Electromagnetismo e Óptica (LEIC)
1ºSem. 2007/2008
http://fisica.ist.utl.pt/~leic_f2

Definições iniciais

Definições Globais

SetDirectory[DirectoryName["WindowTitle"/.NotebookInformation[InputNotebook[]]]];

If[Cases[Contexts[],"Notation`"]=={},Get["Notation`"],NotebookClose[Notebooks[]//Cases[#,z_NotebookObject/;StringMatchQ[ToString[z],"*Notation*"]][[1]]&]]

SetOptions[SelectedNotebook[],WindowSize->{1400,900}];

({{Symbolize[(Subscript[m, _])], Symbolize[(Subscript[(OverVector[u]), _]) ], Symbolize[(Subscript[(OverVector[v]), _])], Symbolize[(_)'], Symbolize[(OverVector[R])], Symbolize[(OverVector[r])], Symbolize[(OverVector[v])], Symbolize[(OverVector[V])], Symbolize[(Subscript[(OverVector[𝕖]), _])]}, {Symbolize[(Subscript[ϵ, o])], Symbolize[(Subscript[μ, o])], Symbolize[(OverVector[E])], Symbolize[(OverVector[F])], Symbolize[(OverVector[B])], Symbolize[(Subscript[(OverVector[e]), _])], \[Placeholder], \[Placeholder], \[Placeholder]}});

SetOptions[SelectedNotebook[],PageFooters->{{Cell[TextData[{"© ",ValueBox["Date"]}],"PageNumber"],Cell[TextData[{"-",CounterBox["Page"],"-"}],"PageNumber"],Cell[TextData[{"© ","A. Rica da Silva, Prof. IST"}],"Header"]},{Cell[TextData[{"© ","A. Rica da Silva,Prof. IST"}],"Header"],Cell[TextData[{"-",CounterBox["Page"],"-"}],"PageNumber"],Cell[TextData[{ValueBox["Date"]}],"Header"]}}]

({{AddInputAlias["p1"->(Subscript[(OverVector[p]), 1])], AddInputAlias["p2"->(Subscript[(OverVector[p]), 2])], AddInputAlias["gP"->P], AddInputAlias["m1"->(Subscript[m, 1])], AddInputAlias["m2"->(Subscript[m, 2])]}, {AddInputAlias[(Subscript[(OverVector[v]), o]),"vo"], AddInputAlias[(Subscript[(OverVector[v]), 1]),"v1"], AddInputAlias[(Subscript[(OverVector[u]), o]),"uo"], AddInputAlias[(Subscript[(OverVector[u]), 1]),"u1"], AddInputAlias[(Subscript[(OverVector[R]), cm]),"R"]}, {AddInputAlias[(Subscript[(OverVector[V]), cm]),"V"], AddInputAlias[(Subscript[(OverVector[r]), 1]),"r1"], AddInputAlias[(Subscript[(OverVector[r]), 2]),"r2"], AddInputAlias["r3"->(Subscript[(OverVector[r]), 3])], AddInputAlias["kc"->(1/(4π (Subscript[ϵ, o])))]}, {AddInputAlias["ex"->(Subscript[(OverVector[e]), x])], AddInputAlias["ey"->(Subscript[(OverVector[e]), y])], AddInputAlias["ez"->(Subscript[(OverVector[e]), z])], AddInputAlias["be"->(OverVector[E])], AddInputAlias["bf"->(OverVector[F])]}, {AddInputAlias["q1"->(Subscript[q, 1])], AddInputAlias["q2"->(Subscript[q, 2])], AddInputAlias["q3"->(Subscript[q, 3])], AddInputAlias["br"->(Subscript[(OverVector[r]), \[SelectionPlaceholder]])], AddInputAlias["0"->(OverVector[O])]}, {AddInputAlias["eo"->(Subscript[ϵ, o])], AddInputAlias["er"->(Subscript[(OverVector[e]), r])], AddInputAlias["ef"->(Subscript[(OverVector[e]), φ])], AddInputAlias["eq"->(Subscript[(OverVector[e]), θ])], AddInputAlias["bb"->(OverVector[B])]}, {AddInputAlias["mo"->(Subscript[μ, o])], AddInputAlias["||"->\[LeftBracketingBar]\[SelectionPlaceholder]\[RightBracketingBar]], AddInputAlias["km"->((Subscript[μ, o])/(4π))], \[Placeholder], \[Placeholder]}});

SIUnits=(Piecewise[{{(Subscript[q, e])-> -1.602176462 ×(10^(-19)), C                    (* Carga do Electrão.; *)}, {(Subscript[ϵ, o])->8.854187817×(10^(-12)), ((A s)/(m V))            (* Permitividade  eléctrica do vácuo.; *)}, {(1/(4π (Subscript[ϵ, o])))-> (10^(-7))×(c^2),  ((m V)/(A s))           (*≈ 9×(10^9)  Constante Eléctrica.; *)}, {c->299792458, (m/s)                 (* Velocidade da luz  no vácuo.; *)}, {(Subscript[μ, o])->4π×(10^(-7)), (N/(A^2))                (* Permeabilidade magnética do vácuo.;*)}}]) //#[[1]]\[Transpose][[1]]&;

Definições Gráficas

As minhas definições gráficas:

RotationMatrix3D[phi_,theta_,psi_]:=(
{{Cos[psi] Cos[phi]-Cos[theta] Sin[phi] Sin[psi],Cos[psi] Sin[phi]+Cos[theta] Cos[phi] Sin[psi],Sin[psi] Sin[theta]},{-Sin[psi] Cos[phi]-Cos[theta] Sin[phi] Cos[psi],-Sin[psi] Sin[phi]+Cos[theta] Cos[phi] Cos[psi],Cos[psi] Sin[theta]},{Sin[theta] Sin[phi],-Sin[theta] Cos[phi],Cos[theta]}})

TransformGraphics3D[obj_,f_]:=(tg3D[obj,f])

cuboidtopolygons[Cuboid[{xmin_,ymin_,zmin_}]]:=cuboidtopolygons[Cuboid[{xmin,ymin,zmin},{xmin,ymin,zmin}+1]]

cuboidtopolygons[Cuboid[{xmin_,ymin_,zmin_},{xmax_,ymax_,zmax_}]]:={Polygon[{{xmax,ymax,zmax},{xmin,ymax,zmax},{xmin,ymin,zmax},{xmax,ymin,zmax}}],Polygon[{{xmax,ymax,zmax},{xmax,ymin,zmax},{xmax,ymin,zmin},{xmax,ymax,zmin}}],Polygon[{{xmax,ymax,zmax},{xmax,ymax,zmin},{xmin,ymax,zmin},{xmin,ymax,zmax}}],Polygon[{{xmin,ymax,zmax},{xmin,ymax,zmin},{xmin,ymin,zmin},{xmin,ymin,zmax}}],Polygon[{{xmin,ymin,zmin},{xmin,ymax,zmin},{xmax,ymax,zmin},{xmax,ymin,zmin}}],Polygon[{{xmin,ymin,zmax},{xmin,ymin,zmin},{xmax,ymin,zmin},{xmax,ymin,zmax}}]}

tg3D[HoldPattern[Graphics3D][g_,o___],f_,p_:0]:=Graphics3D[tg3D[g,f,p],o]

tg3D[d_List,f_,p_:0]:=Map[tg3D[#,f,p]&,d]

tg3D[GraphicsGroup[a_,b___],f_,p_:0]:=GraphicsGroup[tg3D[a,f,p],b]

tg3D[HoldPattern[GraphicsComplex][pts_,prims_,o___],f_,p_:0]:=GraphicsComplex[If[VectorQ[pts],pts,Map[f,pts]],tg3D[prims,f,Length[pts]]]

tg3D[Point[d],f_,p_:0]/;(p===0||!gcq[d,p]):=Point[f[d]]

tg3D[Line[d_List],f_,p_:0]/;(p===0||!gcq[d,p]):=Line[Map[f,d,If[MatrixQ[d[[1]]],{2},{1}]]]

tg3D[Polygon[d_List],f_,p_:0]/;(p===0||!gcq[d,p]):=Polygon[Map[f,d,If[MatrixQ[d[[1]]],{2},{1}]]]

tg3D[Cylinder[d:{{_,_,_},{_,_,_}},r_:1,t___],f_,p_:0]/;numberQ[r]:=Cylinder[f/@d,r,t]

tg3D[Sphere[d_List,r_:1,t___],f_,p_:0]/;numberQ[r]:=Sphere[f[d],r,t]

tg3D[Text[expr_,d_List,opts___],f_,p_:0]:=Text[expr,f[d],opts]

tg3D[c:Cuboid[{_,_,_}],f_,p_:0]:=tg3D[cuboidtopolygons[c],f,p]

tg3D[c:Cuboid[{_,_,_},{_,_,_}],f_,p_:0]:=tg3D[cuboidtopolygons[c],f,p]

tg3D[expr_,f_,p_:0]:=expr

(*transform an entire primitive,not just coordinates.This may be being too clever than necessary... In this case,the transform function needs to have specific knowledge of multipoints,multilines,etc.and of what to do inside a GraphicsComplex;if the f is being applied inside a GC,it will be given two arguments (obj and pointlist),and there will be two variables (newpoints and newcount) that the transform function can depend upon.The transform function must do the work of incrementing the counter,and adding rules of the form pthead[count]->coord to newpoints.(Clearly less than ideal,this needs to be thought about and improved.) Also note global (in this context) symbol pthead used as a wrapper for new coordinates.*)
tg3Ds[HoldPattern[Graphics3D][g_,o___],f_,p_:0]:=Graphics3D[tg3Ds[g,f,p],o]

tg3Ds[d_List,f_,p_:0]:=Map[tg3Ds[#,f,p]&,d]

tg3Ds[GraphicsGroup[a_,b___],f_,p_:0]:=GraphicsGroup[tg3Ds[a,f,p],b]

tg3Ds[HoldPattern[GraphicsComplex][pts_,prims_,o___],f_,p_:0]:=Module[{tg,allpoints=pts,tmppoints},Block[{newpoints={},newcount=0},tg=tg3Ds[prims,f,If[VectorQ[pts]&&p=!=0,p[[pts]],pts]];
tmppoints=newpoints;
If[newpoints≠{},tg=tg/.newpoints;
allpts=allpts~Join~(Last/@newpoints)]];
If[VectorQ[pts]&&p=!=0,tmppoints=Map[(newcount++;pthead[newcount]->Last[#])&,tmppoints];
newpoints=newpoints~Join~tmppoints;
allpoints=pts~Join~(First/@tmppoints),(*else*)allpoints=pts~Join~(Last/@tmppoints)];
GraphicsComplex[allpoints,tg]]

tg3Ds[Point[d],f_,p_:0]/;(p===0||!gcq[d,p]):=f[Point[d]]

tg3Ds[Line[d_List],f_,p_:0]/;(p===0||!gcq[d,p]):=f[Line[d]]

tg3Ds[Polygon[d_List],f_,p_:0]/;(p===0||!gcq[d,p]):=f[Polygon[d]]

tg3Ds[c:Cylinder[d:{{_,_,_},{_,_,_}},r_:1,t___],f_,p_:0]/;numberQ[r]:=f[c]

tg3Ds[s:Sphere[d_List,r_:1,t___],f_,p_:0]/;numberQ[r]:=f[s]

tg3Ds[t:Text[expr_,d_List,opts___],f_,p_:0]:=f[t]

tg3Ds[c:Cuboid[{_,_,_}],f_,p_:0]:=f[c]

tg3Ds[c:Cuboid[{_,_,_},{_,_,_}],f_,p_:0]:=f[c]

tg3Ds[expr_,f_,0]:=expr

tg3Ds[expr_,f_,pts_List]:=f[expr,pts]

(*gcq tests for compliance with a GraphicsComplex indexed primitive*)
gcq[_Integer]:=True
gcq[pts:{__Integer},count_]/;(Max[pts]≤count&&Min[pts]≥1):=True
gcq[pts:{{_Integer..}..},count_]/;(Max[pts]≤count&&Min[pts]≥1):=True
gcq[any_]:=False


RotateShape[shape_,phi_,theta_,psi_]:=(
Block[{rotmat=RotationMatrix3D[N[phi],N[theta],N[psi]]},TransformGraphics3D[shape,(rotmat.#)&]])

TranslateShape[shape_,vec_List]:=(
Block[{tvec=N[vec]},TransformGraphics3D[shape,(tvec+#)&]]/;Length[vec]==3)


AffineShape[shape_,vec_List]:=(
Block[{tvec=N[vec]},TransformGraphics3D[shape,(tvec*#)&]]/;Length[vec]==3)

Cone[(r_)?NumericQ, (h_)?NumericQ, n_Integer] :=N[Table[Polygon[{{r Cos[((2 π i)/n)],r Sin[((2 π i)/n)],-h},{r Cos[((2 π (i+1))/n)],r Sin[((2 π (i+1))/n)],-h},{0,0,h}}],{i,0,n-1}]]/;n>2

Cone[r_:1,h_:1]:=Cone[r,h,Round[20 r]]

SetOptions[Manipulator,{ImageSize->Tiny,AppearanceElements->{ "InputField"}}];

G2=Graphics;
G3=Graphics3D;
TS=TranslateShape;
RS=RotateShape;
LI=Line;
DX=Dashing[{#1,#1,#1}]&;
rd=Red;bl=Blue;gr=Green;
Clear[AS];
AS[sh_,sc_]:=AffineShape[sh,sc]/.Point[z_]:>Point[z sc]

Definição de seta 2D usando o vector e o ponto base como argumentos.

vec2[x_,y_,s_:.1]:={Arrowheads[s],Arrow[{y,y+x}]}

Redefinição dos cones e bases dos vectores 3D de 10 para 6 arestas.

bs[s_]:=TranslateShape[AffineShape[Cone[0.05,0,6],{s,s,s}],{0,0,1-0.2 s}]
cn[s_]:=TranslateShape[AffineShape[Cone[0.05,0.1,6],{s,s,s}],{0,0,1-0.1 s}]

Protótipo de vector unitário (Subscript[(OverVector[e]), y])

RSy[sh_,θ_]:=RS[RS[sh,((-π)/2),0,0],0,θ,(π/2)]
RSz[sh_,ϕ_]:=RS[sh,0,0,ϕ];
RSx[sh_,ψ_]:=RS[sh,0,ψ,0];

(Subscript[e, y])[s_:.5]:={Line[{{0,0,0},{0,1-s .2,0}}],RotateShape[cn[s],0,(π/2),0],RotateShape[bs[s],0,(π/2),0]}

Definição de vector 3D

V3D[x_,s_:0.5]:=Block[{M=Ω[x]},AffineShape[(Subscript[e, y])[s]/.z_Polygon:>Map[M.#1&,z,{2}]/.z_Line:>Map[M.#1&,z,{2}]/.z_Point:>Map[M.#1&,z,{1}],abs[x] {1,1,1}]]
Vect[v_,x_,s_:0.5]:=TranslateShape[V3D[v,(s/(abs[v]))],x]

NL={z_Polygon:>{EdgeForm[],z}};

Definições de versores ud[(OverVector[v])] dum vector (OverVector[v])

TH=AbsoluteThickness[#1]&;
abs[x_]:=FullSimplify[(Sqrt[x.x])]
ud[x_]:=FullSimplify[(x/(Sqrt[x.x]))]

Construção de rotação (Subscript[(OverVector[e]), y])->(OverVector[u]). Designando por u=|(OverVector[u])| e (OverHat[𝕦])=(1/u)(OverVector[u]) o vector de cosenos directores de (OverVector[u]), precisamos de construir dois vectores auxiliares (OverHat[𝕩]) e (OverHat[𝕫]) que formem um triedro ortonormado com (OverHat[𝕦]), de forma que a matriz Ω=((Subscript[(OverVector[e]), j]^′).(Subscript[(OverVector[e]), i])) defina a rotação pretendida. Assim (OverHat[𝕫])=(OverHat[𝕦])×(Subscript[𝕖, y]) e (OverHat[𝕩])=(OverHat[𝕦])×(OverHat[𝕫]). Usando (Subscript[(OverVector[e]), 1]^′)=(OverHat[𝕩]),(Subscript[(OverVector[e]), 2]^′)=(OverHat[𝕦]),(Subscript[(OverVector[e]), 3]^′)=(OverHat[𝕧]), obtemos Ω.

Ω[{0,y_,0}]:=Sign[y]IdentityMatrix[3];
Ω[u_]:=Block[{𝕦,𝕫,𝕩},{ux,uy,uz}={u[[1]],u[[2]],u[[3]]};𝕦={ux,uy,uz};𝕫={-uz,0,ux};𝕩={ux uy,-(ux^2)-(uz^2),uy uz};𝕊={𝕩,𝕦,𝕫};𝕄=Transpose[(((#1)/(abs[#1]))&)/@𝕊]]

Rotações base

𝕏=(({{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}));                    𝕐=(({{0, 0, -1}, {0, 1, 0}, {1, 0, 0}}));                     ℤ=(({{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}));

Sistema de referência {O,(Subscript[(OverVector[e]), x]),(Subscript[(OverVector[e]), y]),(Subscript[(OverVector[e]), z])}

(Subscript[(OverVector[e]), ρ])[φ_]:={Cos[φ],Sin[φ],0}                ;             (Subscript[(OverVector[e]), φ])[φ_]:={-Sin[φ],Cos[φ],0}           ;

({{(Subscript[(OverVector[e]), x])={1,0,0};,  , (Subscript[(OverVector[e]), y])={0,1,0};,  , (Subscript[(OverVector[e]), z])={0,0,1};,  , (OverVector[O])={0,0,0}}});

Definições de arcos 2D e 3D

Clear[arc,arc2D]
arc2D[θ1_:0,θ2_:2 π,n_:100]:=Line[Append[Table[{Cos[q],Sin[q]},{q,θ1,θ1+θ2,((θ2)/n)}],{Cos[θ1+θ2],Sin[θ1+θ2]}]]
arc[θ1_:0,θ2_:2 π]:=arc2D[θ1,θ2]/.{x_,y_}:>{x,y,0}

Definições auxiliares;cyl,cond,halCyl,tube,conn

frame=Graphics3D[{vect[#,(OverVector[O]),.5]&/@{(Subscript[(OverVector[e]), x]),(Subscript[(OverVector[e]), y]),(Subscript[(OverVector[e]), z])},Lgd[{#,(OverVector[O])+(#-.05(({{0, 1, 1}, {1, 0, 1}, {1, 1, 0}})).ud[#])&[ToExpression[#]]}]&/@{"(Subscript[(OverVector[e]), x])","(Subscript[(OverVector[e]), y])","(Subscript[(OverVector[e]), z])"}}/.vect->Vect];

Torus[r1_,r2_,nb_:20]:=ParametricPlot3D[{Cos[t] (r1+r2 Cos[u]),Sin[t] (r1+r2 Cos[u]),r2 Sin[u]},{t,0,2Pi},{u,0,2Pi},PlotPoints->nb,Mesh->False,Boxed->False,Axes->False]

na=20;nb=40;

bend[θ1_:0,θ2_:(π/2)]:=ParametricPlot3D[{Cos[t] (r1+r2 Cos[u]),Sin[t] (r1+r2 Cos[u]),r2 Sin[u]}/.r1->.4/.r2->.05//Release,{t,0,-(π/2)},{u,0,2π},PlotPoints->nb,Mesh->False,Boxed->False,Axes->False]//RS[#,0,θ1,-θ2]&

cyl=RSy[ParametricPlot3D[{.05Cos[u],.05Sin[u],v},{u,0,2π},{v,-1,1},PlotPoints->na,Mesh->False,Boxed->False,Axes->False][[1]],(π/2)];
c1=TS[cyl,{1,0,0}];
c2=TS[AS[RS[c1,0,0,(π/2)],{1,3/2,1}],{2+.4,3+.4,0}];
b1=TS[bend[0,0][[1]],{2,.4,0}];
b2=TS[bend[0,(π/2)][[1]],{2,3+.4,0}];

Definições Texto

(Subscript[tr, z_]):=Sequence@@{FontFamily->"Times New Roman",FontWeight->Plain,FontSlant->Plain,FontSize->z}
Txt[sty__]:=((Text[style[#1,sty],##2]&)@@##1&)/.style:>Style
txt[z_,x_,off_:{0,0},fntf_:"Times New Roman"]:=Text[StyleForm[z,FontSize->12,FontFamily->fntf],x,off]
Lgd[z__List]:=(Txt[(Subscript[tr, 12])]/@{z})

author={GrayLevel[.83],txt["AJRS-"<>(StringJoin@@ToString/@{
{Jan,Fev,Mar,Abr,Mai,Jun,Jul,Ago,Set,Out,Nov,Dez}[[#[[2]]]]," ",#[[3]],", ",#[[1]]}&@Date[]),{0,0},{0,0},"Times"]};
cprt=(Epilog->Inset[Graphics[author],Scaled[{2.1,.5}],Automatic,Automatic,{0,1}]);
SetOptions[Graphics3D,cprt];

Problema 6.3

Uma espira plana, horizontal, tem uma forma rectangular de 30 cm por 20 cm. A espira é percorrida por uma corrente de 1.0 A e está na presença de um campo magnético vertical de 0.10 T. Determine a força em cada um dos lados da espira. Qual é a força total?

Bfield={Opacity[.5],Table[Vect[(Subscript[(OverVector[e]), z]),{i,j,0},1],{i,-2.5,2.5,1},{j,-2.5,1.5,1}]/.NL,Lgd[{"(OverVector[B])",1.2(Subscript[(OverVector[e]), z])}]};
Icurrent={
Vect[(Subscript[(OverVector[e]), x]),-0.5(Subscript[(OverVector[e]), x])+1.2(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),0.7],Lgd[{"ℐ",-0.6(Subscript[(OverVector[e]), x])+1.2(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],
Vect[-(Subscript[(OverVector[e]), x]),0.5 (Subscript[(OverVector[e]), x])-2.2 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),0.7],Lgd[{"ℐ",   0.6(Subscript[(OverVector[e]), x])-2.2(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],
Vect[(Subscript[(OverVector[e]), y]),-2.2(Subscript[(OverVector[e]), x])-1.0(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),0.7],Lgd[{"ℐ",-2.2(Subscript[(OverVector[e]), x])-1.1(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],
Vect[-(Subscript[(OverVector[e]), y]),2.2(Subscript[(OverVector[e]), x])+0.0(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),0.7],Lgd[{"ℐ",    2.2(Subscript[(OverVector[e]), x])+0.1(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}]};
espira=TranslateShape[{TranslateShape[AffineShape[ c1,{.9,1,1}],0.2(Subscript[(OverVector[e]), x])], c2,b1,b2,TranslateShape[c1,3.8(Subscript[(OverVector[e]), y])], 
RSy[{b1,TranslateShape[AffineShape[ c1,{.9,1,1}],0.2(Subscript[(OverVector[e]), x])],c2,b2,TranslateShape[c1,3.8(Subscript[(OverVector[e]), y])]},π]},-2.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])];
Vlabels=Lgd[{"A",-2.4(Subscript[(OverVector[e]), x])-2.4(Subscript[(OverVector[e]), y])},{"B",-2.4(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])},{"C",2.4(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])},{"D",2.4(Subscript[(OverVector[e]), x])-2.4(Subscript[(OverVector[e]), y])}];
Sidedims={{Dashing[{0.005`,0.005`,0.005`}],Line[{-2.4(Subscript[(OverVector[e]), x])+1.5(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),2.4(Subscript[(OverVector[e]), x])+1.5(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],Line[{-2.5(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),-2.5(Subscript[(OverVector[e]), x])-2.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}]},Line[{-2.4(Subscript[(OverVector[e]), x])-2.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),-2.6(Subscript[(OverVector[e]), x])-2.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],Line[{-2.4(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),-2.6(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],Line[{-2.4(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),-2.4(Subscript[(OverVector[e]), x])+1.6(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],Line[{2.4(Subscript[(OverVector[e]), x])+1.4(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),2.4(Subscript[(OverVector[e]), x])+1.6(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],
Lgd[{"ℓ",0(Subscript[(OverVector[e]), x])+1.6(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])},{"𝒸",-2.6(Subscript[(OverVector[e]), x])-0.5(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}]};
Lorentzforce={Red,AbsoluteThickness[2],
Vect[0.5 (Subscript[(OverVector[e]), x]),-2.35(Subscript[(OverVector[e]), x])-0.5 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), AB])",-1.8(Subscript[(OverVector[e]), x])-0.4 (Subscript[(OverVector[e]), y])}],
Vect[-0.5 (Subscript[(OverVector[e]), x]),2.35(Subscript[(OverVector[e]), x])-0.5 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), CD])",1.8(Subscript[(OverVector[e]), x])-0.4 (Subscript[(OverVector[e]), y])}],
Vect[-0.6(Subscript[(OverVector[e]), y]),                 1.35 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), BC])",0.7 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}],
Vect[0.6(Subscript[(OverVector[e]), y]),                 -2.35 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), DA])",-1.6 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])}]};
Battery={AbsoluteThickness[2],Line[{{.2,-2.4,-.1},{.2,-2.4,-1},{.02,-2.4,-1},{.02,-2.4,-1.1},{.02,-2.4,-.9}}],Line[{{-.2,-2.4,-.1},{-.2,-2.4,-1},{-.05,-2.4,-1}}],AbsoluteThickness[4],Line[{{-.05,-2.4,-1.15},{-.05,-2.4,-.85}}],
Lgd[{"-",{.1,-2.4,-1.2}},{"+",{-.1,-2.4,-1.2}}]};
Magm={Green,TH[3],Vect[-1(Subscript[(OverVector[e]), z]),-.5(Subscript[(OverVector[e]), y])-.1(Subscript[(OverVector[e]), z]),1.5]/.NL,Black,Lgd[{"(Subscript[(OverVector[m]), espira])",(#+.1𝕐.(Subscript[(OverVector[e]), z]))&[-(Subscript[(OverVector[e]), z])]}]};

Manipulate[G3[{If[shI,Icurrent,{}],If[shB,Bfield,{}],If[shL,Lorentzforce,{}],If[shM,Magm,{}],Battery,espira,TS[frame[[1]],-0(Subscript[(OverVector[e]), x])-0.5(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])],Vlabels,Sidedims},PlotRange->{{-2.6,2.6},{-2.6,2.6},{-1.2,1}}, BoxRatios->{1,1,1.1/2.6},Axes->False,Boxed->False,ImageSize->550],{{shI,False,"ℐ"},{False,True}},{{shB,False,"(OverVector[B])"},{False,True}},{{shL,False,"(Subscript[(OverVector[F]), m])"},{False,True}},{{shM,False,"(Subscript[(OverVector[m]), esp])"},{False,True}},ControlPlacement->{Right,Right,Right},Paneled->False,SaveDefinitions->True]

Solução

Força de Lorentz sobre cada lado

Uma carga q que se desloque com velocidade (Subscript[(OverVector[v]), q]) num campo eléctrico (OverVector[E])[(OverVector[r])] e magnético (OverVector[B])[(OverVector[r])] irá sentir uma força total, designada Força de Lorentz, dada por

(Subscript[(OverVector[F]), em])[(Subscript[(OverVector[r]), q])]== q (OverVector[E])[(Subscript[(OverVector[r]), q])]+q(Subscript[(OverVector[v]), q])×(OverVector[B])[(Subscript[(OverVector[r]), q])]

A componente magnética da força de Lorentz é o termo (Subscript[(OverVector[F]), m])[(Subscript[(OverVector[r]), q])]=q (Subscript[(OverVector[v]), q])×(OverVector[B])[(Subscript[(OverVector[r]), q])] (onde × indica o produto externo ou vectorial, ver apêndice.)

Mostre que um segmento d(Subscript[(OverVector[r]), c]) de um condutor Γ em repouso, percorrido por uma corrente ℐ e submetido a um campo magnético (OverVector[B])[(Subscript[(OverVector[r]), c])] (onde (Subscript[(OverVector[r]), c]) indica a posição do segmento condutor) sente uma força d(Subscript[(OverVector[F]), m])[(Subscript[(OverVector[r]), c])]==ℐd(Subscript[(OverVector[r]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]

Considere um segmento d(Subscript[(OverVector[r]), c])[ϑ] do condutor, assumidamente parametrizado pelas posições (Subscript[(OverVector[r]), c])[ϑ], onde ϑ∈[(Subscript[ϑ, o]),(Subscript[ϑ, 1])] varia num intervalo  conveniente de ℝ. Para simplificar a leitura vamos deixar cair o argumento [ϑ] de futuro. Se a secção recta do segmento condutor fôr Δ(OverVector[S]) , e representarmos a densidade de cargas de condução (Subscript[q, e]) no condutor por (Subscript[𝒩, c]),  a carga d(Subscript[Q, c]) que no segmento d(Subscript[(OverVector[r]), c]) se vai deslocar é

d(Subscript[Q, c])== (Subscript[𝒩, c])(Subscript[q, e])Δ(OverVector[S])·d(Subscript[(OverVector[r]), c])

Quando percorrido por uma corrente ℐ, se representarmos por (Subscript[(OverVector[v]), c]) a velocidade de deriva média das cargas de condução d(Subscript[Q, c]) do semento d(Subscript[(OverVector[r]), c]) do condutor, devemos ter (Subscript[(OverVector[v]), c])  paralelo a d(Subscript[(OverVector[r]), c]).  Quando submetido a um campo magnético (OverVector[B]), as cargas de condução em movimento vão sentir a força de Lorentz d(Subscript[(OverVector[F]), m]) que irão transmitir ao condutor a que estão ligadas, onde

d(Subscript[(OverVector[F]), m])== d(Subscript[Q, c])(Subscript[(OverVector[v]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]== ((Subscript[𝒩, c])(Subscript[q, e])Δ(OverVector[S])·d(Subscript[(OverVector[r]), c]))(Subscript[(OverVector[v]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]

Como (Subscript[(OverVector[v]), c]) é sempre paralelo a d(Subscript[(OverVector[r]), c]), podemos trocar as suas posições na expressão anterior e obter, tendo em conta que por definição ℐ≡ (Subscript[𝒩, c])(Subscript[q, e])Δ(OverVector[S])·(Subscript[(OverVector[v]), c]) é a carga que consegue passar a secção Δ(OverVector[S]) num segundo,

d(Subscript[(OverVector[F]), m])== d(Subscript[Q, c])(Subscript[(OverVector[v]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]== ((Subscript[𝒩, c])(Subscript[q, e])Δ(OverVector[S])·(Subscript[(OverVector[v]), c]))d(Subscript[(OverVector[r]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]=ℐ d(Subscript[(OverVector[r]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]

Sendo assim a força total sobre um condutor  Γ num campo magnético (OverVector[B]) quando percorrido por uma corrente ℐ é a soma das forças sobre os elementos de corrente representados por ℐd(Subscript[(OverVector[r]), c]):

(Subscript[(OverVector[F]), m])== (∫_Γ)\[DifferentialD](Subscript[(OverVector[F]), m])== ℐ (∫_Γ) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])[(Subscript[(OverVector[r]), c])]== -ℐ(∫_Γ)(OverVector[B])[(Subscript[(OverVector[r]), c])]×\[DifferentialD](Subscript[(OverVector[r]), c])

Quando a parametrização ϑ∈[(Subscript[ϑ, o]),(Subscript[ϑ, 1])]⟶(Subscript[(OverVector[r]), c])[ϑ] é conhecida podemos explicitar este integral de caminho como

(Subscript[(OverVector[F]), m])== ℐ(∫_(Subscript[ϑ, o])^(Subscript[ϑ, 1]))(((d(Subscript[(OverVector[r]), c])[ϑ])/(dϑ))×(OverVector[B])[(Subscript[(OverVector[r]), c])[ϑ]])dϑ

Quando o campo (OverVector[B])[(Subscript[(OverVector[r]), c])] não varia ao longo do condutor Γ este integral simplifica-se

(Subscript[(OverVector[F]), m])==  ℐ ((∫_Γ) d(Subscript[(OverVector[r]), c]))×(OverVector[B])== ℐ((Subscript[(OverVector[r]), f])-(Subscript[(OverVector[r]), i]))×(OverVector[B])

onde (Subscript[(OverVector[r]), f]) e (Subscript[(OverVector[r]), i]) representam as posições inicial e final do condutor.

No caso presente o campo (OverVector[B])=(Subscript[B, z])(Subscript[(OverVector[e]), z]) é constante, pelo que o integral em cada lado se reduz a

Lado AB

(Subscript[(OverVector[F]), AB])== ℐ (∫_AB) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), B])-(Subscript[(OverVector[r]), A]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), B])\[RightBracketingBar](Subscript[B, z])(Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), z])

(Subscript[(OverVector[F]), AB])==ℐ 𝒸 (Subscript[B, z]) (Subscript[(OverVector[e]), x])==1.0×0.2×0.1(Subscript[(OverVector[e]), x])== 0.02(Subscript[(OverVector[e]), x])    (* N *)

Lado BC

(Subscript[(OverVector[F]), BC])== ℐ (∫_BC) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), C])-(Subscript[(OverVector[r]), B]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), C])-(Subscript[(OverVector[r]), B])\[RightBracketingBar](Subscript[B, z])(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), z])

(Subscript[(OverVector[F]), BC])== -ℐ ℓ (Subscript[B, z]) (Subscript[(OverVector[e]), y])==-1.0×0.3×0.1(Subscript[(OverVector[e]), y])== -0.03(Subscript[(OverVector[e]), y])     (* N *)

Lado CD

(Subscript[(OverVector[F]), CD])== ℐ (∫_CD) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), D])-(Subscript[(OverVector[r]), C]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), D])-(Subscript[(OverVector[r]), C])\[RightBracketingBar](Subscript[B, z]) (-(Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), z]))

(Subscript[(OverVector[F]), CD])== -ℐ 𝒸 (Subscript[B, z])(Subscript[(OverVector[e]), x])==-1.0×0.2×0.1(Subscript[(OverVector[e]), x])== -0.02 (Subscript[(OverVector[e]), x])     (* N *)

Lado DA

(Subscript[(OverVector[F]), DA])== ℐ (∫_DA) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), D]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), D])\[RightBracketingBar](Subscript[B, z]) (-(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), z]))

(Subscript[(OverVector[F]), DA])== ℐ ℓ (Subscript[B, z]) (Subscript[(OverVector[e]), y])==1.0×0.3×0.1(Subscript[(OverVector[e]), y])== 0.03(Subscript[(OverVector[e]), y])          (* N *)

Força de Lorentz total sobre a espira

Quando o campo (OverVector[B])[(Subscript[(OverVector[r]), c])] é constante ao longo de um circuito fechado Γ a força total anula-se porque (Subscript[(OverVector[r]), f])≡(Subscript[(OverVector[r]), i]).

(Subscript[(OverVector[F]), m])== ℐ ((∮_esp)\[DifferentialD](Subscript[(OverVector[r]), c]))×(OverVector[B])≡0

Apêndice: Produto Vectorial (OverVector[a])×(OverVector[b])

Definição

O paralelogramo determinado por dois vectores  (OverVector[a])  e (OverVector[b])  em (ℝ^3) é o quadrilátero com lados paralelos a (OverVector[a]) e (OverVector[b]), com comprimentos iguais a a=|(OverVector[a])| e b=|(OverVector[b])|. Escolhendo qualquer lado como a base, e altura a distância até ao lado paralelo oposto, este paralelogramo tem área A=a b Sin[(Subscript[θ, ab])]= (comprimento base)⋆(altura). O produto Vectorial (ou Externo ou Exterior) destes dois vectores é um vector  (OverVector[c])=(OverVector[a])×(OverVector[b]) com direcção ortogonal a ambos  que representa a área orientada (OverVector[c])≡A(OverVector[n]) deste paralelogramo ((OverVector[n]) é a normal ao plano do quadrilátero para o lado em que (OverVector[a]),(OverVector[b]) e (OverVector[n]) formem um triedro directo-regra da mão direita: (OverVector[a])-polegar, (OverVector[b])-indicador, (OverVector[n])-médio).

(OverVector[c])=(OverVector[a])×(OverVector[b])=a b(Subscript[(OverVector[e]), a])×(Subscript[(OverVector[e]), b])             ;            (Subscript[(OverVector[e]), a])×(Subscript[(OverVector[e]), b])=Sin[(Subscript[θ, ab])](OverVector[n]) 	∴	(OverVector[c])=(OverVector[a])×(OverVector[b])=a b Sin[(Subscript[θ, ab])](OverVector[n])

Uma forma mnemónica de calcular (OverVector[a])×(OverVector[b]) em (ℝ^3) consiste em formar uma matriz simbólica de determinante

(OverVector[a])×(OverVector[b])=det(({{(Subscript[(OverVector[e]), x]), (Subscript[(OverVector[e]), y]), (Subscript[(OverVector[e]), z])}, {(Subscript[a, x]), (Subscript[a, y]), (Subscript[a, z])}, {(Subscript[b, x]), (Subscript[b, y]), (Subscript[b, z])}}))=((Subscript[a, y])(Subscript[b, z])-(Subscript[a, z])(Subscript[b, y]))(Subscript[(OverVector[e]), x])+((Subscript[a, z])(Subscript[b, x])-(Subscript[a, x])(Subscript[b, z]))(Subscript[(OverVector[e]), y])+((Subscript[a, x])(Subscript[b, y])-(Subscript[a, y])(Subscript[b, x]))(Subscript[(OverVector[e]), z])

Contudo, na maioria dos casos e em geral, para calcular algébricamente produtos de vectores arbitrários (OverVector[a])×(OverVector[b]) é mais prático usar as propriedades de linearidade e anti-simetria deste produto conjuntamente com o conhecimento do produto vectorial dos vectores de base (Subscript[(OverVector[e]), x]),(Subscript[(OverVector[e]), y]),(Subscript[(OverVector[e]), z]) dois a dois:

(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), y])=(Subscript[(OverVector[e]), z])	;	(Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), z])=(Subscript[(OverVector[e]), x])	;	(Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), x])=(Subscript[(OverVector[e]), y])
(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), x])=0                          ;                          (Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), y])=0                        ;                         (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), z])=0

Propriedades:

(OverVector[a])×(λ (OverVector[b])+μ (OverVector[c]))=λ(OverVector[a])×(OverVector[b])+μ (OverVector[a])×(OverVector[c])

(OverVector[a])×(OverVector[b])=- (OverVector[b])×(OverVector[a])

(OverVector[a])×((OverVector[b])×(OverVector[c]))≠((OverVector[a])×(OverVector[b]))×(OverVector[c])

Problema 6.4

Calcular o momento de força de Laplace-Lorentz na espira do probema anterior.

Solução

O Momento Magnético (Subscript[(OverVector[m]), espira]) que a espira apresenta quando percorrida pela corrente ℐ no sentido apresentado na figura é (usando a regra da mão direita fechar no sentido da corrente quando o polegar aponta na direcção de (Subscript[(OverVector[m]), espira]), neste caso -(Subscript[(OverVector[e]), z]))

(Subscript[(OverVector[m]), espira])== -ℐ 𝒮 (Subscript[(OverVector[e]), z])                        ∴                          (Subscript[(OverVector[m]), espira])==-1.0 ×0.3×0.2(Subscript[(OverVector[e]), z])=-0.06 (Subscript[(OverVector[e]), z])    (* A (m^2)*)

O binário (momento de força) exercido pelo campo magnético (OverVector[B])=(Subscript[B, z])(Subscript[(OverVector[e]), z]) sobre (Subscript[(OverVector[m]), espira])=-ℐ 𝒮 (Subscript[(OverVector[e]), z])   é assim nulo neste caso (porque (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), z])=0)

(OverVector[N])==(Subscript[(OverVector[m]), espira])×(OverVector[B])                               ∴                         (OverVector[N])== -ℐ 𝒮 (Subscript[B, z]) (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), z])≡0

Problema 6.5

Repetir os dois problemas anteriores para o caso em que o campo magnético é horizontal, paralelo ao lado com 20 cm.

Solução

Bfield1={Opacity[.5],Table[Vect[(Subscript[(OverVector[e]), y]),{i,j,0},1],{i,-2.5,2.5,1},{j,-2.5,1.5,1}]/.NL,Lgd[{"(OverVector[B])",{0,2.,0.25}}]};
Lorentzforce1={Red,AbsoluteThickness[2],
Vect[0.6(Subscript[(OverVector[e]), z]),                 1.4 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), BC])",1.4(Subscript[(OverVector[e]), y])+0.6(Subscript[(OverVector[e]), z])}],
Vect[-0.6(Subscript[(OverVector[e]), z]),                 -2.4 (Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z]),.75],Lgd[{"(Subscript[(OverVector[F]), DA])",-2.0 (Subscript[(OverVector[e]), y])-0.6(Subscript[(OverVector[e]), z])}]};
Magm={Green,TH[3],Vect[-1(Subscript[(OverVector[e]), z]),-.5(Subscript[(OverVector[e]), y])-.1(Subscript[(OverVector[e]), z]),1.5]/.NL,Black,Lgd[{"(Subscript[(OverVector[m]), espira])",(#+.1𝕐.(Subscript[(OverVector[e]), z]))&[-(Subscript[(OverVector[e]), z])]}]};
Manipulate[G3[{If[shI,Icurrent,{}],If[shB,Bfield1,{}],If[shL,Lorentzforce1,{}],If[shM,Magm,{}],Battery,espira,TranslateShape[frame[[1]],-0(Subscript[(OverVector[e]), x])-0.5(Subscript[(OverVector[e]), y])-0.1(Subscript[(OverVector[e]), z])],Vlabels,Sidedims},PlotRange->{{-2.6,2.6},{-2.6,2.6},{-1.2,1}}, BoxRatios->{1,1,1.1/2.6},Axes->False,Boxed->False,ImageSize->550],{{shI,False,"ℐ"},{False,True}},{{shB,False,"(OverVector[B])"},{False,True}},{{shL,False,"(Subscript[(OverVector[F]), m])"},{False,True}},{{shM,False,"(Subscript[(OverVector[m]), esp])"},{False,True}},ControlPlacement->{Right,Right,Right},Paneled->False,SaveDefinitions->True]

Força de Lorentz sobre cada lado

No caso em que (OverVector[B])=(Subscript[B, y])(Subscript[(OverVector[e]), y]) as formulas anteriores dão

Lado AB

(Subscript[(OverVector[F]), AB])== ℐ (∫_AB) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), B])-(Subscript[(OverVector[r]), A]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), B])\[RightBracketingBar](Subscript[B, y])(Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), y])

(Subscript[(OverVector[F]), AB])==0    (* N *)

Lado BC

(Subscript[(OverVector[F]), BC])== ℐ (∫_BC) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), C])-(Subscript[(OverVector[r]), B]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), C])-(Subscript[(OverVector[r]), B])\[RightBracketingBar](Subscript[B, y])(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), y])

(Subscript[(OverVector[F]), BC])== ℐ ℓ (Subscript[B, y]) (Subscript[(OverVector[e]), z])==0.03(Subscript[(OverVector[e]), z])     (* N *)

Lado CD

(Subscript[(OverVector[F]), CD])== ℐ (∫_CD) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), D])-(Subscript[(OverVector[r]), C]))×(OverVector[B])=-ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), D])-(Subscript[(OverVector[r]), C])\[RightBracketingBar](Subscript[B, y]) (-(Subscript[(OverVector[e]), y])×(Subscript[(OverVector[e]), y]))

(Subscript[(OverVector[F]), CD])== 0       (* N *)

Lado DA

(Subscript[(OverVector[F]), DA])== ℐ (∫_DA) \[DifferentialD](Subscript[(OverVector[r]), c])×(OverVector[B])== ℐ((Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), D]))×(OverVector[B])=ℐ \[LeftBracketingBar](Subscript[(OverVector[r]), A])-(Subscript[(OverVector[r]), D])\[RightBracketingBar](Subscript[B, y]) (-(Subscript[(OverVector[e]), x])×(Subscript[(OverVector[e]), y]))

(Subscript[(OverVector[F]), DA])== -ℐ ℓ (Subscript[B, z]) (Subscript[(OverVector[e]), z])==- 0.03(Subscript[(OverVector[e]), z])          (* N *)

Momento de Força sobre a espira

O momento magnético da espira continua a ser (Subscript[(OverVector[m]), espira])=-ℐ𝒮(Subscript[(OverVector[e]), z]) como anteriormente, mas agora o campo (OverVector[B])==(Subscript[B, y])(Subscript[(OverVector[e]), y]) tem uma orientação diferente.

(OverVector[N])==(Subscript[(OverVector[m]), espira])×(OverVector[B]) ==(-ℐ𝒮(Subscript[(OverVector[e]), z]))×((Subscript[B, y])(Subscript[(OverVector[e]), y])  )

O produto externo (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), y])== -(Subscript[(OverVector[e]), x]) pelo que

(OverVector[N])== -ℐ 𝒮 (Subscript[B, y]) (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), y])≡ ℐ 𝒮 (Subscript[B, y]) (Subscript[(OverVector[e]), x]) == 1.0×0.3×0.2×0.1(Subscript[(OverVector[e]), x])=0.006 (Subscript[(OverVector[e]), x])   (* N m *)

Problema 7.2

Calcular, recorrendo à lei de Ampère, o campo (OverVector[B]) criado por um cabo rectilíneo de comprimento infinito e raio R, percorrido pela corrente I distribuída uniformemente na sua secção, para:

r>R

r<R

Solução

a) Caso em que r>R

De acordo com a lei de Ampère a circulação do campo magnético (OverVector[B])[(OverVector[r])] ao longo de um caminho fechado qualquer Γ é proporcional à soma algébrica das correntes que atravessam qualquer superfíce 𝒮 cujo bordo seja o caminho Γ, ou seja ∂𝒮≡Γ. Simbólicamente, se (Subscript[(OverVector[r]), Γ]) designar a posição dos pontos do circuito Γ,

(∮_Γ)(OverVector[B])[(Subscript[(OverVector[r]), Γ])]·\[DifferentialD](Subscript[(OverVector[r]), Γ])== (Subscript[μ, o])(Subscript[ℐ, int])

A constante de proporcionalidade é a permeabilidade magnética do vácuo

(Subscript[μ, o])=4π ×(10^(-7))   (* (N/(A^2)) =(H/m)*)

Pela lei de Biot-Savart, um segmento de corrente ℐd(Subscript[(OverVector[r]), c]) na posição (Subscript[(OverVector[r]), c]) contribui para o campo no ponto (OverVector[r]) através da formula

\[DifferentialD](OverVector[B])[(OverVector[r]) ;ℐ\[DifferentialD](Subscript[(OverVector[r]), c]),(Subscript[(OverVector[r]), c])]== ((Subscript[μ, o])/(4π))((ℐ\[DifferentialD](Subscript[(OverVector[r]), c])×((OverVector[r])-(Subscript[(OverVector[r]), c])))/((\[LeftBracketingBar](OverVector[r])-(Subscript[(OverVector[r]), c])\[RightBracketingBar])^3))

Se d(Subscript[(OverVector[r]), c])=dz(Subscript[(OverVector[e]), z]) então o produto externo deve ser perpendicular ao plano vertical passando pelo condutor e o ponto (OverVector[r]). Se usarmos "coordenadas cilíndricas" com eixo (Subscript[(OverVector[e]), z]) alinhado com o condutor, isto equivale a afirmar o seguinte:

Manipulate[SH[{frame/.NL, G3[{{Opacity[.3],RSy[AffineShape[cyl,{2,.5,.5}],π/2],Opacity[.3],{#,Lgd[{"φ",1.1#[[1,Length[#[[1]]]/2]]}]}&[AffineShape[arc[0,φ],.3{1,1,1}]],TS[{AffineShape[arc[0,2π],{ρ,ρ,1}],Vect[(Subscript[(OverVector[e]), ρ])[φ],(OverVector[O])]/.NL,Vect[(Subscript[(OverVector[e]), φ])[φ],(OverVector[O])]/.NL},Z(Subscript[(OverVector[e]), z])],Line[{Z(Subscript[(OverVector[e]), z]),ρ(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z]),ρ(Subscript[(OverVector[e]), ρ])[φ],(OverVector[O])}]},
Vect[ρ(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z]),(OverVector[O])],Vect[z (Subscript[(OverVector[e]), z]),(OverVector[O])],{AbsoluteThickness[2],Blue,Vect[ρ(Subscript[(OverVector[e]), ρ])[φ]+(Z-z)(Subscript[(OverVector[e]), z]),z (Subscript[(OverVector[e]), z])],Vect[(.3)(Subscript[(OverVector[e]), z]),z(Subscript[(OverVector[e]), z]),.5],
Vect[(1/((((ρ^2)+(((Z-z))^2)))^(3/2)))Cross[(Subscript[(OverVector[e]), z]),ρ(Subscript[(OverVector[e]), ρ])[φ]],ρ(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z])],Lgd[{"(OverVector[r])-(Subscript[(OverVector[r]), c])",z(Subscript[(OverVector[e]), z])+(.5#+.1𝕏.#)&[ρ(Subscript[(OverVector[e]), ρ])[φ]+(Z-z)(Subscript[(OverVector[e]), z])]},{"ℐ\[DifferentialD](Subscript[(OverVector[r]), c])",(#+.14𝕏.(Subscript[(OverVector[e]), z]))&[(z+.3)(Subscript[(OverVector[e]), z])]},{"\[DifferentialD](OverVector[B])",(#+.05ℤ.#)&[ρ(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z])+(1/((((ρ^2)+(((Z-z))^2)))^(3/2)))Cross[(Subscript[(OverVector[e]), z]),ρ(Subscript[(OverVector[e]), ρ])[φ]]]}]},
Lgd[{"(Subscript[(OverVector[e]), ρ])",(#+.05ℤ.#)&[(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z])]},{"(Subscript[(OverVector[e]), φ])",(#+.05ℤ.#)&[(Subscript[(OverVector[e]), φ])[φ]+Z(Subscript[(OverVector[e]), z])]},{"(OverVector[r])",(.8#-.1ℤ.#)&[ρ(Subscript[(OverVector[e]), ρ])[φ]+Z(Subscript[(OverVector[e]), z])]}
,{"(Subscript[(OverVector[r]), c])",(#+.12𝕏.(Subscript[(OverVector[e]), z]))&[z(Subscript[(OverVector[e]), z])]}]}]},ImageSize->600,PlotRange->2{{-1.5,1.5},{-1.5,1.5},{-.75,1.25}},ViewAngle->.21,ViewPoint->{2.5,-1.1,1.2},Boxed->False],{{ρ,.8},0,2},{{Z,.5},-2,2},{{z,1.5},-2,2,.01},{{φ,π/4},0.0001,2π},Paneled->False,ControlPlacement->Right,SaveDefinitions->True]

(Subscript[(OverVector[r]), c])=(Subscript[z, c])(Subscript[(OverVector[e]), z])	                             posição no condutor à cota z 	;
ℐ\[DifferentialD](Subscript[(OverVector[r]), c])=ℐ\[DifferentialD](Subscript[z, c])(Subscript[(OverVector[e]), z])                               elemento de corrente de comprimento \[DifferentialD]z 	;    
(OverVector[r])=ρ(Subscript[(OverVector[e]), ρ])[φ]+z(Subscript[(OverVector[e]), z])                              posição geral em coordenadas cilíndricas	;

\[DifferentialD](OverVector[B])[(OverVector[r]) ;ℐ\[DifferentialD](Subscript[(OverVector[r]), c]),(Subscript[(OverVector[r]), c])]== ((Subscript[μ, o])/(4π))((ℐ\[DifferentialD](Subscript[z , c])(Subscript[(OverVector[e]), z])×(ρ(Subscript[(OverVector[e]), ρ])[φ]+(z-(Subscript[z, c]))(Subscript[(OverVector[e]), z])))/((((ρ^2)+(((z-(Subscript[z, c])))^2)))^(3/2)))== ((Subscript[μ, o])/(4π))((ℐ ρ\[DifferentialD](Subscript[z , c]))/((((ρ^2)+(((z-(Subscript[z, c])))^2)))^(3/2))) (Subscript[(OverVector[e]), z])×(Subscript[(OverVector[e]), ρ])[φ]

\[DifferentialD](OverVector[B])[(OverVector[r]) ;ℐ\[DifferentialD](Subscript[(OverVector[r]), c]),(Subscript[(OverVector[r]), c])]== ((Subscript[μ, o])/(4π))((ℐ ρ\[DifferentialD](Subscript[z , c]))/((((ρ^2)+(((z-(Subscript[z, c])))^2)))^(3/2))) (Subscript[(OverVector[e]), φ])[φ]

O campo total (OverVector[B])[(OverVector[r])] é a soma de contribuições d(OverVector[B])[(OverVector[r]) ;ℐd(Subscript[(OverVector[r]), c]),(Subscript[(OverVector[r]), c])], todas apenas tendo componente segundo (Subscript[(OverVector[e]), φ])[φ], donde se conclui que

(OverVector[B])[(OverVector[r])]==((Subscript[μ, o])/(4π))ℐ ρ ((∫_(-∞)^∞)  (1/((((ρ^2)+(((z-(Subscript[z, c])))^2)))^(3/2)))\[DifferentialD](Subscript[z , c]))(Subscript[(OverVector[e]), φ])[φ]== (Subscript[B, φ])[ρ](Subscript[(OverVector[e]), φ])[φ]

Concluímos assim que (OverVector[B]) apenas possui componente (Subscript[B, φ])[ρ] diferente de zero. A circulação deste campo numa circunferência horizontal Γ de raio r centrada no condutor, no sentido do fecho dos dedos da mão direita quando o polegar aponta na direcção de ℐ, deve ser

(Subscript[(OverVector[r]), Γ])== r (Subscript[(OverVector[e]), ρ])[φ]           	 φ∈[0,2π]	;                                  \[DifferentialD](Subscript[(OverVector[r]), Γ])=r ((\[DifferentialD](Subscript[(OverVector[e]), ρ])[φ])/(\[DifferentialD]φ))\[DifferentialD]φ== r (Subscript[(OverVector[e]), φ])[φ]\[DifferentialD]φ

(∮_Γ)(OverVector[B])[(Subscript[(OverVector[r]), Γ])]·\[DifferentialD](Subscript[(OverVector[r]), Γ])== (Subscript[μ, o]) ℐ                          ∴                           (∫_0^(2π))(Subscript[B, φ])[r]r\[DifferentialD]φ== 2π r (Subscript[B, φ])[r]== (Subscript[μ, o])ℐ

Daqui se conclui que, em coordenadas cilíndricas,

(OverVector[B])[(OverVector[r])]==(( (Subscript[μ, o]))/(2π))(ℐ/ρ)(Subscript[(OverVector[e]), φ])[φ]

b) Caso em que r<R

Quando se pretende calcular o campo dentro do condutor podemos usar os mesmos argumentos que anteriormente para calcular a circulação do campo (OverVector[B]) ao longo de uma circunferência horizontal Γ de raio r<R, centrada no condutor. Contudo agora apenas parte da corrente ℐ passa dentro desta circunferência. Assumindo uma densidade uniforme de corrente (OverVector[J])=(ℐ/(π (R^2)))(Subscript[(OverVector[e]), z]), temos que

(Subscript[ℐ, r])== (OverVector[J])·Δ(Subscript[(OverVector[S]), r])== (ℐ/( π (R^2)))π (r^2)                ∴                     (Subscript[ℐ, r])== ((((r/R)))^2)ℐ

A aplicação da lei de Ampère agora dá

(∮_Γ)(OverVector[B])[(Subscript[(OverVector[r]), Γ])]·\[DifferentialD](Subscript[(OverVector[r]), Γ])== (Subscript[μ, o]) (Subscript[ℐ, r])                          ∴                           (∫_0^(2π))(Subscript[B, φ])[r]r\[DifferentialD]φ== 2π r (Subscript[B, φ])[r]== (Subscript[μ, o]) ((((r/R)))^2)ℐ

Daqui se conclui que, em coordenadas cilíndricas  dentro do condutor

(OverVector[B])[(OverVector[r])]==(( (Subscript[μ, o]))/(2π))((((ρ/R)))^2)(ℐ/ρ)(Subscript[(OverVector[e]), φ])[φ]== (( (Subscript[μ, o]))/(2π))(1/(R^2))ℐ ρ (Subscript[(OverVector[e]), φ])[φ]