文档介绍：Chapter 6
ostatic Fields in Matter
Problem
A A A A A A ml A
N B B B
= m2 X 1; 1 = 4_1rr"3 [3(mi. r ) r - ml ] ; r = y; ml = mlZ; m2 = m2Y' 1 = --41rr3z,
A A 2 2 (abI)2A . . .
N =-- ( ) H 1 b I S I F I
47rr~ yxz = --47rr~x. ereml =1ra , m2 = . a IN = -- 4 ~ ma onentatIOn:
'downward I (-z).
Problem
elF= I dl X B; tIN = r X elF = Ir X (dI X B). Now (Prob. ): r X(dI X B) + dl X (B X r) + B X
(r X ill) = O. But d[r X (r X B)] = dr X (r X B) + r X (dr X B) (since B is constant), and dr = dI, so
dl X (B X r) = r X (dI X B) - d[r X (r X B)]. Hence 2r X (dI X B) = d[r X (r X B)] - B X (r X dI).
dN =!I {d[r X (r X B)] '- B X (r X dI)}. :. N = !I {§d[rX (r X B)] - B X §(r X dI)}. But the first term
is zero (§d(... ) = 0), and the second integral is 2a (Eq. ). So N = -I(B X a) = m X B. qed
Problem
(a)
I~
Accordingto Eq. , F = 27rIRBcos(). But B =
l!".[3(~r-mtlr ' and B cas () = B. YA, so B cas () =
~~ [3(ml .i)(i. y) - (mi' y)]. But ml . Y = 0 and
i . Y = sinq" while ml . i = ml cos(). :. Bcos() =
~~3ml sin q, cas q,.
F =27rIR~~3ml sin q,cos q,. Now sinq, = ~, cosq, = yr2 - R2/r, so F = 3~mlIR2~.
But IR27r = m2, so F = :!.!!!!.mlm2 ~, while for a dipole, R« r, so F = 3 m2.
2". r 21r r 4
(b)F = V(m2. B) = (m2. V)B = (m2:z) [~z\(~(ml. z)z - ml)] = ~mlm2z d~(z\),
~ ~
2ml -3:\ z
21r r4 z.
or, sincez = r: I F = - m2 A I
113
114 CHAPTER 6. OSTATIC FIELDS IN MATTER
Problem
dF = J {(dyy) X B(O,y,O) + (dzz) X B(O,t,z) - (dyy) X B(O,y,t) - (dzz) X B(O,O,z)}
= J {-(dY y) X JB(O, y, t) -~ B(O, y, On+(dz z) X JB(O, t, z) -y B(O, 0, zn}
~ t8B ~ t8B
8z 8y
B
:::}Jt2 zx8 .
B_YX8 NotethatJdy~Blo O ~t~Bl o ooandJdz~B ~t~B .
{ 8Y 8z } [ z ,y, z, , y I O,O,z y I 0,0,0 ]
X Y z x y Z
F = mOO 1 - 0 1 0 =m y8Bx - x8By- x8Bz - z8Bx
8Bz ~!!..!i... 8Bz ~!!..!i... {8y 8y 8z 8z }
{ 8y 8y 8y 8z 8z 8z }
~ 8Bx ~ 8Bx ~ 8Bx
=m x-+y-+z- =O