Introduction to Electrodynamics - ch01

Introduction to Electrodynamics - ch01 - INSTRUCTOR'S...

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INSTRUCTOR'S SOLUTIONS MANUAL INTRODUCTION to ELECTRODYNAMICS Third Edition David J. Griffiths
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TABLE OF CONTENTS Chapter 1 Vector Analysis 1 Chapter 2 Electrostatics 22 Chapter 3 Special Techniques 42 Chapter 4 Electrostatic Fields in Matter 73 Chapter 5 Magnetostatics 89 Chapter 6 Magnetostatic Fields in Matter 113 Chapter 7 Electrod ynamics 125 Chapter 8 Conservation Laws 146 Chapter 9 Electromagnetic Waves 157 Chapter 10 Potentials and Fields 179 Chapter 11 Radiation 195 Chapter 12 Electrodynamics and Relativity 219
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Chapter 1 Vector Analysis Problem 1.1 (a) From the diagram, IB + CI COSO3 = IBI COSO1 + ICI COSO2' Multiply by IAI. IAIIB + CI COSO3 = IAIIBI COSO1 + IAIICI COSO2. So: A.(B + C) = A.B + A.C. (Dot product is distributive.) Similarly: IB + CIsin 03 = IBI sin 01 + ICIsin O2, Mulitply by IAI n. IAIIB + CI sin 03 n = IAIIBI sin 01 n + IAIICI sin O2 n. If n is the unit vector pointing out of the page, it follows that Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.) (b) For the general case, see G. E. Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product). Problem 1.2 The triple cross-product is not in general associative. For example, suppose A = ~ and C is perpendicular to A, as in the diagram. Then (B XC) points out-of-the-page, and A X(B XC) points down, and has magnitude ABC. But (AxB) = 0,so (Ax B) xC = 0 :f. Ax(BxC). ICI sin 82 IBlsin81 A k-hB BxC iAx(Bxe) Problem 1.3 z A = + 1x + 1Y- H; A = /3; B = 1x + 1Y+ Hi B = /3. A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso => cosO = ~. 10 = COS-1(t) ~ 70.5288° I x y Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, we might pick the base (A) and the left side (B): A = -1 x + 2 y + 0 z; B = -1 x + 0 Y + 3 z. 1
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2 CHAPTER 1. VECTOR ANALYSIS x y Z AxB = I -1 2 0 1= 6x + 3y + 2z. -1 0 3 This has the' right direction, but the wrong magnitude. length: IAxBI=v36+9+4=7. Problem 1.5 To make a unit vector out of it, simply divide by its ft - AXB 16 A 3 2 I - IAXBI = '7X+ '7y+ '7z . x y Z Ax Ay Az (ByCz - BzCy) (BzCx - BxCz) (BxCy - ByCx) = x[Ay(BxCy - ByCx) - Az(BzCx - BxCz)] + yO + zO (I'll just check the x-component; the others go the same way.) = x(AyBxCy - AyByCx - AzBzCx + AzBxCz) + yO + zOo B(A.C) - C(A.B) = [Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)] x + 0 y + 0 z = x(AyBxCy + AzBxCz - AyByCx - AzBzCx) + yO + zOoThey agree. Ax(BxC) = Problem 1.6 Ax(BXC)+Bx(CxA)+Cx(A-xB) = B(A.C)-C(A.B)+C(A.B)-A(C.B)+A(B.C)-B(C.A) = o. So: Ax(BxC) - (AxB)xC = -Bx(CxA) = A(B.C) - C(A.B). If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or one is zero), or else B.C = B.A = 0, in whichcase B is perpendicular to A and C (including the case B = 0).
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This note was uploaded on 04/18/2008 for the course PHYS 333 taught by Professor Smith during the Spring '08 term at Loyola Chicago.

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Introduction to Electrodynamics - ch01 - INSTRUCTOR'S...

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