3 Gabarito Prova 3 { Eletromagnetismo I { Noturno ?· 3a Gabarito Prova 3 { Eletromagnetismo I { Noturno…

Download 3 Gabarito Prova 3 { Eletromagnetismo I { Noturno ?· 3a Gabarito Prova 3 { Eletromagnetismo I { Noturno…

Post on 24-Nov-2018

212 views

Category:

Documents

0 download

TRANSCRIPT

<ul><li><p>3a Gabarito Prova 3 Eletromagnetismo I Noturno</p><p>Q1 [4.0]</p><p>(a) [0.5]</p><p>~B(t, r, ) ' 0k2c</p><p>4</p><p>eikr</p><p>rr ~p = 0 k</p><p>2 c</p><p>4</p><p>eikr</p><p>rp0eitr </p><p>(p0ze</p><p>it)=0k</p><p>2cp04</p><p>ei(krt)</p><p>r(z cos + sen ) z</p><p>= 0k2cp0</p><p>4</p><p>ei(krt)</p><p>rsen </p><p>(b) [1.5]</p><p>~ ~B = 1r</p><p>[1</p><p>sen </p><p> (B sen )</p><p>r (rB)</p><p>r</p><p>]= 0k</p><p>2cp04</p><p>1</p><p>r</p><p>[ei(krt)</p><p>r2sen </p><p>(sen2 </p><p>)</p><p>r sen (ei(krt)</p><p>)r</p><p>]</p><p> i0k3cp0</p><p>4</p><p>ei(krt)</p><p>rsen </p><p>Entao</p><p>~E = c2</p><p>(~ ~B)dt</p><p>~E = c20k</p><p>3p04</p><p>ei(krt)</p><p>rsen </p><p>~E =0k</p><p>3c3p04</p><p>ei(krt)</p><p>rsen </p><p>(c) [1.0]</p><p>~Sp =1</p><p>0~E ~B</p><p>= 10</p><p>0k3c3p0</p><p>4</p><p>ei(krt)</p><p>rsen </p><p>0k2cp0</p><p>4</p><p>ei(krt)</p><p>rsen ( )</p><p>=0k</p><p>5c4p20162</p><p>e2i(krt)</p><p>r2sen2 r =</p><p>00</p><p>05c4p20</p><p>162c5</p><p>e2i(krt)</p><p>r2sen2 r</p><p>=4p20</p><p>1620c3e2i(krt)</p><p>r2sen2 r</p><p>(d) [1.0]</p><p>~Spt =1</p><p>T</p><p> T0</p><p>1</p><p>0Re[ ~E]Re[ ~B]dt</p><p>=4p20</p><p>1620c3sen2 </p><p>r2r</p><p>2</p><p> 2/0</p><p>cos2(kr t)dt</p><p>=4p20</p><p>3220c3sen2 </p><p>r2r</p><p>Q2 [3.0]</p><p>1</p></li><li><p>a) [1,0] Condicoes de contorno</p><p>~Etang1 = ~Etang2 (E1p E1p) cos 1 = E2p cos 2</p><p>~Htang1 = ~Htang2 n11c</p><p>(E1p + E1p) =</p><p>n22c</p><p>E2p</p><p>Segue que</p><p>E1p E1p =cos 2cos 1</p><p>E2p; E1p + E1p =</p><p>n21n12</p><p>E2p</p><p>entao</p><p>E1p =</p><p>n21n12</p><p> cos 2cos 1n21n12</p><p>+ cos 2cos 1</p><p>E1p; E2p =2</p><p>cos 2cos 1</p><p>+ n21n12</p><p>E1p</p><p>Da</p><p>rp =E1E1</p><p>=</p><p>n21n12</p><p> cos 2cos 1n21n12</p><p>+ cos 2cos 1</p><p>=</p><p>n22</p><p>cos 1 n11 cos 2n22</p><p>cos 1 +n11</p><p>cos 2</p><p>e</p><p>tp =E2E1</p><p>=2</p><p>cos 2cos 1</p><p>+ n21n12</p><p>=2n11 cos 1</p><p>n11</p><p>cos 2 +n22</p><p>cos 1</p><p>b) [0,5]</p><p>rp = 0n22</p><p>cos 1 n11</p><p>cos 2 = 0</p><p>(n21n12</p><p>)1 sin2 B </p><p>1 n</p><p>21</p><p>n22sin2 B = 0(</p><p>n21n12</p><p>)2(n12n22</p><p>)2sin2 B 1 +</p><p>n21n22</p><p>sin2 B = 0</p><p>sin2 B =</p><p>1(n21n12</p><p>)2n21n22(n12n21</p><p>)2c) [0,5]</p><p>tp =2</p><p>cos 2cos 1</p><p>+ n21n12</p><p>=2</p><p>1n22n1</p><p>+ n21n12=2n11n2</p><p>Os coeficientes associados a conservacao de energia sao o R e T (e nao r e p).</p><p>d) [0,5]</p><p>T =~S2 kt~S1 kt</p><p>=(2/v2)E</p><p>2p2 cos 2</p><p>(1/v1)E2p1 cos 1=2v11v2</p><p>t2pcos 2cos 1</p><p>=2v11v2</p><p>(2n11n2</p><p>)2 1n22n1</p><p>=22v1n111v2n2</p><p>n1n2</p><p>=v2v1</p><p>v1/c</p><p>v2/c= 1</p><p>2</p></li><li><p>e) [0,5]</p><p>rs = 0n11</p><p>cos 1 n22</p><p>cos 2 = 0(n12n21</p><p>)1 sin2 1 </p><p>1 n</p><p>21</p><p>n22sin2 1 = 0(</p><p>n21n12</p><p>)2(n12n22</p><p>)2sin2 1 1 +</p><p>n21n22</p><p>sin2 1 = 0</p><p>sin2 1 =</p><p>1(n21n12</p><p>)2n21n22(n21n12</p><p>)2</p><p>Q3 [3.0]</p><p>a) [0,5]</p><p> ~E = ~B</p><p>t ey</p><p>y[Eei(kyt)ez</p><p>]= ikEei(kyt)ex = </p><p> ~B</p><p>t</p><p> ~B</p><p>t= ikEei(kyt)ex ~B =</p><p>k</p><p>Eei(kyt)ex</p><p>b) [0,5] Para que a onda possa se propagar, o vetor de onda precisa ser real; como</p><p>k =</p><p>2 2pc2</p><p>entao e preciso que &gt; p.</p><p>c) [0,5]</p><p>(k2) =</p><p>(2 2pc2</p><p>)</p><p>2k</p><p>k = 2</p><p>c2 </p><p>k</p><p>k= vgvf = c</p><p>2</p><p>d)</p><p>~S =1</p><p>40</p><p>[~E ~B + ~E ~B</p><p>]=</p><p>1</p><p>40</p><p>[Eei(kyt)z k</p><p>Eei(kyt)x+ Eei(kyt)z k</p><p>Eei(kyt)x</p><p>]=</p><p>1</p><p>20</p><p>k</p><p>|E|2y</p><p>e) [0,5] Se &lt; p, entao,</p><p>k = i</p><p>2p 2</p><p>c2=i</p><p>; c</p><p>2p 2</p><p>e~E = Eei(kyt)z = Eeiey/z</p><p>3</p></li><li><p>f) [0,5]</p><p>~E = Eey/ei z ~B = i|k|Eey/zeitex</p><p>~S =1</p><p>40</p><p>[~E ~B + ~E ~B</p><p>]=</p><p>1</p><p>40</p><p>[Eeitey/ez (i)</p><p>|k|Eeitey/ex + E</p><p>eitey/ez i|k|Eeitey/ex</p><p>]= ey</p><p>1</p><p>20</p><p>|k||E|2e2y/[i+ i] = 0</p><p>4</p></li></ul>