{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Introduction to Electrodynamics - ch01

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

This preview shows pages 1–5. Sign up to view the full content.

INSTRUCTOR'S SOLUTIONS MANUAL INTRODUCTION to ELECTRODYNAMICS Third Edition David J. Griffiths

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 + ICI sin 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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). Conclusion: Ax(BxC) = (Ax B)xC <=:=} either A is parallel to C, or B is perpendicular to A and C. Problem 1.7 ~= (4x+6y+8z) - (2x+8y+7z) = !2x-2y+ zl ~ = yl4 + 4 + 1 = @J . ~ 12A 2A lAI ~ = ;Z; = 3x - 3Y + 3z :rroblem 1.8 (a) A.yBy + A.zBz = (cos cpAy + sin cpAz)(coscpBy + sin cpBz) + (- sin cpAy + cos cpAz)( - sin cpBy + cos cpBz) = COS2 cpAyBy + sincpcoscp(AyBz + AzBy) + sin2 cpAzBz + sin2 cpAyBy - sin cpcoscp(AyBz + AzBy) + COS2 cpAzBz = (COS2 cp + sin2 cp)AyBy + (sin2 cp + COS2 cp)AzBz = AyBy + AzBz. ./ - 2 - 2 - 2 - 3 -- - 3 ( 3 ) ( 3 ) .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}