\/b><\/span><\/span><\/sup>=18.10<\/b><\/span><\/span>-26<\/b><\/span><\/span><\/sup>\/\u03bb<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 \u03bb<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3,75.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03bb < 3,7.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/p>\n

 <\/p>\n

36-<\/b><\/span><\/span><\/span> <\/b>Um dos processos pr\u00e1ticos para se determinar a dire\u00e7\u00e3o e o sentido do vetor indu\u00e7\u00e3o magn\u00e9tica\u00a0\"\"\u00a0ou vetor campo magn\u00e9tico\"\", \u00e9 a regra da m\u00e3o direita. Esse sentido de\"\"\u00a0depende do sentido da corrente que o origina.<\/b><\/span><\/span><\/p>\n

Voc\u00ea coloca o polegar no sentido da corrente com a m\u00e3o espalmada (primeira figura), em seguida <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

voc\u00ea fecha a m\u00e3o para pegar o\u00a0fio (segunda figura) e o sentido da \u201cfechada\u201d de m\u00e3o \u00e9 o sentido do vetor\"\"\u00a0(terceira figura). Observe na terceira figura que\u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O p\u00f3lo norte de uma b\u00fassola indica sempre o sentido das linhas de indu\u00e7\u00e3o fornecido pela regra da m\u00e3o direita.<\/b><\/span><\/span><\/p>\n

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

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Um dos processos pr\u00e1ticos para se determinar a dire\u00e7\u00e3o e o sentido do vetor indu\u00e7\u00e3o magn\u00e9tica\u00a0\"\"\u00a0ou vetor campo magn\u00e9tico\"\", \u00e9 a regra da m\u00e3o direita\u00a0 —\u00a0 esse sentido de\"\"\u00a0depende do sentido da corrente que o origina\u00a0 —\u00a0 voc\u00ea coloca o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

polegar no sentido da corrente com a\u00a0m\u00e3o espalmada (primeira figura), em seguida voc\u00ea fecha a m\u00e3o para pegar o fio (segunda figura) e o sentido da \u201cfechada\u201d de m\u00e3o \u00e9 o sentido do vetor\"\"\u00a0(terceira figura)\u00a0 —\u00a0 observe na terceira figura que\u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto\u00a0 —\u00a0 no caso do exerc\u00edcio, utilizando essa regra os campos magn\u00e9ticos gerados ter\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

os aspectos indicados na figura e, se voc\u00ea observ\u00e1-los atentamente verificar\u00e1 que a \u00fanica alternativa correta \u00e9 a D.<\/b><\/span><\/span><\/p>\n

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

38-(UEPA)<\/b><\/span><\/span><\/span><\/p>\n

Campo magn\u00e9tico\"\"\u00a0criado por um fio condutor reto percorrido por corrente el\u00e9trica\u00a0 —\u00a0 na figura 2, o fio condutor da figura 1 est\u00e1 colocado perpendicularmente \u00e0 folha a esta folha de papel.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O sentido do campo depende do sentido da corrente no fio. A agulha da b\u00fassola se alinha com esse campo.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0Um dos processos pr\u00e1ticos para se determinar a dire\u00e7\u00e3o e o sentido do vetor indu\u00e7\u00e3o magn\u00e9tica\u00a0\"\"\u00a0ou vetor campo magn\u00e9tico\"\", \u00e9 a regra da m\u00e3o direita. Esse sentido de\"\"\u00a0depende do sentido da corrente que o origina.<\/b><\/span><\/span><\/p>\n

Voc\u00ea coloca o polegar no sentido da corrente com a m\u00e3o espalmada (primeira figura), em seguida voc\u00ea fecha a m\u00e3o para pegar o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0fio\u00a0(segunda figura) e o sentido da \u201cfechada\u201d de m\u00e3o \u00e9 o sentido do vetor\"\"\u00a0(terceira figura). Observe na terceira figura que\u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto.<\/b><\/span><\/span><\/p>\n

Comprova-se experimentalmente que a intensidade do campo magn\u00e9tico\"\"\u00a0depende da intensidade da corrente el\u00e9trica i, da dist\u00e2ncia r do fio at\u00e9 o ponto (P) onde se quer o campo magn\u00e9tico e do meio onde o condutor se encontra. Essa depend\u00eancia de\"\"\u00a0com o meio \u00e9 fornecida pela constante \u03bc que recebe o nome de permeabilidade magn\u00e9tica do meio e no v\u00e1cuo ela vale\u00a0\u03bc<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>T.m\/A. Matematicamente:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio\u00a0 —\u00a0 \u03bc==4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>T.m\/A\u00a0 —\u00a0 r=16m\u00a0 —\u00a0 i=2,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0A\u00a0 —\u00a0 B=\u03bc.i\/2\u03c0r=4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>.2,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/2\u03c0.16\u00a0 —\u00a0 B=(2,4\/8).10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>T\u00a0 —\u00a0 B=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>T\u00a0 —\u00a0 B\/B<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0 B\/B<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Campo magn\u00e9tico originado por um condutor retil\u00edneo extenso percorrido por corrente el\u00e9trica 01-\u00a0Usando a regra da m\u00e3o direita\u00a0 —\u00a0 R- D 02-\u00a0Observe na express\u00e3o B=\u03bc.i\/2\u03c0r que B \u00e9 inversamente proporcional a r\u00a0 —\u00a0\u00a0R- E 03-\u00a0B=\u03bci\/2\u03c0r=4\u03c0.10-7.5\/2\u03c00,2\u00a0 —\u00a0 B=5,0.10-6T\u00a0 —\u00a0 regra da m\u00e3o direita\u00a0 —\u00a0 penetrando na folha de papel\u00a0 —\u00a0\u00a0R- E 04-\u00a0O polo norte<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1914,"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-1918","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1918","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=1918"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1918\/revisions"}],"predecessor-version":[{"id":10571,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1918\/revisions\/10571"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1914"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1918"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}