%% This document created by Scientific Word (R) Version 2.0 \documentclass[12pt]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{sw20jart} %TCIDATA{TCIstyle=Article/ART4.LAT,jart,sw20jart} \input tcilatex \addtolength{\topmargin}{-0.4in} \addtolength{\textheight}{1in} \addtolength{\oddsidemargin}{-0.2in} \addtolength{\textwidth}{0.6in} \thispagestyle{empty} \QQQ{Language}{ American English } \begin{document} \begin{center} \textbf{Phys 742 - EM I} \textbf{Final exam - 6 May 1996} \end{center} \smallskip\ \noindent \textbf{1.} \textbf{(a) }Find the mutual inductance $M$ of two coaxial circular loops of radii $R_1$ and $R_2$ separated by a vertical distance $h.$ The result can be left in the form of a single integral to be evaluated numerically or approximated. \textbf{(b) }Find the force $F$ between the two loops when currents $I_1$ and $I_2$ flow in them. Again, the result can be left in the form of a single integral. When the currents are parallel, is there attraction or repulsion? \smallskip\ \noindent \textbf{2.} \textbf{(a) }Consider two planar loops of areas\textbf{\ }$A_1$ and $A_2$ not too different in shape from circles and stacked above each other at a distance $h$ such that $h^2>>$\textbf{\ }$A_1$ and $h^2>>$\textbf{\ }$A_2$. Find the force $F$ between the two loops when currents $I_1$ and $I_2$ flow in them. Also find the mutual inductance of the two loops. Explicit formulas are required. \textbf{(b)} Check that for two circular loops the results of part (a) agree with those of problem 1 when $h>>$\textbf{\ }$R_1$ and $h>>$\textbf{\ }$R_2.$ \textbf{(c)} Find the force between a circular loop of radius $R$ and a much smaller loop of area $A$ located on its axis, when currents $I_1$ and $I_2$ flow in the two loops. An explicit formula is required, and it must be valid for all values of the distance $z$ (or $h)$ along the axis. Sketch the result. \smallskip\ \ \noindent \textbf{3.} Recall that the Stokes parameters are defined as \begin{eqnarray*} s_0 &=&a_1^2+a_2^2 \\ s_1 &=&a_1^2-a_2^2 \\ s_2 &=&2a_1a_2\cos (\delta _2-\delta _1) \\ s_3 &=&2a_1a_2\sin (\delta _2-\delta _1) \end{eqnarray*} where you are expected to supply the definitions of $a_1,\,a_2,\,\delta _1,\,\delta _2$ (and also of $\mathbf{\epsilon }_1$ here below). Taking $% s_0=1$ for convenience, what are the values of the other Stokes parameters for radiation that is \textbf{(a)} linearly polarized at 45$^{\circ }$ to the $\mathbf{\epsilon }% _1 $ vector \textbf{(b)} circularly polarized, counterclockwise (left handed in the usual optics language) \textbf{(c)} totally unpolarized \textbf{(d) }elliptically polarized, counterclockwise, with semiaxes $a$ and $b$ in the ratio $a/b=2$ and the major axis at 30$^{\circ }$ from $\mathbf{% \epsilon }_1.$ \newpage\ \smallskip\ \ \noindent \textbf{4.} It is possible to put on a surface an optical coating of thickness $w$ and refractive index $n,$ such that no light will be reflected, at normal incidence. Solve the problem of such a coating on a slab having refractive index $n^{\prime },$ taking the thickness of the slab to be effectively infinite. What must the thickness of the coating be if $n^{\prime }=2n=4$ and the incident wavelength (in air) is 5000\AA ? Why is this called a quarter\thinspace -\thinspace wavelength optical coating? (Air can be taken to have refractive index $1$ for this problem) \smallskip\ \ \noindent \textbf{5.} Solve the boundary-value problem for a superconducting sphere of radius $a$ in a weak external magnetic field $\mathbf{B}_0.$ Assume that the magnetic flux is totally expelled from the volume occupied by the sphere. (You may wish to describe this by saying that in the sphere $\mathbf{M}=-\mathbf{H}% /4\pi ,$ or that a surface current $\mathbf{K}$ flows on the sphere). \textbf{(a)} Find the total $\mathbf{B}$ outside the sphere \textbf{(b)} Choose polar coordinates in such a way that $\mathbf{K}$ has only the component $K_\phi $ and give the explicit expression for $K_\phi .$ \textbf{(c) }Find a vector potential $\mathbf{A}$ that describes $\mathbf{B}$ everywhere. \smallskip\ \ \noindent \textbf{6.} True or false? If the assertion is false, explain briefly why, or give a counterexample. \smallskip\ \textbf{(a)} The dielectric constant $\varepsilon $ of a ``passive'' medium is always $\geq 1$ at zero frequency \textbf{(b)} The dielectric constant $\varepsilon $ of a ``passive'' medium is always $\geq 1$ at all frequencies \textbf{(c) }The magnetic permeability\textbf{\ }$\mu $ of any medium is always $\geq 1$ at zero frequency \textbf{(d) }An isolated, uncharged body at rest is always attracted to a region of stronger $E$ \textbf{(e) }An isolated, uncharged body at rest is always attracted to a region of stronger $B$ \textbf{(f) }The Lorentz gauge condition is\textbf{\ }$\nabla \cdot \mathbf{A% }=0$ \textbf{(g)} Any analytic function $f(z),$ where $z=x+iy,$ satisfies $\dfrac{% \partial ^2f}{\partial x^2}+\dfrac{\partial ^2f}{\partial y^2}=0$ \newpage\ \textbf{Answer 1} \smallskip\ \noindent \textbf{(a) }Use $MI_1I_2=(1/c)\int \mathbf{A}_1\mathbf{\cdot J}% _2\,d^3x=(2\pi /c)R_2I_2A_1$ with \[ A_1=\frac{I_1}c\int_0^{2\pi }\frac{R_1\cos \phi ^{\prime }\,d\phi ^{\prime }% }{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}} \] as in Jackson's eq. (5.36). Obtain \[ M=\frac{2\pi }{c^2}R_1R_2\int_0^{2\pi }\frac{\cos \phi ^{\prime }\,d\phi ^{\prime }}{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}} \] \noindent \textbf{(b)} $F=I_1I_2\partial M/\partial h$ (note the sign!) gives \[ F=-\frac{2\pi }{c^2}I_1I_2R_1R_2\int_0^{2\pi }\frac{h\cos \phi ^{\prime }\,d\phi ^{\prime }}{\left( h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }\right) ^{3/2}} \] $F$ is attractive when the currents are parallel, i.e., when $I_1I_2>0.$ \hrulefill \smallskip\ \textbf{Answer 2} \smallskip\ \noindent \textbf{(a) }The magnetic moment of a loop is $m=I\mathcal{A}/c.$ Then \[ F=-6m_1m_2/h^4=-6I_1I_2\mathcal{A}_1\mathcal{A}_2/c^2h^4 \] \[ M=2\mathcal{A}_1\mathcal{A}_2/c^2h^3 \] \noindent \textbf{(b)} Use the expansion \[ \frac 1{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}}\simeq \frac 1h\left( 1-\frac 12\frac{R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}{h^2}% \right) \] in the answer 1(a) to obtain \[ M\simeq \frac{2\pi R_1^2R_2^2}{c^2h^3}\int_0^{2\pi }\cos ^2\phi ^{\prime }\,d\phi ^{\prime }=\frac{2\pi ^2R_1^2R_2^2}{c^2h^3} \] \smallskip\ \noindent \textbf{(c)} On-axis, the field of the large loop is \[ B_1=\frac{2\pi R^2I_1}{c\left( R^2+z^2\right) ^{3/2}} \] The magnetic moment of the small loop is $m_2=I_2\mathcal{A}/c.$ Then \[ F=m_2\frac{\partial B_1}{\partial z}=-\frac{6\pi R^2\mathcal{A}}{c^2}\frac z{\left( R^2+z^2\right) ^{5/2}} \] This agrees with the result of (a) when $z>>R.$ and $\frac z{\left( 1+z^2\right) ^{5/2}}$ \[ \FRAME{itbpF}{4.8118in}{1.9994in}{0in}{}{}{Plot }{\special{language "Scientific Word";type "MAPLEPLOT";width 4.8118in;height 1.9994in;depth 0in;display "USEDEF";plot_snapshots TRUE;function \TEXUX{$\frac{-z}{\left( 1+z^2\right) ^{5/2}}$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "-5";xmax "5";xviewmin "-5";xviewmax "5";yviewmin "-0.2975";yviewmax "0.2974";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle "boxed";xis \TEXUX{z};var1name \TEXUX{$x$};valid_file "T";tempfilename 'C:/SW20/DOCS/DR3IYP7N.wmf';tempfile-properties "XP";}} \] \hrulefill \smallskip\ \textbf{Answer 3} \smallskip\ \noindent \textbf{(a)} $s_1=0,\,s_2=1,\,s_3=0.$ \noindent \textbf{(b) }$s_1=0,\,s_2=0,\,s_3=1$ \noindent \textbf{(c) }$s_1=0,\,s_2=0,\,s_3=0$ \noindent \textbf{(d) }$s_1=3/10,\,s_2=3\sqrt{3}/10\simeq .5196152,\,s_3=4/5. $ One way to solve part (d) is to go to circularly polarized components, where the Stokes parameters are given by \begin{eqnarray*} s_0 &=&a_{+}^2+a_{-}^2 \\ s_1 &=&2a_{+}a_{-}\cos (\delta _{-}-\delta _{+}) \\ s_2 &=&2a_{+}a_{-}\sin (\delta _{-}-\delta _{+}) \\ s_3 &=&a_{+}^2-a_{-}^2 \end{eqnarray*} For counterclockwise polarization, $a_{+}=\left( a+b\right) /\sqrt{2}$ and $% a_{-}=\left( a-b\right) /\sqrt{2}.$ Taking for convenience $a=2$ and $b=1$ we find $a_{+}=3/\sqrt{2}$ and $a_{-}=1/\sqrt{2}.$ Further, $\delta _{-}-\delta _{+}=2\alpha =60^{\circ }.$ Then $s_0=5,\,s_1=3/2,\,s_2=3\sqrt{3}% /2,\,s_3=4,$ and we get the answer by normalizing $s_0$ to $1.$ $.$\hrulefill \smallskip\ \textbf{Answer 4} \smallskip\ Let the electric fields be \[ \begin{array}{lll} E_{0+}\,e^{ik_0z}+E_{0-}\,e^{-ik_0z} & \text{for} & z<0 \\ E_{+}\,e^{ikz}+E_{-}\,e^{-ikz} & \text{for} & 0w \end{array} \] with $k_0=\omega /c,\,k=nk_{0,}\,k^{\prime }=n^{\prime }k_0.$ The boundary conditions are \[ \left\{ \begin{array}{l} E_{0+}\,+E_{0-}\,=E_{+}\,+E_{-}\, \\ E_{0+}\,-E_{0-}\,=n\left( E_{+}\,+E_{-}\right) \, \end{array} \right. \] \[ \left\{ \begin{array}{l} \,E_{+}\,e^{ikw}+E_{-}\,e^{-ikw}=E_{+}^{\prime }\,e^{ik^{\prime }w} \\ \,n\left( E_{+}\,e^{ikw}-E_{-}\,e^{-ikw}\right) =n^{\prime }E_{+}^{\prime }\,e^{ik^{\prime }w} \end{array} \right. \] The first two equations give \[ \left( \begin{array}{l} E_{0+} \\ E_{0-} \end{array} \right) =\frac 12\left( \begin{array}{ll} 1+n & 1-n \\ 1-n & 1+n \end{array} \right) \left( \begin{array}{l} E_{+} \\ E_{-} \end{array} \right) \] and the last two give similarly \[ \left( \begin{array}{l} E_{+} \\ E_{-}\, \end{array} \right) =\frac 1{2n}\left( \begin{array}{l} \left( n+n^{\prime }\right) e^{-ikw} \\ \left( n-n^{\prime }\right) e^{ikw} \end{array} \right) E_{+}^{\prime }\,e^{ik^{\prime }w} \] From this we get directly \[ \left( \begin{array}{l} E_{0+} \\ E_{0-} \end{array} \right) =\frac 1{4n}\,\left( \begin{array}{l} \left( 1+n\right) \left( n+n^{\prime }\right) e^{-ikw}+\left( 1-n\right) \left( n-n^{\prime }\right) e^{ikw} \\ \left( 1-n\right) \left( n+n^{\prime }\right) e^{-ikw}+\left( 1+n\right) \left( n-n^{\prime }\right) e^{ikw} \end{array} \right) E_{+}^{\prime }\,e^{ik^{\prime }w} \] The reflection coefficient vanishes when $E_{0-}/E_{0+}=0.$ For $n>1$ and $% n^{\prime }>n,$ this can happen only if \[ e^{2ikw}=-1 \] i.e. for $w=(2j+1)(\lambda /4),$ with $j$ integer. Even then, we must have the condition \[ \left( 1-n\right) \left( n+n^{\prime }\right) +\left( 1+n\right) \left( n^{\prime }-n\right) =0 \] which simplifies to give $n^{\prime }=n^2.$ This is satisfied by the problem's data, $n=2,\,n^{\prime }=4.$ A quarter-wavelength coating corresponds to $j=0.$ Note that $\lambda $ is the wavelength in the coating, $\lambda =\lambda _0/n.$ In our case, $% \lambda =2500$ \negthinspace \AA . No reflection occurs for $w[$\AA $% ]=2500/4=625$. \end{document}