Introduction to Electrodynamics
Introduction to Electrodynamics
4th Edition
ISBN: 9781108420419
Author: David J. Griffiths
Publisher: Cambridge University Press
Students have asked these similar questions
Problem 7.21 Imagine a uniform magnetic field, pointing in the z direction and filling all space (B = Bo 2). A positive charge is at rest, at the origin. Now somebody turns off the magnetic field, thereby inducing an electric field. In what direction does the charge move? 16
Assume MKS units... Let Q be an open subset of R³. Let B: :Q - R³ be a continuous vector .field, representing a magnetic field in 3-D space. 7 Let P be a particle with charge q E R and mass m > 0. If p is at position (x. y, z) in Q and R³ is the velocity of p, at time t, then p feels a force 7(7,7) given by - 7(7,J) := q V × B (7) . Suppose that p moves along a curve C as time t varies from a to b, and that p has position vector (t) and instantaneous velocity (t) at time t. ř (1) Explain why the two vectors 7'(t) × È(7(t)) and 7'(t) are perpen- dicular at every time t = [a, b]. (2) Using Part (1), calculate W := the work done on the particle p by the force as p moves from D = 7(a) to E = √ (b) along C. F (3) Prove that ((t)||²)=27' (t) • F(t), at each time t. (4) Using Parts (2) and (3), and Newton's Second Law, prove that if the magnetic force - ₹(7,7) is the total force on p at every time t, then p moves along C at a constant speed. dt
Problem 6.16A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility Xm. A current I flows down the inner conductor and returns along the outer one; in each case, the current dis- tributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field. FIGURE 6.24 | NOTE: (PLEASE SOLVE BY EXPLAİNİNG İNTERMEDİATE STEPS İN ANSWERS.) (ANSWERED) Problem 6.16 FH•dl = Ifane = I, so H = ộ. B = 40(1+ Xm)H = | Ho(1 + Xm); I -ộ. M = XmH 2TS XmI 2ns %3D 2ns XmI s ds at s= a; J, = V×M = 0. K, = Mxân: 2na 2n8 Xmlâ, at s = b. 2nb Total enclosed current, for an amperian loop between the cylinders: Xm! - 2xa = (1+ Xm)I, I+ B. dl = Holenc = Ho(1+ Xm)I >B = 4o(1+Xm)I 2 %3D SO 2ла
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON