(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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As pointed out by a student \ in class on Wednesday, the correct expression is obtained as follows: \ \>", "Text"], Cell[TextData[{ StyleBox["We know that the magnetic field H of a wire located at the origin \ and carrying current ii in the + z direction is ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`H\_\[Phi]\)]], StyleBox[" = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`ii\/\(2\ \[Pi]\ \[Rho]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[", where \[Rho] is the radial coordinate in cylindtical \ coordinates measured from the wire. The scalar potential satisfies ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ StyleBox[\(1\/\[Rho]\), FontColor->RGBColor[1, 0, 0]], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[PartialD]\_\[Phi]\ \[CapitalPhi]\_M\ = \ -\), "TraditionalForm"], \(H\_\[Phi]\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox[" = -", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`ii\/\(2\ \[Pi]\ \[Rho]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[". From this, it follows that ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\_M\)]], StyleBox[" = -", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`ii\/\(\(2\)\(\ \)\(\[Pi]\)\(\ \)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["\[Phi], where \[Phi] is the azimuthal angle. We measure this \ angle counterclockwise from an origin centered on the wire. For a wire \ located at (x,y) = (d/2,0), the angle is \[Phi] = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(\(\(tan\^\(-1\)\)(y\/\(x - d/2\))\)\(.\)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" The potential for the two wires is, therefore, \n", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\_M\)]], StyleBox[" = -", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(ii\/\(\(2\)\(\ \)\(\[Pi]\)\(\ \)\)\) (\)], FontColor->RGBColor[1, 0, 0]], StyleBox["\[Phi][d/2] - \[Phi][-d/2] ) = -", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{\(ii\/\(\(2\)\(\ \)\(\[Pi]\)\(\ \)\)\), RowBox[{"(", " ", FormBox[\(\(\(tan\^\(-1\)\)(y\/\(x - d/2\))\)\(\ \)\), "TraditionalForm"]}]}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox["- ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(\(\(tan\^\(-1\)\)(y\/\(x + d/2\))\)\()\)\), "TraditionalForm"]}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], ". Let's proceed to expand this potential in powers of d:" }], "Text", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(\[CapitalPhi]M\ := \ \(-\((ii/\((2 \[Pi])\))\)\) \((ArcTan[\[Rho]\ \ Sin[\[Phi]]/\((\[Rho]\ Cos[\[Phi]]\ - d/2)\)] - ArcTan[\[Rho]\ Sin[\[Phi]]/\((\[Rho]\ Cos[\[Phi]]\ + d/2)\)]\ )\)\)], "Input"], Cell[TextData[StyleBox["Make the small d approximation.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(\[CapitalPhi]app\ = \ FullSimplify[Normal[Series[\[CapitalPhi]M, {d, 0, 2}]]]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["part (b)", "Subsubsection"], Cell[TextData[StyleBox["Introduce a cylindrical shield and solve the \ resulting BV problem:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[TextData[{ StyleBox["Let \[Kappa] = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(-\(ii\/\(2 \[Pi]\)\)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["in the equal. Given that the \"driving\" term from part (a) has \ Sin[\[Phi]] symmetry, we may assume that the magnetic scaled potential has \ the general analytic form ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Phi]\_m\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" = (A \[Rho]+ B/\[Rho]) Sin[\[Phi]]. Thus,\n\[Phi]in = (\ \[Alpha] \[Rho]+ \[Kappa]/\[Rho]) Sin[\[Phi]] \[Rho] < a\n\[Phi]mid = (\ \[Beta] \[Rho]+ \[Gamma]/\[Rho]) Sin[\[Phi]] a < \[Rho] RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(eqs\ := {\[Kappa]/a^2\ + \ \[Alpha]\ \[Equal] \ \[Beta] + \[Gamma]/ a^2, \ \[Kappa]/ a^2 - \[Alpha]\ \[Equal] \ \[Mu] \((\(-\[Beta]\) + \[Gamma]/ a^2)\), \ \[Beta] + \[Gamma]/b^2 \[Equal] \[Delta]/ b^2, \ \[Mu] \((\(-\[Beta]\) + \[Gamma]/b^2)\)\ == \ \[Delta]/ b^2}\)], "Input"], Cell[BoxData[ \(vars := \ {\[Alpha], \[Beta], \[Gamma], \[Delta]}\)], "Input"], Cell[BoxData[ \(sol\ := \ Solve[eqs, vars] // First\)], "Input"], Cell[BoxData[ \(A1 := \ FullSimplify[\[Alpha] /. \ sol]\)], "Input"], Cell[BoxData[ \(B1 := \ FullSimplify[\[Beta] /. \ sol]\)], "Input"], Cell[BoxData[ \(C1 := \ \ FullSimplify[\[Gamma] /. \ sol]\)], "Input"], Cell[BoxData[ \(D1 := \ \ FullSimplify[\[Delta] /. \ sol]\)], "Input"], Cell[TextData[{ " ", StyleBox["Set up the potential in the three regions", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(\[Phi]in\ = \ \((A1\ \[Rho] + \[Kappa]/\[Rho])\)\ Sin[\[Phi]]\)], \ "Input"], Cell[BoxData[ \(\[Phi]mid\ = \ \((B1\ \[Rho] + C1/\[Rho])\)\ Sin[\[Phi]]\)], "Input"], Cell[BoxData[ \(\[Phi]out\ = \ \((D1/\[Rho])\)\ Sin[\[Phi]]\)], "Input"], Cell[TextData[StyleBox["Find the shielding factor", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(F = \ D1\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Part (c) ", "Subsubsection"], Cell[TextData[StyleBox["Find an approximate formula for small t = (b-a) and \ large \[Mu]", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Fapp\ = \ Normal[Series[D1, {\[Mu], Infinity, 1}]]\)], "Input"], Cell[BoxData[ \(Fapp\ = \ Fapp\ /. \ a \[Rule] \ b - t\)], "Input"], Cell[BoxData[ \(Fapp = \ Normal[Series[Fapp, {t, 0, 1}]]\)], "Input"], Cell[BoxData[ \(\(\(Fapp\)\(\ \)\(=\)\(\ \)\(Fapp[\([2]\)]\)\(\ \)\)\)], "Input"], Cell[BoxData[ \(Fapp\ /. \ {\[Mu] \[Rule] \ 200, b \[Rule] \ 1.25, \ t \[Rule] \ 0.3}\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Prob 5.24", "Subsection"], Cell[BoxData[ \(Clear["\<`*\>"]\)], "Input"], Cell[TextData[StyleBox["Mixed BV problem for a hole of radius a in a \ conducting plane.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[CellGroupData[{ Cell["\<\ Preliminary setup. Illustrate solution to integral equation\ \>", "Subsubsection"], Cell[TextData[{ " ", StyleBox["Evaluate some special cases of discontinuous integrals:", FontColor->RGBColor[1, 0, 0]], "\n\n", StyleBox["I1 = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Infinity]\(\( j\_1\)( ka)\)\ \(\(J\_1\)(k\[Rho])\) \[DifferentialD]k\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" & I2 = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\(\[Infinity]\)\(\ \)\)k\ \(\(j\_1\ \)(ka)\)\ \(\(J\_1\)(k\[Rho])\) \[DifferentialD]k\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" \n\nNote that ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`j\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox["(z) = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\@\(\[Pi]\/\(2\ z\)\)\), "TraditionalForm"], \(J\_\(3/2\)\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox["(z)", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell["\<\ I1 = {Integrate[Sqrt[Pi/(2*k*a)]*BesselJ[3/2, k*a]*BesselJ[1, k*\[Rho]], {k, 0, Infinity}, Assumptions -> {\[Rho] > 0, a > \[Rho]}], Integrate[Sqrt[Pi/(2*k*a)]*BesselJ[3/2, k*a]*BesselJ[1, k*\[Rho]], {k, 0, Infinity}, Assumptions -> {a > 0, \[Rho] > a}]}\ \>", "Input"], Cell["\<\ I2 = {Integrate[Sqrt[(k*Pi)/(2*a)]*BesselJ[3/2, k*a]*BesselJ[1, k*\[Rho]], {k, 0, Infinity}, Assumptions -> {\[Rho] > 0, a > \[Rho]}], Integrate[Sqrt[(k*Pi)/(2*a)]*BesselJ[3/2, k*a]*BesselJ[1, k*\[Rho]], {k, 0, Infinity}, Assumptions -> {a > 0, \[Rho] > a}]}\ \>", "Input"], Cell[TextData[StyleBox["Here is the magnetic scalar potential on the surface \ ( a factor Sin[\[Phi]] is omitted)", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(\[CapitalPhi]m\ = \ Simplify[\((2\ H\_0\ a^2/\[Pi])\)\ Piecewise[{{I1[\([1]\)], \[Rho] < a}, {I1[\([2]\)], \[Rho] > a}}]]\)], "Input"], Cell[TextData[StyleBox["Here is the normal component of H on the surface z=0. \ ", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Hn\ = \ \ Simplify[\((2\ H\_0\ a^2/\[Pi])\)\ Piecewise[{{I2[\([1]\)], \ \[Rho] < a}, {I2[\([2]\)], \[Rho] > a}}]]\)], "Input"], Cell[TextData[StyleBox["Plot the result for a=1.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Plot[{\[CapitalPhi]m /. \ {H\_0 \[Rule] \ 1, a \[Rule] \ 1}, Hn /. \ {H\_0 \[Rule] \ 1, a \[Rule] \ 1}}, \[IndentingNewLine]{\[Rho], 0, 2}, PlotRange \[Rule] \ {0, 2}, PlotStyle \[Rule] \ {{Thickness[0.01], Hue[0.95]}, {Thickness[0.01], Hue[0.6]}}, TextStyle \[Rule] {FontFamily \[Rule] "\", FontSize \[Rule] 12}, Ticks \[Rule] {{0, 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2}, {0, 0.5, 1, 1.5, 2}}, \[IndentingNewLine]PlotLabel \[Rule] \ \ "\<\[CapitalPhi]m (red) & Hn (blue)\>"]\)], "Input"], Cell[TextData[{ StyleBox["Examine the potential asymptotically (let \"a\" be small!) ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(j\_1\)(ka)\ = \(1\/3\) ka\)], FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(\[CapitalPhi]m\ = \ \((2 a^2\ H\_0/\[Pi])\) \((a/3)\) Integrate[\ k\ \ BesselJ[1, k\ \[Rho]]\ Exp[\(-k\)\ z]\ , {k, 0, Infinity}, Assumptions \[Rule] \ {z > 0, \[Rho]\ \[Element] \ Reals}]\)], "Input"], Cell[BoxData[ \(\[CapitalPhi]m\ = \ Simplify[\[CapitalPhi]m\ Sin[\[Phi]]\ /. \ {z^2 + \[Rho]^2 \[Rule] \ r^2, \[Rho] \[Rule] r\ Sin[\[Theta]]\ }, Assumptions \[Rule] \ r > 0]\)], "Input"], Cell[TextData[{ StyleBox["This is the potential of a dipole that is centered on the hole \ and directed along the y-axis with moment \nm = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(8 a\^3\)\/3\), "TraditionalForm"], \(H\_0\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Solution to Rob 5.24 ", "Subsubsection"], Cell[TextData[StyleBox["Determine the potential for z=0 in the region \ \[Rho]>a.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(\[Phi]out\ = \ \((2 a^2\ H\_0/\[Pi])\)\ Integrate[\ Sqrt[\ \[Pi]/\((2\ k\ a)\)]\ BesselJ[3/2, k\ a]\ BesselJ[1, k\ \[Rho]]\ , {k, 0, Infinity}, Assumptions \[Rule] \ {a\ > 0, \ \[Rho] > \ a}\ ]\)], "Input"], Cell[TextData[StyleBox["Find and simplify the cartesian components of the \ field:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(H\[Rho] = \ \(-D[\[Phi]out, \[Rho]]\) Sin[\[Phi]]\)], "Input"], Cell[BoxData[ \(H\[Phi]\ = \ \(-\[Phi]out\)\ Cos[\[Phi]]/\[Rho]\)], "Input"], Cell[BoxData[ \(Hx\ = \ Simplify[Cos[\[Phi]]\ H\[Rho] - Sin[\[Phi]]\ H\[Phi]] /. \ Sin[2 \[Phi]] \[Rule] \ 2\ x\ y/\[Rho]^2\)], "Input"], Cell[BoxData[ \(Hy\ = Expand[Sin[\[Phi]]\ H\[Rho] + Cos[\[Phi]]\ H\[Phi]] /. \ {Sin[\[Phi]] \[Rule] \ y/\[Rho], Cos[\[Phi]] \[Rule] \ x/\[Rho]}\)], "Input"], Cell[BoxData[ \(Hy\ = \ Expand[Hy\ /. \ x^2\ \[Rule] \ \[Rho]^2 - y^2]\)], "Input"], Cell[BoxData[ \(Hyy\ = \ Simplify[Coefficient[Hy, y^2]]\)], "Input"], Cell[BoxData[ \(Hy0 = \ Expand[Hy - \ Coefficient[Hy, y^2] y^2]\)], "Input"], Cell[BoxData[ \(\(\(\ \)\(Hy\ = \ Hy0 + Hyy\ y^2\)\)\)], "Input"], Cell[TextData[StyleBox["Determine the currents on the upper surface. The \ components of the current are \n(Kx, Ky) = (-Hy-H0, Hx)", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Kx\ = \ FullSimplify[\(-H\_0\) - Hy\ /. \ {H\_0 \[Rule] \ 1, \ a \[Rule] 1, \ \[Rho] \[Rule] \ Sqrt[x^2 + y^2]}]\)], "Input"], Cell[BoxData[ \(Ky\ = \ Simplify[Hx\ /. \ {H\_0 \[Rule] \ 1, \ a \[Rule] 1, \ \[Rho] \[Rule] \ Sqrt[x^2 + y^2]}]\)], "Input"], Cell[BoxData[ \(Kxx = \ Piecewise[{{0, x^2 + y^2 < 1}, {Kx, x^2 + y^2 > 1}}]\)], "Input"], Cell[BoxData[ \(Kyy = \ Piecewise[{{0, x^2 + y^2 < 1}, {Ky, x^2 + y^2 > 1}}]\)], "Input"], Cell[TextData[StyleBox["Plot the currents", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(<< Graphics`PlotField`\)], "Input"], Cell[BoxData[ \(ff\ = \ PlotVectorField[{Kxx, Kyy}, {x, \(-1.5\), 1.5}, {y, \(-1.5\), 1.5}, PlotPoints \[Rule] \ 10, ColorFunction \[Rule] \ Hue, DisplayFunction \[Rule] \ Identity]\)], "Input"], Cell[BoxData[ \(c1\ = \ Graphics[{Thickness[0.008], Circle[{0, 0}, 1]}]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(Kup = \ Show[ff, c1, DisplayFunction \[Rule] \ $DisplayFunction]\)], "Input"], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Owing to the fact that the signs of Hx and Hy are changed on the \ lower surface and that ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`H\_0\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" vanishes, the current flow on the lower surface is (Klx, Kly) = \ (-Hy, Hx).", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Klx\ = \ FullSimplify[\(-Hy\)\ /. \ {H\_0 \[Rule] \ 1, \ a \[Rule] 1, \ \[Rho] \[Rule] \ Sqrt[x^2 + y^2]}]\)], "Input"], Cell[BoxData[ \(Kly\ = Ky\)], "Input"], Cell[BoxData[ \(Klxx = \ Piecewise[{{0, x^2 + y^2 < 1}, {Klx, x^2 + y^2 > 1}}]\)], "Input"], Cell[BoxData[ \(Klyy = \ Piecewise[{{0, x^2 + y^2 < 1}, {Kly, x^2 + y^2 > 1}}]\)], "Input"], Cell[BoxData[ \(gg\ = \ PlotVectorField[{Klxx, Klyy}, {x, \(-1.5\), 1.5}, {y, \(-1.5\), 1.5}, PlotPoints \[Rule] \ 10, ColorFunction \[Rule] \ Hue, DisplayFunction \[Rule] \ Identity]\)], "Input"], Cell[BoxData[ \(Klo\ = \ Show[{gg, c1}, DisplayFunction \[Rule] \ $DisplayFunction]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Prob 5.25", "Subsection"], Cell[BoxData[ \(Clear["\<`*\>"]\)], "Input"], Cell[TextData[StyleBox["Interaction energy of a long wire with a rectangular \ or circular loop.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[CellGroupData[{ Cell["part (a)", "Subsubsection"], Cell[TextData[{ StyleBox["A wire carrying current ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_\(\(2\)\(\ \)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["runs parallel to y and is located at (x,z) = d ( Cos[\[Alpha]], \ Sin[\[Alpha]] ). \nA rectangular loop with sides a and b is in the (xy) \ plane centered at the origin. The sides of length a are parallel to the y \ axis and those of length b are parallel to the x axis. The loop carries \ current ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox[". The current in the side of the loop that is located at x=b/2 \ runs in the same direction as ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_2\)], FontColor->RGBColor[1, 0, 0]], StyleBox[". Determine the interaction energy ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`W\_12\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" and the force exerted on the loop by the wire.", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[TextData[{ StyleBox["Express ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`W\_12\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Integral]A\ . \ \[DifferentialD]\)], FontColor->RGBColor[1, 0, 0]], StyleBox["l = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\(I\_2\), "TraditionalForm"], \(\(a\)\(\ \)\([\)\(\ \)\(A\_y\)\)}], "TraditionalForm"], \((b/2, 0)\)}], "-", \(A\_y\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox["(-b/2,0)]", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Ay[x_, z_]\ = \ \(-\ \((I\_2\ \[Mu]\_0)\)\)/\((4\ \[Pi])\)\ Log[\((x - \ d\ Cos[\[Alpha]])\)^2 + \((z - d\ Sin[\[Alpha]])\)^2]\)], "Input"], Cell[BoxData[ \(W12\ = \ I\_1\ a\ \((Ay[b/2, 0] - Ay[\(-b\)/2, 0])\)\)], "Input"], Cell[TextData[StyleBox["The force on the circuit (method 1)", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Fd\ = \ \(-Simplify[D[W12, d]]\)\)], "Input"], Cell[BoxData[ \(Fa\ = \ \(-Simplify[D[W12, \[Alpha]]/d]\)\)], "Input"], Cell[BoxData[ \(Fxx\ = \ Simplify[Fd\ Cos[\[Alpha]] - \ Fa\ Sin[\[Alpha]]]\)], "Input"], Cell[BoxData[ \(Fzz\ = \ Simplify[Fd\ Sin[\[Alpha]] + \ Fa\ Cos[\[Alpha]]]\)], "Input"], Cell[TextData[StyleBox["Examine some limits: \[Alpha]= 0 or \[Alpha] = \ \[Pi]/2.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(FullSimplify[{Fxx, Fzz} /. \ \[Alpha] \[Rule] \ 0]\)], "Input"], Cell[BoxData[ \(FullSimplify[{Fxx, Fzz} /. \ \[Alpha] \[Rule] \ \[Pi]/2]\)], "Input"], Cell[TextData[StyleBox["One can easily verify the correctness of these \ results.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[TextData[{ StyleBox["Here is an alternative direct calculation of the force using F = \ ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\([\[DifferentialD]l\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" x B]", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(\(\(\ \)\(Bz\ = \ D[Ay[x, z], x] /. \ z \[Rule] 0\)\)\)], "Input"], Cell[BoxData[ \(\(\(\ \)\(Bx\ = \ \(-D[Ay[x, z], z]\) /. \ z \[Rule] 0\)\)\)], "Input"], Cell[BoxData[ \(Fx\ = \ \(I\_1\) a\ Simplify[\((\ Bz /. \ {x \[Rule] \ b/2})\)\ - \ \((Bz /. {\ x \[Rule] \ \(-b\)/2})\)]\)], "Input"], Cell[BoxData[ \(Fz\ = \ \(-I\_1\)\ a\ Simplify[\((\ Bx /. \ {x \[Rule] \ b/2})\)\ - \ \((Bx /. {\ x \[Rule] \ \(-b\)/2})\)]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Circle: Energy calculation by Two methods", "Subsubsection"], Cell[TextData[{ "Method 1) ", Cell[BoxData[ \(TraditionalForm\`W\_12\)]], " = ", Cell[BoxData[ \(TraditionalForm\`I\_1\)]], Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\((Flux)\)\ \ = \ I\_\(\(1\)\(\ \)\)\ \(\[Integral]\_0\%a\ \[Rho]\ \[DifferentialD]\ \[Rho]\ \(\[Integral]\_0\%\(2 \[Pi]\)\ \ \ \[DifferentialD]\[Phi]\ \ B\_z[\ \[Rho]\ Cos[\[Phi]]\ , 0]\)\)\)\)\)]] }], "Text"], Cell[TextData[{ StyleBox["Evaluate ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`B\_z\)], FontColor->RGBColor[1, 0, 0]], StyleBox["(x,z) ", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Bz\ = \ D[Ay[x, z], x]\)], "Input"], Cell[TextData[{ StyleBox["Evaluate ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(B\_n\), "TraditionalForm"], "=", \(\(B\_z\)\((\)\(\[Rho]\ Cos\)\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox["[\[Phi]],0)", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Bn\ = \ Bz /. \ {x \[Rule] \ \[Rho]\ Cos[\[Phi]]\ , z \[Rule] 0}\)], "Input"], Cell[BoxData[ \(Nt\ = \ 10\)], "Input"], Cell[TextData[{ StyleBox["Expand ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`B\_n\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" in an asymptotic series in d.", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Bn\ = \ Expand[Simplify[Normal[Series[Bn, {d, Infinity, Nt}]]]]\)], "Input"], Cell[BoxData[ \(W\_12\ = \ I\_1\ Integrate[\ \[Rho]\ Integrate[ Bn, {\[Phi], 0, 2 \[Pi]}], {\[Rho], 0, a}]\)], "Input"], Cell[TextData[StyleBox["Simplify this result", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(\(\(\ \)\(W\_12\ = \ \((\[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ \ a)\)\ Expand[ W\_12/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ \ a)\)]\)\)\)], "Input"], Cell[TextData[{ StyleBox["Here is an interesting comparison: (1 - ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\@\(1 - z\^2\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[")/z = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ StyleBox[\(z\/2 + z\^3\/8 + z\^5\/16 + \(5\ z\^7\)\/128 + \(7\ z\^9\)\/256\), FontFamily->"Times New Roman", FontSlant->"Italic"]], FontColor->RGBColor[1, 0, 0]], StyleBox[". \nIt follows that the term in (..) above is Re{(1 - ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\@\(1 - z\^2\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[")/z} with z = a ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`e\^\(i\ \[Alpha]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["/d.", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(S[u_] = \ Normal[Series[1 - Sqrt[1 - u^2], {u, 0, Nt}]/u]\)], "Input"], Cell[BoxData[ \(u\/2 + u\^3\/8 + u\^5\/16 + \(5\ u\^7\)\/128 + \(7\ u\^9\)\/256\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Here is the real part of Re{(1 - ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\@\(1 - z\^2\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[")/z} with z = a ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`e\^\(i\ \[Alpha]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["/d for comparison", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Sreal\ = \ Expand[\((ExpToTrig[S[a\ Exp[I\ \[Alpha]]/d]] + ExpToTrig[S[a\ Exp[\(-I\)\ \[Alpha]]/d]])\)/2]\)], "Input"], Cell[TextData[{ StyleBox["As an alternative, we may evaluate the interaction energy as ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`W\_12\)], FontColor->RGBColor[1, 0, 0]], StyleBox["= ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`a \(\[Integral]\_0\%\(2 \[Pi]\)\ Cos[\[Phi]]\ \(\(A\_y\)(a\ Cos[\[Phi]], 0)\)\ \[DifferentialD]\[Phi]\)\)], FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Ayy\ = \ Ay[a\ Cos[\[Phi]], 0]\)], "Input"], Cell[TextData[StyleBox["Expand Ay in a series", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Nt\ = \ 10\)], "Input"], Cell[BoxData[ \(Aexp\ = \ Expand[Simplify[Normal[Series[Ayy, {d, Infinity, Nt}]]]]\)], "Input"], Cell[BoxData[ \(W\_12 = I\_1\ a\ Integrate[Cos[\[Phi]]\ Aexp, {\[Phi], 0, 2\ \[Pi]}]\)], "Input"], Cell[TextData[StyleBox["Simplify this result", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(W\_12\ = \ \((\[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ a)\)\ \ Expand[W\_12/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ a)\)]\)], \ "Input"], Cell[TextData[{ StyleBox["Note that both methods give the same result, which can be \ rewritten: \nW= ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{\(\[Mu]\_0\), FormBox[\(\(\ \)\(I\_1\)\), "TraditionalForm"], \(I\_2\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox[" a Re{(1 - ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\@\(1 - z\^2\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[")/z} with z = a ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`e\^\(i\ \[Alpha]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["/d.", FontColor->RGBColor[1, 0, 0]] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Force on circular circuit", "Subsubsection"], Cell[TextData[StyleBox["Direct calculation:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Bz = \ D[Ay[x, z], x] /. \ {x \[Rule] \ a\ Cos[\[Phi]], z \[Rule] \ 0}\)], "Input"], Cell[BoxData[ \(Bx = \ \(-D[Ay[x, z], z]\) /. \ {x \[Rule] \ a\ Cos[\[Phi]], z \[Rule] \ 0}\)], "Input"], Cell[TextData[StyleBox["Expand in a series", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Nt\ = \ 10\)], "Input"], Cell[BoxData[ \(Bxa\ := \ Simplify[Normal[Series[Bx, {d, Infinity, Nt}]]]\)], "Input"], Cell[BoxData[ \(Bza\ := \ Simplify[Normal[Series[Bz, {d, Infinity, Nt}]]]\)], "Input"], Cell[BoxData[ \(Fx = \ \(I\_1\) a\ Integrate[Cos[\[Phi]]\ Bza, {\[Phi], 0, 2 \[Pi]}]\)], "Input"], Cell[BoxData[ \(Fz = \ \(-I\_1\) a\ Integrate[Cos[\[Phi]]\ Bxa, {\[Phi], 0, 2 \[Pi]}]\)], "Input"], Cell[TextData[StyleBox["Simplify the above:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Fzs\ = \ \ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ Expand[ Fz/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0)\)]\)], "Input"], Cell[BoxData[ \(Fxs\ = \ \ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0\ Expand[ Fx/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \[Mu]\_0)\)]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(Normal[Series[{1/Sqrt[1 - z^2] - 1}, {z, 0, Nt}]]\)], "Input"], Cell[BoxData[ \({z\^2\/2 + \(3\ z\^4\)\/8 + \(5\ z\^6\)\/16 + \(35\ z\^8\)\/128 + \(63\ \ z\^10\)\/256}\)], "Output"] }, Open ]], Cell[TextData[StyleBox["Alternative:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Fd\ = \ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \(\[Mu]\_0\) Expand[\(-D[W\_12, d]\)/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \ \[Mu]\_0)\)]\)], "Input"], Cell[BoxData[ \(Fa\ = \ \ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \(\[Mu]\_0\) Expand[\(-D[ W\_12/d, \[Alpha]]\)/\((\ \[ImaginaryI]\_1\ \[ImaginaryI]\_2\ \ \[Mu]\_0)\)]\)], "Input"], Cell[BoxData[ \(Fxx\ = \ Simplify[Fd\ Cos[\[Alpha]] - \ Fa\ Sin[\[Alpha]]]\)], "Input"], Cell[BoxData[ \(Fxx\ = \ Expand[Fxx/\((I\_1\ I\_2\ \[Mu]\_0)\)] I\_1\ I\_2\ \[Mu]\_0\)], "Input"], Cell[BoxData[ \(Fzz\ = \ Expand[Simplify[Fd\ Sin[\[Alpha]] + \ Fa\ Cos[\[Alpha]]]]\)], "Input"], Cell[BoxData[ \(c1\ = Sum[Coefficient[ Expand[Fzz/\((I\_1\ I\_2\ \[Mu]\_0)\)], \((a/d)\)^\((2 k)\)]\ \((a/d)\)^\((2 k)\), {k, 1, 5}]\)], "Input"], Cell[BoxData[ \(Fzz\ = \ I\_1\ I\_2\ \(\[Mu]\_0\) \((c1[\([1]\)] + Sum[Simplify[Coefficient[c1, \((a/d)\)^k]] \((a/d)\)^k, {k, 4, 10, 2}])\)\)], "Input"], Cell[BoxData[ \(Fxs\ \[Equal] \ Fxx\)], "Input"], Cell[BoxData[ \(Fzs\ \[Equal] \ Fzz\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Prob 5.34", "Subsection"], Cell[TextData[StyleBox["Interaction between two circular loops", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Clear["\<`*\>"]\)], "Input"], Cell[CellGroupData[{ Cell["parts (a & b)", "Subsubsection"], Cell[TextData[{ StyleBox["The expression for the mutual inductance of two identical \ co-axial rings of radius a separated by distance R is\n", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`M\_12\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\((\[Mu]\_0\)\)], FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(\(\[Pi]a\^2\)\()\)\)\ \(e\^\(-kR\)\) \ \[DifferentialD]k\)], FontColor->RGBColor[1, 0, 0]], StyleBox[". We are required to develop an asymptotic series for ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`M\_12\)], FontColor->RGBColor[1, 0, 0]], StyleBox[". For this purpose, we expand ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\([\(J\_1\)(ka)\), "TraditionalForm"], "]"}], "2"], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox["in a Taylor series.", FontColor->RGBColor[1, 0, 0]] }], "Text", TextAlignment->Left], Cell[BoxData[ \(Nt\ = \ 8\)], "Input"], Cell[BoxData[ \(Ser[k_] = \ \ Normal[Series[BesselJ[1, k\ a]^2, {a, 0, 8}]]\)], "Input"], Cell[TextData[StyleBox["Integrate the series expression:", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(M12\ = \ \((\(\[Mu]\_\(\(0\)\(\ \)\)\) \[Pi]\ a/2)\)\ Expand[ 2\ a\ Integrate[Ser[k]\ \ E^\((\(-k\)\ R)\), {k, 0, Infinity}, Assumptions \[Rule] \ R > 0]]\)], "Input"], Cell[TextData[StyleBox["The above expression agrees with that given in \ Jackson.", FontColor->RGBColor[1, 0, 0]]], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["part (c)", "Subsubsection"], Cell[TextData[{ StyleBox["Obtain an expression for the mutual inductance of two co-planar \ rings. For this purpose, we start with a scalar potential for B of the ring \ centered at the origin. (For simplicity, we omit a factor ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\[Mu]\_0\)], FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox["/2 everywhere below)", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(\[CapitalPhi]m\ = \ \((1 - z/Sqrt[a^2 + z^2])\)\)], "Input"], Cell[BoxData[ \(Baxis\ = \ \(-Simplify[D[\[CapitalPhi]m, z]]\)\)], "Input"], Cell[TextData[StyleBox["Obtain an asymptotic series for \[CapitalPhi]m", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Nt = 10\)], "Input"], Cell[BoxData[ \(\[CapitalPhi]ser\ = \ Normal[Series[\[CapitalPhi]m, {z, Infinity, Nt}]]\)], "Input"], Cell[TextData[StyleBox["Analytically continue the series", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(t\ = \ Table[Coefficient[\[CapitalPhi]ser, z^\((\(-k\))\)], {k, 2, Nt, 2}]\)], "Input"], Cell[BoxData[ \(\[CapitalPhi]ser\ = \ Sum[t[\([k]\)]\ LegendreP[2 k - 1, \[Mu]]/r^\((2 k)\), {k, 1, Nt/2}]\)], "Input"], Cell[TextData[{ StyleBox["Now find the field ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`B\_z\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" in the plane (\[Mu]=0). ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(Note : \ \ \ B\_z\)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(-B\_\[Theta]\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" = + (1/r)\[PartialD]\[CapitalPhi]/\[PartialD]\[Theta] = -(1/r)\ \[PartialD]\[CapitalPhi]/\[PartialD]\[Mu]", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Bz[r_]\ = \ \(-Expand[ D[\[CapitalPhi]ser, \[Mu]]/r\ ]\) /. \ \[Mu] \[Rule] 0\)], "Input"], Cell[TextData[StyleBox["Introduce coordinates radial and angular coordinates \ \[Rho] and \[Xi] relative to center of second loop.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Bzrel\ = \ Bz[Sqrt[R^2 + \[Rho]^2 + 2\ \[Rho]\ R\ Cos[\[Xi]]]]\)], "Input"], Cell[BoxData[ \(Bzrel = \ Normal[Series[Bzrel, {R, Infinity, 10}]]\)], "Input"], Cell[TextData[StyleBox["Evaluate the Flux through the second loop (introduce \ a - sign since flux is downward)", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Flux\ = \ \(-Integrate[\ \[Rho]\ Integrate[ Bzrel, {\[Xi], 0, 2\ \[Pi]}], {\[Rho], 0, a}]\)\)], "Input"], Cell[TextData[{ StyleBox["Now use relation Flux = ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(M\_21\), "TraditionalForm"], \(I\_1\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox[" to determine ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`M\_21\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" (resupply omitted factor)", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Flux\ = \ \((\(\[Mu]\_0\) I\_1\ a\ \[Pi])\)/4\ \ Expand[ 2\ Flux/\((a\ \[Pi])\)]\)], "Input"], Cell[TextData[{ StyleBox["Factoring ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`I\_1\)], FontColor->RGBColor[1, 0, 0]], StyleBox["leads to a result that agrees with the expression for ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`M\_\(\(12\)\(\ \)\)\)], FontColor->RGBColor[1, 0, 0]], StyleBox["in the text.", FontColor->RGBColor[1, 0, 0]] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Part (d)", "Subsubsection"], Cell[TextData[StyleBox["Method (1) Look at forces in the co-planar case by \ direct calculation", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Fx1\ = \ Expand[a\ Integrate[ Cos[\[Xi]]\ Bzrel, {\[Xi], 0, 2\ \[Pi]}] /. \ \[Rho] \[Rule] \ a]\)], "Input"], Cell["\<\ Fy1 = Expand[Integrate[Sin[\[Xi]]Bzrel, {\[Xi], 0, 2 \[Pi]}] /. \[Rho] -> \ a]\ \>", "Input"], Cell[TextData[{ StyleBox["Reintroduce omitted factors ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\((I\_1\), "TraditionalForm"], \(I\_2\)}], "TraditionalForm"], \(\[Mu]\_0\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox[")/2", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Fx1\ = \ \((\[Mu]\_0\ \(I\_1\) I\_2\ /2)\)\ \ Fx1\)], "Input"], Cell[TextData[StyleBox["The force on co-planar loops is repulsive and \ directed along line of centers.", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[TextData[StyleBox["Now, consider co-axial case", FontColor->RGBColor[1, 0, 0]]], "Text"], Cell[BoxData[ \(Br := \ \(-D[\[CapitalPhi]ser, r]\) /. \ {r \[Rule] \ Sqrt[R^2 + a^2], \ \[Mu] \[Rule] \ R/Sqrt[R^2 + a^2]}\)], "Input"], Cell[BoxData[ \(B\[Theta] := \ D[\[CapitalPhi]ser, \[Mu]]\ a/r^2\ /. \ \ {r \[Rule] \ Sqrt[R^2 + a^2], \ \[Mu] \[Rule] \ R/Sqrt[R^2 + a^2]}\)], "Input"], Cell[BoxData[ \(Fz\ := \ \(-R\)/Sqrt[R^2 + a^2]\ B\[Theta]\ - \ a/Sqrt[R^2 + a^2]\ Br\)], "Input"], Cell[BoxData[ \(Fz = \ Normal[Series[Fz, {R, Infinity, 10}]]\)], "Input"], Cell[TextData[{ StyleBox["Reintroduce omitted factors ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\((I\_1\), "TraditionalForm"], \(I\_2\)}], "TraditionalForm"], " ", "2", "\[Pi]", " ", "a", " ", \(\[Mu]\_0\)}], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], StyleBox[")/2", FontColor->RGBColor[1, 0, 0]] }], "Text"], Cell[BoxData[ \(Fz\ = \ \((\(\[Mu]\_\(\(0\)\(\ \)\)\) I\_1/2)\) I\_2\ \((2\ \[Pi]\ a)\)\ Fz\)], "Input"], Cell[TextData[StyleBox["The force on co-axial loops is attractive and \ directed along line of centers.", FontColor->RGBColor[1, 0, 0]]], "Text"] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 685}}, WindowSize->{1016, 651}, WindowMargins->{{-4, Automatic}, {Automatic, 93}}, Magnification->1.5 ] (******************************************************************* Cached data follows. 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