Vector Fields

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Vector fields Let F = (Fx;Fy;Fz) be a vector field. The divergence of F is @F @F @F r · F = x + y + z ; @x @y @z while the curl of F is @F @F @F @F @F @F r × F = z − y ; x − z ; y − x : @y @z @z @x @x @y A vector field F is conservative if there is a scalar field φ such that F = rφ. Such a φ is called a potential function for F. The potential function is unique up to addition of a constant. 1. Consider r = (x; y; z). (a) Show that r · r = 3 and r × r = 0. (b) Find a potential function φ for which rφ = r. 2. Let a and b be constant vectors. Compute the divergence and curl of a(r · b) and a × r: 3. Define 1 1 φ(r) = = : krk px2 + y2 + z2 (a) Let E = −∇φ. Show that 0x1 r 1 E(r) = = y : krk3 2 2 2 3=2 @ A (x + y + z ) z (b) Compute r · E and r × E. 4. Define A(r) = −2 log(krk) z^ = (0; 0; − log(x2 + y2)): (a) Let B = r × A. Show that 2 2y 2x B(r) = θ^ = − ; ; 0 : krk x2 + y2 x2 + y2 (b) Compute r · B and r × B. 5. A two-dimensional vector field F = (Fx;Fy) can also be regarded as a three-dimensional vector field F = (Fx;Fy; 0) restricted to the xy-plane. In this setting, compute r · F and r × F in terms of Fx and Fy. 1 6. (a) Let φ be a scalar field. Show that r × rφ = 0. Remark: a vector field is \irrotational" if it has curl 0. This problem shows that any conservative vector field is irrotational. In a simply connected domain, the two are equivalent. (b) Let A be a vector field. Show that r · (r × A) = 0. Remark: a vector field is \solenoidal" if it has divergence 0. This problems shows that any vector field which is the curl of another is solenoidal. When the domain is R3, the two are equivalent. 7. Verify directly that the vector field u(r) = (ex(x cos y + cos y − y sin y); ex(−x sin y − sin y − y cos y); 0) is irrotational and express it as the gradient of a scalar field φ. Also check that u is solenoidal and show that it can be written as the curl of the vector field v = (0; 0; ) for some function that you should determine. 8. Consider the two-dimensional vector field y x F(x; y) = − ; x2 + y2 x2 + y2 defined on D = f(x; y) j (x; y) 6= (0; 0)g. (a) Let f(x; y) = arctan(y=x). Show that rf = F on f(x; y) j x > 0g. (b) (?) Explain why there is no (differentiable) function g : D ! R such that rg = F on all of D. Hint: using f, define g on f(x; y) j x > 0g and f(x; y) j x < 0g. 9. (?, Maxwell's equations) In electromagnetism, Maxwell's equations in a vacuum relate the electric field E and magnetic field B to the electric scalar potential φ and magnetic vector potential A by r · E = 0 r · B = 0 1 @B 1 @E r × E = − r × B = : c @t c @t Let 1 c U = (E · E + B · B) and S = (E × B): 8π 4π Using these definitions together with Maxwell's equations, give a proof of the energy conservation law @U + r · S = 0: @t 2.

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Electromagnetic Fields and Energy
MIT OpenCourseWare http://ocw.mit.edu Haus, Hermann A., and James R. Melcher. Electromagnetic Fields and Energy. Englewood Cliffs, NJ: Prentice-Hall, 1989. ISBN: 9780132490207. Please use the following citation format: Haus, Hermann A., and James R. Melcher, Electromagnetic Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed [Date]). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1989. ISBN: 9780132490207. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms 8 MAGNETOQUASISTATIC FIELDS: SUPERPOSITION INTEGRAL AND BOUNDARY VALUE POINTS OF VIEW 8.0 INTRODUCTION MQS Fields: Superposition Integral and Boundary Value Views We now follow the study of electroquasistatics with that of magnetoquasistat ics. In terms of the flow of ideas summarized in Fig. 1.0.1, we have completed the EQS column to the left. Starting from the top of the MQS column on the right, recall from Chap. 3 that the laws of primary interest are Ampère’s law (with the displacement current density neglected) and the magnetic flux continuity law (Table 3.6.1). � × H = J (1) � · µoH = 0 (2) These laws have associated with them continuity conditions at interfaces. If the in terface carries a surface current density K, then the continuity condition associated with (1) is (1.4.16) n × (Ha − Hb) = K (3) and the continuity condition associated with (2) is (1.7.6). a b n · (µoH − µoH ) = 0 (4) In the absence of magnetizable materials, these laws determine the magnetic field intensity H given its source, the current density J.
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