大物习题第10章答案

=
10π −3
cos(π/3)
=
5π −3
=
−5.23
10.3
10.4
10.5 ²¤
x=d ¢ Ä »
B
=
µ0I 2πx
,
x+a
Φ = BdS =
Badx
x
= µ0Ia x+a dx 2π x x
=
µ0I a 2π
ln
x
+ x
a
εi(x)
=
dΦ − dt
=
µ0I a2 2πx(x +
a)
·
dx dt
Ψ′
=
N1Φ′
=
µ0πN1N2R22I L
,
M′
=
Ψ′ I
=
µ0πN1N2R22 L
7
,
dΦ
=
BdS
=
Bhdr
=
µ0N I 2πr
h
dr
Φ=
dΦ
=
µ0N Ih 2π
b dr ar
=
µ0N Ih 2π
ln
b a
L
=
Ψ I
=
NΦ I
=
µ0N 2h 2π
ln
b a
10.18 ²¤ Ç (1) «­ª
B1
=
µ0I 2R
,
Φ1
=
B1S1
=
πr2B1
=
µ0πr2I 2R
,
M1
=
Φ1 = µ0πr2 . I 2R
=
µ0πr2R2ωIm − 2(R2 + d2)3/2
cos ωt.
B=
µ0I 2π
11 x − x+a
Baidu Nhomakorabea
,
Φ = BdS = Bbdx
=
µ0Ib 2a 2π a
11 x − x+a
dx
=
µ0I b 2π
ln
4 3
.
M
=
Φ I
=
µ0b 2π
ln
4 3
.
10.20 ²¤ ÃÉ
R1
> R2,
n1 =
N1 L
10 Æ Å
10.1
10.2 ²¤
N = 103, a = 0.1, S = a2 = 10−2,
Φ = BS cos 60◦ = 10−4 sin(100πt/3)
Ψ = N Φ = 10−1 sin(100πt/3)
εi
=
dΨ − dt
=
10π −3
cos(100πt/3)
t = 0.01s, εi
´ ¨ ¢ «­ª (2)
(9.17)
d
B2
=
µ0I R2 2(R2 + d2
)3/2
,
Φ2
=
B2S2
=
πr2B2
=
µ0πr2 2(R2 +
R2I d2)3/2
,
M1
=
Φ2 I
=
µ0πr2R2 2(R2 + d2)3/2
.
5
(3)
10.19 ²¤
dI dt
=
ωIm cos ωt,
εi
=
−M2
dI dt
−L/3
=
−
2L/3 −L/3
ωBrdr
=
−
1 6
ωB
L2
10.11
10.12 ²¤
εi = (v × B) · dr
=
−ωB
R 0
rdr
=
−
1 2
ωBR2
¶ Æ ³ ¥È­ªÊ Ë ¡
4
10.13
10.14
10.15
10.16
10.17 ²¤ ´ ¨Ê (9.31) ¢½ ¿§­ª
B
=
µ0N I 2πr
=
µ0I a2 v 2πx(x + a)
Ii
=
εi(d) R
=
µ0I a2v 2πd(d + a)R
3
10.6 10.7 10.8
10.9 ²¤
10.10
v × B = −vBer,
(v × B) · dr = −vBdr = −ωBrdr,
εi =
2L/3
(v × B) · dr = −
vBdr
,
n2
=
N2 L
,
» I, ¿­ª
B
=
µ0n1I
=
µ0N1I , L
¬ ¿ ­ ­º
¦
Φ
=
BS2
=
µ0πN1R22I L
,
Ψ
=
N2Φ
=
µ0πN1N2R22 L
I
,
M
=
Ψ I
=
µ0πN1N2R22 L
6
° £µ¼¿ »© ­ª
B′
=
µ0n2I
=
µ0N2I L
,
Φ′
=
B′S2
=
µ0πN2R22I L
,