Introduction to Electrodynamics
INSTRUCTOR'S SOLUTIONS MANUAL
INTRODUCTION to ELECTRODYNAMICS
Third Edition
David J. Griffiths
TABLE OF CONTENTS
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12
Vector Analysis Electrostatics Special Techniques Electrostatic Fields in Matter Magnetostatics Magnetostatic Fields in Matter
1 22 42 73 89 113 125 146 157 179 195 and Relativity 219
Electrod ynamics Conservation Electromagnetic Laws Waves
Potentials and Fields Radiation Electrodynamics
Chapter
1
Vector
Problem 1.1
Analysis
(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.) IAIIB + CI sin 03 n = IAIIBI sin 01 n + IAIICI sin O2n. If n is the unit vector pointing out of the page, it follows that Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.)
ICI sin 82
Similarly: IB + CI sin 03 = IBI sin 01 + ICI sin O2, Mulitply by IAI n.
IBlsin81
A
(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).
k-hB
BxC iAx(Bxe) z Problem 1.3
A
= + 1x + 1Y - H; A = /3;
B
= 1x + 1Y+
Hi B = /3. =>cosO= ~. x y
A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso
10 = COS-1(t) ~ 70.5288° Problem 1.4
I
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
2
CHAPTER
1. VECTOR ANALYSIS
x
AxB
y
Z
2 0 1= 6x + 3y + 2z. -1 0 3 This
INTRODUCTION to ELECTRODYNAMICS
Third Edition
David J. Griffiths
TABLE OF CONTENTS
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12
Vector Analysis Electrostatics Special Techniques Electrostatic Fields in Matter Magnetostatics Magnetostatic Fields in Matter
1 22 42 73 89 113 125 146 157 179 195 and Relativity 219
Electrod ynamics Conservation Electromagnetic Laws Waves
Potentials and Fields Radiation Electrodynamics
Chapter
1
Vector
Problem 1.1
Analysis
(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.) IAIIB + CI sin 03 n = IAIIBI sin 01 n + IAIICI sin O2n. If n is the unit vector pointing out of the page, it follows that Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.)
ICI sin 82
Similarly: IB + CI sin 03 = IBI sin 01 + ICI sin O2, Mulitply by IAI n.
IBlsin81
A
(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).
k-hB
BxC iAx(Bxe) z Problem 1.3
A
= + 1x + 1Y - H; A = /3;
B
= 1x + 1Y+
Hi B = /3. =>cosO= ~. x y
A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso
10 = COS-1(t) ~ 70.5288° Problem 1.4
I
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
2
CHAPTER
1. VECTOR ANALYSIS
x
AxB
y
Z
2 0 1= 6x + 3y + 2z. -1 0 3 This