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Unformatted text preview: he quantities):
dΦ B
dB L2 dB
ε=
= Aloop
=
dt
dt 4 π dt
where the rate of change of the field is dB/dt = 0.0100 T/s. Consequently, we obtain ε2 2 ( L2 / 4π ) 2 (dB / dt ) 2 d 2 L3 ⎛ dB ⎞ (1.00 × 10−3 m) 2 (0.500 m)3
2
P=
=
=
( 0.0100 T/s )
⎜
⎟=
−8
2
R
64πρ ⎝ dt ⎠
64π (1.69 ×10 Ω ⋅ m)
ρ L /(π d / 4)
= 3.68 ×10−6 W . 32. Noting that ΔB = B, we find the thermal energy is 1180 CHAPTER 30 38. From the “kink” in the graph of Fig. 3055, we conclude that the radius of the circular
region is 2.0 cm. For values of r less than that, we have (from the absolute value of Eq.
3020)
d Φ B d ( BA)
dB
E (2π r ) =
=
=A
= π r 2a
dt
dt
dt
which means that E/r = a/2. This corresponds to the slope of that graph (the linear
portion for small values of r) which we estimate to be 0.015 (in SI units). Thus,
a = 0.030 T/s.
39. The magnetic field B can be expressed as bg b g B t = B0 + B1 sin ωt + φ 0 ,
where B0 = (30.0 T + 29.6 T)/2 = 29.8 T and B1 = (30.0 T – 29.6 T)/2 = 0.200 T. Then
from Eq. 3025
E= FG IJ
HK b g b g 1 dB
rd
1
r=
B0 + B1 sin ωt + φ 0 = B1ωr cos ωt + φ 0 .
2 dt
2 dt
2 We note that ω = 2πf and that the factor in front of the cosine is the maximum value of
the field. Consequently,
1
1
Emax = B1 ( 2π f ) r = ( 0.200 T )( 2π )(15 Hz ) (1.6 ×10−2 m ) = 0.15 V/m.
2
2
40. Since NΦB = Li, we obtain c hc h 8.0 × 10−3 H 5.0 × 10−3 A
Li
=
= 10 × 10−7 Wb.
ΦB =
.
400
N
41. (a) We interpret the question as asking for N multiplied by the flux through one turn: c h b gc hb gb g 2 Φ turns = NΦ B = NBA = NB πr 2 = 30.0 2.60 × 10−3 T π 0100 m = 2.45 × 10−3 Wb.
. (b) Equation 3033 leads to
L= NΦ B 2.45 × 10−3 Wb
=
= 6.45 × 10−4 H.
380 A
.
i 42. (a) We imagine dividing the oneturn solenoid into N small circular loops placed
along the width W of the copper strip. Each loop carries a current Δi = i/N. Then the
magnetic field inside the solenoid is 1182 CHAPTER 30 or  di / dt  = 5.0A/s. We might, for example, uniformly reduce the current from 2.0 A to
zero in 40 ms.
45. (a) Speaking anthropomorphically, the coil wants to fight the changes—so if it wants
to push current rightward (when the current is already going rightward) then i must be in
the process of decreasing.
(b) From Eq. 3035 (in absolute value) we get
L= ε
di / dt 17 V
= 6.8 × 10−4 H.
2.5kA / s = 46. During periods of time when the current is varying linearly with time, Eq. 3035 (in
absolute values) becomes  ε = L  Δi / Δt  . For simplicity, we omit the absolute value
signs in the following.
(a) For 0 < t < 2 ms, ε=L
(b) For 2 ms < t < 5 ms, ε=L b gb g 4.6 H 7.0 A − 0
Δi
.
=
= 16 × 104 V.
−3
Δt
2.0 × 10 s b gb g 4.6 H 5.0 A − 7.0 A
Δi
= 31 × 103 V.
.
=
−3
Δt
5.0 − 2.0 10 s (c) For 5 ms < t < 6 ms, ε=L b b g gb g 4.6 H 0 − 5.0 A
Δi
=
= 2.3 × 104 V.
−3
Δt
6.0 − 5.0 10 s b g 47. (a) Voltage is proportional to inductance (by Eq. 3035) just as, for resistors, it is
proportional to resistance. Since the (independent) voltages for series elements add (V1 +
V2), then inductances in series must add, Leq = L1 + L2 , just as was th...
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 Fall '08
 schuller
 Magnetism, Work

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