MIT OpenCourseWare http://ocw.mit.edu Continuum Electromechanics For any use or distribution of this textbook, please cite as follows: Melcher, James R. Continuum Electromechanics. Cambridge, MA: MIT Press, 1981. Copyright Massachusetts Institute of Technology. ISBN: 9780262131650. Also available online from MIT OpenCourseWare at http://ocw.mit.edu (accessed MM DD, YYYY) under Creative Commons license Attribution-NonCommercial-Share Alike. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Problems for Chapter 2For Section 2.3:Prob. 2.3.1 Perfectly conducting plane parallel plates are shorted at z = 0 and driven by a distributedcurrent source at z = -Z, as shown in Fig. P2.3.1.i(t)Fig. P2.3.1(a)Apply the normalization of Eq. 4b to Maxwell's equations used to represent the fields between theplates. There is no material between the plates, so magnetization, polarization and conductionbetween the plates are ignorable.(b) Simplify these equations by assuming that "= E (Z,t)l and = H (z,t)i y(c)The driving current is i(t) = Re I1 exp jut. Find E , H , the surface current and surface chargeon the lower plate to second order.(d) Convert the results of (c)to dimensional expressions.(e)Solve for the exact fields and expand in a to check the results of (d).Prob. 2.3.2source v(t).Prob. 2.3.1.The parallel plates of Prob. 2.3.1 are now driven along their left edges by a voltageThey are open along their right edges. Carry out the steps analogous to those ofA normalization that makes the EQS limit the zero order approximation is appropriate.Prob. 2.3.3 Perfectly conducting plane parallel electrodes in the planes x = a and x = 0 "sandwich"and make electrical contact with a layer of material having conductivity a and thickness a. Theseplates are driven along their edges so that the surface current is Re K exp(jwt)_ in the lower plateat z = -k and the negative of this in the upper plate. The edges of the plates at z = 0 are "open-circuit." In the conductor, fields take the form Ex(z,t), Hy(z,t).(a)Show that all of Maxwell's equations are satisfied if2dfi+k2H = 0;2 ydzdHk/2o-o -1 dHk 0 -eJoW1;Ex (a+ JWEo) dz(b) Show thatSe-jkz ejkz jWtH = Re K e e eY jk2 -jkP.y e -e-jkz jkz WtE Re Kjk(e + e )eJtx (+ jW+e)(e -e )o(q) In Fig. 2.3.1, T -+ l/W and provided Te: Tm, there are two possibilities:(i) WT << 1 and WT << 1. Show that in this case kk << 1 andem mK ejtE KRex (Re + jo)so that the system is equivalent to a capacitor shorted by a resistor (what values?).(ii) WTem << 1, WTe << 1. Show that in this case k + (-1 + j)/6m, where the skin depthProblems for Chap. 22.476 E/ 2/Wo, and that Hy is the superposition of "skin-effect" waves decaying in the direction ofphase propagation.(d) Now, consider the EQS model from the outset. Under what conditions are the laws (Eqs. 23a -27a)valid? Show that the solution for Ex is consistent with part (c).(e) Consider the magnetoquasistatic laws (Eqs. 23b -27b) from the outset and show that the result isconsistent with part (c). For what conditions are these laws valid?Prob. 2.3.4 Given the EQS laws, Eqs. 23a -25a, together with conduction and polarization constitutivelaws and the material motions, E, and pf can be determined. This is generally possible because theconstitutive laws do not typically involve H. Then, if ý is required, Eqs. 26a and 26b, together witha magnetization constitutive law-can be used. It is clear that these relations uniquely define it,because they stipulate both V x H and V * I. Consider now the analogous question of uniquely deter-mining iin an MQS system. In such a system the conduction and magnetization constitutive lawsrespectively take the formJf = (r,t)(E + vx1 H) ; M=(H,)and Eqs. 23b -25b together with a knowledge of the material motion can be used to find H and M.Show that 1 is then uniquely specified and that recourse to Gauss' Law is made only to make an"after the fact" evaluation of the charge density.For Section 2.4:Prob. 2.4.1 A material suffers a rigid-body rotation about the z axis with constant angular velocity0. The particle at the position (ro, 0) when t = 0 is found at(ro,6o,t) = r cos(tO+ 6)i + r sin(Gt + o)iyat a subsequent time t. This Lagrangian description is pictured in Fig. P2.4.1. Use Eqs. 2.4.1and2.4.2 to show that the velocity and acceleration are respectively÷t -tv r=~ [-sin(t + eo)ix + cos(Qt + eo) y]- _ 22a = -_QYFig. P2.4.1. Specific examplein which rigid-body steadyrotation isrepresented in(a) Lagrangiancoordinates andI-LE1 iYXk ) u er ancoordinates. (a)Prob. 2.4.2 One incentive for using an Eulerian representation is that motions which are timedependent in Lagrangian coordinates can become independent of time. To illustrate, consider thealternative representation of the rigid body rotation of Prob. 2.4.1.The material velocity at a given point (r,6) or (x,y) isv = ir = (-r sin ei + r cos 6iy) Q(-yi + xiy0 x y x yi.e., the velocity is independent of time. Clearly the acceleration is not obtained by taking thepartial derivative with respect to time, as might be suggested by the misuse of Eq. 2.4.2. UseEq. 2.4.4 to find a and compare to the result of Prob. 2.4.1.Problems of Chap. 22.48For Section 2.5:Prob. 2.5.1 A scalar function takes the traveling-wave form 1 = ReO(x,y) expj(At-kz) in the frameof reference (T,t). The primed frame moves in the z direction relative to the unprimed frame withthe velocity U. Use the convective derivative to find the rate of change of 0 for an observer movingwith the velocity Ui .Compute this same time rate of change by expressing ( = O(x',y',z',t') andfinding 3/Dt'. Usezthese results to deduce the transformation W' = W -kU. If W' = 0, W = kU.Explain in physical terms.Prob. 2.5.2 A vector function A(x,y,z,t) can also be evaluated as A(x',y',z',t') where the primecoordinates are related to the unprimed ones by Eq. 2.5.1. Show that Eq. 2.5.2b holds.For Section 2.6:Prob. 2.6.1 The one-dimensional form of Leibnitz' rule pertains to taking an integral between end-points (b) and (a)which are themselves a function of time, as sketched in Fig. P2.6.1.Fig. P2.6.1. One-dimensional form of db doLeibnitz' rule specifies how derivative dtcan be taken of the integral between I Xtime-varying endpoints. b(t) a(t)Define A = f(x,t)i and use Eq. 2.6.4 with a suitable surface to show that, for the one-dimensional case, Leibnitz' rule becomesa(t) a-d )f(x,t)dx a dx + f(a,t)ý -f(b,t)ddt fat dt dtb(t) bProb. 2.6.2 The following
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