Introduction to Electrodynamics
4th Edition
ISBN: 9781108420419
Author: David J. Griffiths
Publisher: Cambridge University Press
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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
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