{"id":1932,"date":"2016-02-11T23:30:10","date_gmt":"2016-02-11T23:30:10","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1932"},"modified":"2024-08-20T10:12:33","modified_gmt":"2024-08-20T10:12:33","slug":"forca-magnetica-sobre-uma-carga-movel-imersa-num-campo-magnetico","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletromagnetismo\/forca-magnetica-sobre-uma-carga-movel-imersa-num-campo-magnetico\/","title":{"rendered":"For\u00e7a magn\u00e9tica sobre uma carga m\u00f3vel imersa num campo magn\u00e9tico"},"content":{"rendered":"

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

\u00a0Quando uma <\/b><\/span><\/span><\/span>carga el\u00e9trica<\/b><\/span><\/span><\/span> que se move <\/b><\/span><\/span><\/span>com velocidade\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0no <\/b><\/span><\/span><\/span>interior de um campo magn\u00e9tico\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

sobre ela surge uma for\u00e7a de origem magn\u00e9tica\u00a0\"\"\u00a0(<\/b><\/span><\/span><\/span> denominada for\u00e7a de Lorentz), <\/b><\/span><\/span><\/span>com as seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Dire\u00e7\u00e3o e sentido de\u00a0\"\" \"\" fornecidos pela regra da m\u00e3o esquerda<\/b><\/span><\/span><\/span> conforme mostrado na <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura acima.<\/b><\/span><\/span><\/span><\/p>\n

Observe na <\/b><\/span><\/span><\/span>figura da direita que\u00a0\"\"\u00a0\u00e9 perpendicular a\u00a0\"\"\u00a0e a\u00a0\"\",<\/b><\/span><\/span><\/span> o que <\/b><\/span><\/span><\/span>imp\u00f5e a condi\u00e7\u00e3o de que\u00a0\"\" e\"\" <\/b><\/span><\/span><\/span>devem <\/b><\/span><\/span><\/span>pertencer a um<\/b><\/span><\/span><\/span> <\/b>mesmo plano.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Observe tamb\u00e9m que <\/b><\/span><\/span><\/span>\u03b8 \u00e9 o \u00e2ngulo entre\u00a0\"\" e\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Conhecendo a dire\u00e7\u00e3o e o sentido de\u00a0\"\" e de\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>voc\u00ea pode, pela <\/b><\/span><\/span><\/span>regra da m\u00e3o esquerda,<\/b><\/span><\/span><\/span> <\/b>determinar a dire\u00e7\u00e3o e o sentido da for\u00e7a magn\u00e9tica\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0como voc\u00ea pode observar no <\/b><\/span><\/span><\/span>esquema abaixo,<\/b><\/span><\/span><\/span> onde uma <\/b><\/span><\/span><\/span>carga positiva<\/b><\/span><\/span><\/span> move-se <\/b><\/span><\/span><\/span>com velocidade\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0no <\/b><\/span><\/span><\/span>interior<\/b><\/span><\/span><\/span> de um <\/b><\/span><\/span><\/span>campo magn\u00e9tico uniforme\u00a0.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que a <\/b><\/span><\/span><\/span>dire\u00e7\u00e3o e sentido das linhas de indu\u00e7\u00e3o magn\u00e9tica e consequentemente de\u00a0\"\"\u00a0\u00e9 do<\/b><\/span><\/span><\/span> <\/b>polo norte para o polo sul.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Adaptando <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>indicador e o m\u00e9dio nas dire\u00e7\u00f5es de\u00a0\"\" e de\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0voc\u00ea <\/b><\/span><\/span><\/span>determina a dire\u00e7\u00e3o e sentido de\"\".<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o:<\/b><\/span><\/span><\/span> Se <\/b><\/span><\/span><\/span>q for negativa voc\u00ea deve inverter o sentido da for\u00e7a magn\u00e9tica\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

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

O que voc\u00ea deve saber, informa\u00e7\u00f5es e dicas<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como a <\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span> \"\" tem sempre <\/b><\/span><\/span><\/span>dire\u00e7\u00e3o perpendicular ao vetor velocidade<\/b><\/span><\/span><\/span> \"\" e como a <\/b><\/span><\/span><\/span>pot\u00eancia de uma for\u00e7a<\/b><\/span><\/span><\/span> \u00e9 fornecida por <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V.cos\u03b8,<\/b><\/span><\/span><\/span> ent\u00e3o \u03b8 = 90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0e cos 90<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sup><\/span>= 0\u00a0\"\" P<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.V.0\u00a0 \"\"\u00a0 P<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= 0 \"\"\u00a0 se a <\/b><\/span><\/span><\/span>pot\u00eancia \u00e9 nula o trabalho tamb\u00e9m ser\u00e1<\/b><\/span><\/span><\/span>, pois P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = W\/\u2206t \"\" 0 = W\/\u2206t \"\" <\/b><\/span><\/span><\/span><\/p>\n

W = 0.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

O trabalho realizado pela for\u00e7a magn\u00e9tica \u00e9 sempre nulo ou, a for\u00e7a magn\u00e9tica nunca realiza trabalho.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Conven\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Alguns exemplos:<\/b><\/span><\/span><\/span><\/p>\n

Nas <\/b><\/span><\/span><\/span>figuras <\/b><\/span><\/span><\/span>abaixo est\u00e3o <\/b><\/span><\/span><\/span>representadas<\/b><\/span><\/span><\/span> as for\u00e7as <\/b><\/span><\/span><\/span>magn\u00e9ticas e a disposi\u00e7\u00e3o final da m\u00e3o esquerda.<\/b><\/span><\/span><\/span> Observe que nos <\/b><\/span><\/span><\/span>casos b e c<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica \"\"<\/b><\/span><\/span><\/span> tem <\/b><\/span><\/span><\/span>sentido oposto<\/b><\/span><\/span><\/span> ao dado pela <\/b><\/span><\/span><\/span>regra da m\u00e3o esquerda, pois a carga el\u00e9trica \u00e9 negativa.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

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

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

\u00a0Voc\u00ea j\u00e1 viu que as <\/b><\/span><\/span><\/span>principais caracter\u00edsticas da for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span> \"\" que atua sobre uma <\/b><\/span><\/span><\/span>carga el\u00e9trica imersa num campo magn\u00e9tico tem as seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Dire\u00e7\u00e3o e sentido de<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0 \"\" fornecidos pela <\/b><\/span><\/span><\/span>regra da m\u00e3o esquerda <\/b><\/span><\/span><\/span>conforme <\/b><\/span><\/span><\/span>mostrado na<\/b><\/span><\/span><\/span> <\/b>figura abaixo.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\"\"<\/p>\n

Observe na <\/b><\/span><\/span><\/span>figura da direita<\/b><\/span><\/span><\/span> que\u00a0\"\"\u00a0\u00e9 <\/b><\/span><\/span><\/span>perpendicular a\u00a0\"\"\u00a0e a\u00a0\"\"<\/b><\/span><\/span><\/span>, o que imp\u00f5e a condi\u00e7\u00e3o de que\u00a0\"\" e\"\"devem pertencer a um mesmo plano. Observe tamb\u00e9m que \u03b8 \u00e9 o \u00e2ngulo entre\u00a0\"\" e\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Intensidade de\u00a0\"\" \"\"<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>proporcional a q, V, B<\/b><\/span><\/span><\/span> e ao <\/b><\/span><\/span><\/span>sen\u03b8,<\/b><\/span><\/span><\/span> obedecendo \u00e0 equa\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C<\/b><\/span><\/span><\/span>arga em repouso (V = 0) ou lan\u00e7ada com velocidade\u00a0\"\" paralelamente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme \"\"<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

Observe que, <\/b><\/span><\/span><\/span>neste caso<\/b><\/span><\/span><\/span> o \u00e2ngulo entre\u00a0\"\" e\u00a0\"\" \u00e9 <\/b><\/span><\/span><\/span>90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0(s\u00e3o perpendiculares) e que sen90<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sup><\/span>= 1.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

O que voc\u00ea deve saber, informa\u00e7\u00f5es e dicas<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observe na <\/b><\/span><\/span><\/span>express\u00e3o do per\u00edodo (T)<\/b><\/span><\/span><\/span> que ele <\/b><\/span><\/span><\/span>n\u00e3o depende da velocidade da part\u00edcula, nem do raio da circunfer\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

O trabalho realizado por\"\"\u00a0sobre q \u00e9 nulo<\/b><\/span><\/span><\/span>, pois\"\"<\/b><\/span><\/span><\/span>\u00a0\u00e9 perpendicular ao plano formado por\"\" e \"\"<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>W =<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d.cos90<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span> = F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d.0<\/b><\/span><\/span><\/span> <\/b>= 0<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Campos magn\u00e9tico e el\u00e9trico agindo sobre uma part\u00edcula ou feixe de part\u00edculas eletrizadas com carga q<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Confira os exerc\u00edcios com resolu\u00e7\u00e3o comentada<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

For\u00e7a magn\u00e9tica sobre uma carga m\u00f3vel imersa num campo magn\u00e9tico \u00a0Quando uma carga el\u00e9trica que se move com velocidade\u00a0\u00a0no interior de um campo magn\u00e9tico\u00a0\u00a0 sobre ela surge uma for\u00e7a de origem magn\u00e9tica\u00a0\u00a0( denominada for\u00e7a de Lorentz), com as seguintes caracter\u00edsticas: \u00a0Dire\u00e7\u00e3o e sentido de\u00a0 fornecidos pela regra da m\u00e3o esquerda conforme mostrado na figura acima. Observe na figura da direita<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":243,"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-1932","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1932","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=1932"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1932\/revisions"}],"predecessor-version":[{"id":1941,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1932\/revisions\/1941"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/243"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}