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Unformatted text preview: refore, we need to change the voltage difference across the capacitor at the rate of
7.5 ×105 V/s .
14. Consider an area A, normal to a uniform electric field E . The displacement current
density is uniform and normal to the area. Its magnitude is given by Jd = id/A. For this
situation , id = ε 0 A(dE / dt ) , so
dE
dE
1
Jd = ε 0 A
= ε0
.
A
dt
dt
15. The displacement current is given by id = ε 0 A(dE / dt ), where A is the area of a plate
and E is the magnitude of the electric field between the plates. The field between the
plates is uniform, so E = V/d, where V is the potential difference across the plates and d is
the plate separation. Thus,
ε A dV
id = 0
.
d dt
Now, ε0A/d is the capacitance C of a parallelplate capacitor (not filled with a dielectric),
so
dV
id = C
.
dt
16. We use Eq. 3214: id = ε 0 A(dE / dt ). Note that, in this situation, A is the area over
which a changing electric field is present. In this case r > R, so A = πR2. Thus,
i
i
dE
2.0 A
V
.
= d = d 2=
= 7.2 ×1012
2
m ⋅s
dt ε 0 A ε 0π R
π ( 8.85 × 10−12 C2 /N ⋅ m 2 ) ( 0.10 m ) 17. (a) Using Eq. 2710, we find E = ρJ = ρi
A .
c162 × 10
= −8 hb Ω ⋅ m 100 A
−6 5.00 × 10 m 2 g = 0.324 V m . (b) The displacement current is id = ε 0 dΦE
dE
d ⎛ ρi ⎞
di
= ε0 A
= ε 0 A ⎜ ⎟ = ε 0 ρ = 8.85 × 10−12 F/m 1.62 × 10−8 Ω ( 2000 A s )
dt
dt
dt ⎝ A ⎠
dt ( )( ) = 2.87 × 10−16 A.
(c) The ratio of fields is B ( due to id ) μ0id 2π r id 2.87 ×10−16 A
=
==
= 2.87 ×10−18.
B ( due to i ) μ0i 2π r i
100 A 18. From Eq. 2811, we have i = (ε / R ) e−
the capacitor. Equation 3216 gives t/ τ since we are ignoring the selfinductance of 1249
B= μ0id r
.
2π R 2 Furthermore, Eq. 259 yields the capacitance
C= ε 0π (0.05 m) 2
0.003 m = 2.318 × 10−11 F , so that the capacitive time constant is τ = (20.0 × 106 Ω)(2.318 × 10−11 F) = 4.636 × 10−4 s.
At t = 250 × 10−6 s, the current is
i= 12.0 V
t/
e− τ = 3.50 × 10−7 A .
20.0 x 106 Ω Since i = id (see Eq. 3215) and r = 0.0300 m, then (with plate radius R = 0.0500 m) we
find
μ i r (4π× 10−7 T ⋅ m/A)(3.50 × 10−7 A)(0.030 m)
B= 0d2 =
= 8.40 × 10−13 T .
2
2π R
2π (0.050 m)
2 19. (a) Equation 3216 (with Eq. 265) gives, with A = πR ,
B= μ0id r μ0 J d Ar μ0 J d (π R 2 )r 1
=
=
= μ0 J d r
2π R 2
2π R 2
2π R 2
2 1
(4π× 10−7 T ⋅ m/A)(6.00 A/m 2 )(0.0200 m) = 75.4 nT .
2
μi
μ J π R2
= 67.9 nT .
(b) Similarly, Eq. 3217 gives B = 0 d = 0 d
2π r
2π r
= 20. (a) Equation 3216 gives B =
(b) Equation 3217 gives B = μ0id r
= 2.22 μ T .
2π R 2 μ0id
= 2.00 μ T .
2π r 21. (a) Equation 3211 applies (though the last term is zero) but we must be careful with
id,enc . It is the enclosed portion of the displacement current, and if we related this to the
displacement current density Jd , then
r
r
⎛1
r3 ⎞
id enc = ∫ J d 2π r dr = (4.00 A/m 2 )(2π ) ∫ (1 − r / R ) r dr = 8π ⎜ r 2 −
⎟
0
0
3R ⎠
⎝2...
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 Fall '08
 schuller
 Magnetism, Work

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