{"id":1899,"date":"2016-02-11T22:53:03","date_gmt":"2016-02-11T22:53:03","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1899"},"modified":"2024-08-20T10:07:30","modified_gmt":"2024-08-20T10:07:30","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-magnetica-sobre-um-condutor-retilineo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletromagnetismo\/forca-magnetica-sobre-um-condutor-retilineo-imerso-num-campo-magnetico-uniforme\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-magnetica-sobre-um-condutor-retilineo\/","title":{"rendered":"For\u00e7a magn\u00e9tica sobre um condutor retil\u00edneo – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

For\u00e7a magn\u00e9tica sobre um condutor retil\u00edneo<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0Caso(a)\u00a0 —\u00a0 F(a)=B.i.L.sen45<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>=(\u221a2\/2).BiL\u00a0 —\u00a0 caso (b)\u00a0 —\u00a0 F(b)=BiL.sen90<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>=BiL\u00a0 —\u00a0 caso (c)\u00a0 —\u00a0 F(c)=BiLsen0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span> <\/b>R- B<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> O campo magn\u00e9tico\u00a0\"\"\u00a0\u00e9 dirigido do p\u00f3lo norte para o p\u00f3lo sul\u00a0 —\u00a0 nos trechos BC e AD a for\u00e7a magn\u00e9tica \u00e9 nula (B e i s\u00e3o paralelos)\u00a0 —\u00a0 usando a regra da m\u00e3o esquerda as for\u00e7as nos trechos AB (F<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>) e CD (F<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>) est\u00e3o indicadas na figura<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A intensidade das for\u00e7as que agem sobre BC e AD \u00e9 zero (B e i s\u00e3o paralelos, \u03b8=0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 Intensidade de F<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=BiLsen\u03b8=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10.0,2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=4,0.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>T\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Para a invers\u00e3o no sentido da for\u00e7a que atua sobre o fio deve-se inverter o sentido da corrente OU o sentido do campo magn\u00e9tico\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Para que a barra met\u00e1lica condutora PQ esteja em equil\u00edbrio, seu peso (vertical e para baixo) <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

deve ser anulado pela for\u00e7a magn\u00e9tica que deve ser vertical e para cima\u00a0 —\u00a0 usando a regra da m\u00e3o esquerda,\u00a0\"\"\u00a0est\u00e1 indicado na figura acima, saindo do p\u00f3lo norte e indo para o p\u00f3lo sul\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>X \u00e9 um p\u00f3lo norte e Y \u00e9 um p\u00f3lo sul.<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0 C\u00e1lculo da intensidade da corrente el\u00e9trica no nos fios e no condutor cil\u00edndrico\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,10 + 0,02\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,12\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 0,12=4,8\/i\u00a0 —\u00a0 i=4,8\/0,12\u00a0 —\u00a0 i=40 A\u00a0 —\u00a0 como o condutor cil\u00edndrico fica em equil\u00edbrio, a for\u00e7a magn\u00e9tica F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0deve ser vertical e para cima para equilibrar o peso que \u00e9 vertical e para baixo\u00a0 — F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 BiLsen\u03b8=mg\u00a0 —\u00a0 B.40.0,10.1=0,005.10\u00a0 —\u00a0 B=0,05\/4\u00a0 —\u00a0 B=0,0125T\u00a0 —<\/b><\/span><\/span>\u00a0 R- A\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0For\u00e7a magn\u00e9tica vertical e para baixo\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=BiLsen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=0,1.10.0,4.1\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=0,4N\u00a0 —\u00a0 for\u00e7a gravitacional vertical e para baixo\u00a0 —\u00a0 P=mg=0,03.10\u00a0 —\u00a0 P=0,3N\u00a0 —\u00a0 for\u00e7a el\u00e1stica vertical e para cima\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=k.x\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=10x\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0+ P = F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

0,4 + 0,3=10x\u00a0 —\u00a0 x=0,7\/10=0,07m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0x=7,0cm<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0a) Se a chave est\u00e1 aberta n\u00e3o passa corrente pelo fio e a for\u00e7a magn\u00e9tica sobre ele \u00e9 nula e assim, a for\u00e7a de tra\u00e7\u00e3o no fio condutor r\u00edgido (indica\u00e7\u00e3o do dinam\u00f4metro) \u00e9 igual ao peso do fio\u00a0 —\u00a0 P=m.g=0,2.10\u00a0 —\u00a0 o dinam\u00f4metro indica\u00a0<\/b><\/span><\/span>T=2,0N<\/b><\/span><\/span><\/p>\n

b) Com a chave fechada, circula no fio condutor r\u00edgido uma corrente el\u00e9trica i e surge sobre o fio uma for\u00e7a magn\u00e9tica para cima, que deve anular o peso do fio, pois o dinam\u00f4metro indica zero\u00a0 — F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 i e B s\u00e3o perpendiculares \u2013 \u03b8=90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 B.i.\u2113.sen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=2\u00a0 —\u00a0 1.i.0,2=2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>i=10A<\/b><\/span><\/span>\u00a0 —\u00a0 regra da m\u00e3o esquerda F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0para cima, B entrando,<\/b><\/span><\/span>\u00a0i deve ter dire\u00e7\u00e3o horizontal e sentido para a direita.<\/b><\/span><\/span><\/p>\n

c) R=U\/i\u00a0 —\u00a0 6=U\/10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=60V<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Com a tens\u00e3o el\u00e9trica desligada a for\u00e7a el\u00e1stica aplicada por cada mola deve equilibrar o peso do fio\u00a0 —\u00a0 2F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 2.k.x=P\u00a0 —\u00a0 2.5.2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 P=20.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=0,02N<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja o esquema abaixo onde foi usada a regra da m\u00e3o esquerda<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como as intensidades de B e de i s\u00e3o constantes, ent\u00e3o quem varia \u00e9 o comprimento L de cada trecho do fio, tracionados\u00a0 por for\u00e7as diferentes, as quais, em cada trecho tem a dire\u00e7\u00e3o e sentido fornecidos pela regra da m\u00e3o esquerda e esquematizados na<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Bi4u\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=Bi3u\u00a0 —\u00a0 F<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=Bi3u\u00a0 —\u00a0 F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=Bi2u\u00a0 —\u00a0 F<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=Biu\u00a0 —\u00a0 na dire\u00e7\u00e3o X\u00a0 —\u00a0 F<\/b><\/span><\/span>RX<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0\u2013F<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=4Biu \u2013 3Biu \u2013 Biu\u00a0 —\u00a0 F<\/b><\/span><\/span>RX<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 na dire\u00e7\u00e3o Y\u00a0 —\u00a0 F<\/b><\/span><\/span>RY<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=3Biu \u2013 2Biu\u00a0 —\u00a0 F<\/b><\/span><\/span>RY<\/b><\/span><\/span><\/sub>=Biu\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Quando o gerador \u00e9 ligado, com a chave A aberta e a chave C fechada, a corrente el\u00e9trica circula apenas no trecho OO\u2019P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 quando a massa M=0,008kg \u00e9 pendurada no ponto K, meio do segmento P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>, para o sistema permanecer em equil\u00edbrio , a for\u00e7a magn\u00e9tica F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0deve equilibrar o peso da massa M=0,008kg. ou seja, dever\u00e1 ser vertical e para cima (veja figura) com<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

dire\u00e7\u00e3o e sentido fornecidos pela regra da m\u00e3o esquerda —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,008.10\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,08N<\/b><\/span><\/span><\/p>\n

b) Ainda nas condi\u00e7\u00f5es anteriores \u00a0—\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=BiL\u00a0 —\u00a0 0,08=B.2.0,20\u00a0 —<\/b><\/span><\/span>\u00a0 B=0,2T<\/b><\/span><\/span><\/p>\n

c) Fechando a chave A e abrindo a chave C a corrente el\u00e9trica circular\u00e1 nos trechos P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>, com as for\u00e7as\u00a0\"\"\u00a0e\u00a0 –\"\"\u00a0colocadas nos pontos m\u00e9dios K (de P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>) e N (de P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) e que foram obtidas pela regra da m\u00e3o esquerda\u00a0 —\u00a0 nessas condi\u00e7\u00f5es a espira tender\u00e1 a girar em torno de OO\u2019\u00a0 —\u00a0\u00a0 como nos trechos P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0B, L e i s\u00e3o os mesmos, as intensidades de\u00a0\"\"\u00a0, –\"\"\u00a0e\u00a0\"\"\u00a0s\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

iguais e valem F=0,08N\u00a0 —\u00a0 para que se equilibre o bin\u00e1rio (\"\"\u00a0e\u00a0 –\"\") voc\u00ea deve provocar um torque no sentido oposto, ou seja,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0deve pendurar no ponto K uma massa M<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>\u00a0que seja o dobro de M1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,16kg\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 (veja figura)<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0A corrente i no circuito principal (no sentido anti-hor\u00e1rio) faz surgir na barra condutora uma for\u00e7a magn\u00e9tica de intensidade F, vertical e para baixo (fornecida pela regra da m\u00e3o esquerda) que comprime para baixo as molas, estabelecendo contato com o circuito secund\u00e1rio anexo, fechando o circuito e acendendo a l\u00e2mpada\u00a0 —\u00a0 a for\u00e7a eletromotriz m\u00ednima E faz com que as for\u00e7as el\u00e1sticas das molas e a for\u00e7a magn\u00e9tica se equilibrem\u00a0 —\u00a0 for\u00e7a magn\u00e9tica\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=BiLsen\u03b8=BiLsen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=BiL\u00a0 —\u00a0 for\u00e7a el\u00e1stica de uma mola\u00a0 —\u00a0 F\u2019<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=kx\u00a0 —\u00a0 como s\u00e3o duas molas F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>= 2F\u2019<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=2kx\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=2kx\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 BiL=2kh (I) \u00a0—\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

2R=E\/i\u00a0 —\u00a0 i=E\/2R (II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 B(E\/2R)L=2kh\u00a0 —\u00a0 0,01.E.2.0,05.0,3=2.2.0,03\u00a0 —\u00a0 E=4,0V<\/b><\/span><\/span>\u00a0 —\u00a0 R- E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Quando o m\u00fasculo card\u00edaco se contrai surge uma corrente el\u00e9trica no fio\u00a0 —\u00a0 no caso, i saindo do plano da p\u00e1gina\u00a0 — o campo magn\u00e9tico\u00a0\"\"\u00a0\u00e9 vertical e para baixo\u00a0 —\u00a0 regra da m\u00e3o esquerda\u00a0 —\u00a0 a for\u00e7a magn\u00e9tica\u00a0\"\"\u00a0ser\u00e1 horizontal e para a direita\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=BiL (direta mente proporcional a i)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Utilizando a regra da m\u00e3o esquerda\u00a0 —\u00a0 indicador \u2013 indica o campo magn\u00e9tico\u00a0\"\", horizontal para a direita (do p\u00f3lo norte para<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0o p\u00f3lo sul)\u00a0 —\u00a0 m\u00e9dio \u2013 indica a corrente el\u00e9trica i (entrando na folha de papel)\u00a0 —\u00a0 polegar \u2013 indica a for\u00e7a magn\u00e9tica\u00a0\"\"\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

observe na figura que o polegar est\u00e1 indicando que\u00a0\"\"\u00a0\u00e9 vertical e para baixo<\/b><\/span><\/span><\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe no exerc\u00edcio anterior que a for\u00e7a magn\u00e9tica\u00a0\"\"\u00a0que age sobre o fio \u00e9 vertical e para baixo, juntamente com o peso\u00a0\"\"\u00a0\u00a0—a for\u00e7a el\u00e1stica\u00a0\"\"\u00a0age sobre o fio puxando-o para cima\u00a0 —\u00a0 intensidade de\u00a0\"\"\u00a0\u00a0—\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=B.i.\u2113.sen\u03b8=5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.4.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.2.sen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=40.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.1\u00a0 — \u00a0F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=0,4N\u00a0 —\u00a0 intensidade do peso\u00a0\"\"\u00a0\u00a0—\u00a0 P=mg=6.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 P=6N\u00a0 —\u00a0 a intensidade da for\u00e7a resultante sobre o fio, vertical e para baixo vale\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=6.0 + 0,4\u00a0 —\u00a0 F\u2019<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=6,4N\u00a0 —\u00a0 a for\u00e7a el\u00e1stica\u00a0\"\", vertical e para cima vale\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=kx\u00a0 — \u00a0F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.x\u00a0 —\u00a0 como o sistema est\u00e1 em equil\u00edbrio a intensidade da for\u00e7a resultante \u00e9 nula, ou seja\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=F\u2019<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>x=6,4\u00a0 —\u00a0 x=6,4\/2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 x=3,2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

17-<\/b><\/span><\/span><\/span> a) Como a barra met\u00e1lica possui resist\u00eancia \u00f4hmica R=5\u2126 e o gerador resist\u00eancia \u00f4hmica interna r=5\u2126 e eles est\u00e3o associados em s\u00e9rie, a resist\u00eancia equivalente\u00a0 vale R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5 + 5=10\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 10 = U\/5\u00a0 —\u00a0 U=\u03b5=50V\u00a0 —\u00a0 pot\u00eancia el\u00e9trica dissipada pela barra met\u00e1lica\u00a0 —\u00a0 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>barra<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 5.5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=125W<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) Conhecendo a dire\u00e7\u00e3o e sentido do campo magn\u00e9tico (perpendicular \u00e0 folha de papel e penetrando nela) e da corrente<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

el\u00e9trica i (horizontal e para a direita), utilizando a regra da m\u00e3o esquerda voc\u00ea localiza a for\u00e7a magn\u00e9tica (vertical e para cima \u2013 veja figura acima)\u00a0 —\u00a0 a intensidade dessa for\u00e7a magn\u00e9tica vale\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=B.i.\u2113.sen\u03b8=0,4.5.0,5.sen90<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=1.1\u00a0 —\u00a0 F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1N\u00a0 —\u00a0 sobre a barra agem as for\u00e7as\u00a0 —\u00a0 for\u00e7a magn\u00e9tica\u00a0\"\"\u00a0(vertical e para cima), for\u00e7a el\u00e1stica\u00a0\"\"\u00a0aplicada pela mola (vertical e para cima) e a for\u00e7a peso\u00a0\"\"\u00a0(vertical e para baixo)\u00a0 —\u00a0 como o enunciado afirma que o sistema est\u00e1 em equil\u00edbrio, a intensidade da for\u00e7a resultante\u00a0\"\"\u00a0sobre a barra deve ser nula\u00a0 —\u00a0 assim, F<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/sub>+ F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0= P\u00a0 —\u00a0\u00a0 k.x + 1 = m.g\u00a0 —\u00a0 80.x + 1 = 5\u00a0 —\u00a0 x=4\/80\u00a0 —\u00a0 x=0,05m ou\u00a0<\/b><\/span><\/span>x=5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre For\u00e7a magn\u00e9tica sobre um condutor retil\u00edneo 01- 02-\u00a0Caso(a)\u00a0 —\u00a0 F(a)=B.i.L.sen450=(\u221a2\/2).BiL\u00a0 —\u00a0 caso (b)\u00a0 —\u00a0 F(b)=BiL.sen900=BiL\u00a0 —\u00a0 caso (c)\u00a0 —\u00a0 F(c)=BiLsen0o=0\u00a0 —\u00a0\u00a0 R- B 03- O campo magn\u00e9tico\u00a0\u00a0\u00e9 dirigido do p\u00f3lo norte para o p\u00f3lo sul\u00a0 —\u00a0 nos trechos BC e AD a for\u00e7a magn\u00e9tica \u00e9 nula (B e i s\u00e3o paralelos)\u00a0 —\u00a0 usando<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1895,"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-1899","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1899","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=1899"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1899\/revisions"}],"predecessor-version":[{"id":10581,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1899\/revisions\/10581"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1895"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}