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Unformatted text preview: = ε 0 μ0π r 2 ( 0.00450 V/m ⋅ s ) . For r = 0.0200 m, this gives
1
B = ε 0 μ0 r (0.00450 V/m ⋅ s)
2
1
= (8.85 × 10−12 C2 /N ⋅ m 2 )(4π ×10−7 T ⋅ m/A)(0.0200 m)(0.00450 V/m ⋅ s)
2
= 5.01×10−22 T .
(b) With r > R, the expression above must replaced by
B (2π r ) = ε 0 μ0π R 2 ( 0.00450 V/m ⋅ s ) . Substituting r = 0.050 m and R = 0.030 m, we obtain B = 4.51 × 10−22 T.
10. (a) Here, the enclosed electric flux is found by integrating
r
r⎛
⎛1
r⎞
r3 ⎞
Φ E = ∫ E 2π rdr = t (0.500 V/m ⋅ s)(2π ) ∫ ⎜1 − ⎟ rdr = tπ ⎜ r 2 −
⎟
0
0
3R ⎠
⎝ R⎠
⎝2 with SI units understood. Then (after taking the derivative with respect to time) Eq. 323
leads to
⎛1
r3 ⎞
B (2π r ) = ε 0 μ0π ⎜ r 2 −
⎟.
3R ⎠
⎝2 1247
⎛ μ ε R 2 dE ⎞
⎛ μ ε R 2 dV ⎞
⎛ μ ε R2
⎞
Bmax = ⎜ 0 0
=⎜ 0 0
= ⎜ 0 0 Vmaxω cos (ωt ) ⎟
⎟
⎟
⎝ 2r dt ⎠max ⎝ 2rd dt ⎠ max ⎝ 2rd
⎠ max
= μ0ε 0 R 2Vmaxω
2rd ( for r ≥ R ) (note the B ∝ r–1 dependence — see also Eqs. 3216 and 3217). The plot (with SI units
understood) is shown below. 12. From Sample Problem — “Magnetic field induced by changing electric field,” we
know that B ∝ r for r ≤ R and B ∝ r–1 for r ≥ R. So the maximum value of B occurs at r =
R, and there are two possible values of r at which the magnetic field is 75% of Bmax. We
denote these two values as r1 and r2, where r1 < R and r2 > R.
(a) Inside the capacitor, 0.75 Bmax/Bmax = r1/R, or r1 = 0.75 R = 0.75 (40 mm) =30 mm.
(b) Outside the capacitor, 0.75 Bmax/Bmax = (r2/R)–1, or
r2 = R/0.75 = 4R/3 = (4/3)(40 mm) = 53 mm.
(c) From Eqs. 3215 and 3217, μi
μ i ( 4π ×10 T ⋅ m A ) ( 6.0 A )
Bmax = 0 d = 0 =
= 3.0 ×10−5 T.
2π R 2π R
2π ( 0.040 m )
−7 13. Let the area plate be A and the plate separation be d. We use Eq. 3210:
id = ε 0 FG IJ
HK FG IJ
HK dV
dΦ E
d
ε A dV
AE = ε 0 A
= ε0
=0
,
dt d
d
dt
dt
dt bg or dV id d id
15 A
.
=
==
= 7.5 × 105 V s .
dt ε 0 A C 2.0 × 10−6 F CHAPTER 32 1248 The...
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 Fall '08
 schuller
 Magnetism, Work

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