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Unformatted text preview: the current flowing through the wire is
V 3.00 × 10−9 V
= 1.115 × 10−4 A .
i= =
−5
R 2.69 × 10 Ω
Therefore, in 3.00 ms, the amount of charge drifting through a cross section is
ΔQ = iΔt = (1.115 × 10−4 A)(3.00 × 10−3 s) = 3.35 × 10−7 C .
30. We use R ∝ L/A. The diameter of a 22gauge wire is 1/4 that of a 10gauge wire.
Thus from R = ρL/A we find the resistance of 25 ft of 22gauge copper wire to be
R = (1.00 Ω)(25 ft/1000 ft)(4)2 = 0.40 Ω.
31. (a) The current in each strand is i = 0.750 A/125 = 6.00 × 10–3 A. 1049
36. Since the current spreads uniformly over the hemisphere, the current density at any
given radius r from the striking point is J = I / 2π r 2 . From Eq. 2610, the magnitude of
the electric field at a radial distance r is
ρI
E = ρw J = w 2 ,
2π r
where ρ w = 30 Ω ⋅ m is the resistivity of water. The potential difference between a point at
radial distance D and a point at D + Δr is ΔV = − ∫ D +Δr D Edr = − ∫ D +Δr D ρw I
ρ I⎛ 1
ρI
1⎞
Δr
dr = w ⎜
− ⎟=− w
,
2
2π r
2π ⎝ D + Δr D ⎠
2π D( D + Δr ) which implies that the current across the swimmer is
i=  ΔV  ρ w I
Δr
=
.
R
2π R D( D + Δr ) Substituting the values given, we obtain
(30.0 Ω ⋅ m)(7.80 × 104 A)
0.70 m
i=
= 5.22 × 10−2 A .
3
2π (4.00 ×10 Ω)
(35.0 m)(35.0 m + 0.70 m) 37. From Eq. 2625, ρ ∝ τ–1 ∝ veff. The connection with veff is indicated in part (b) of
Sample Problem —“Mean free time and mean free distance,” which contains useful
insight regarding the problem we are working now. According to Chapter 20, veff ∝ T .
Thus, we may conclude that ρ ∝ T . 38. The slope of the graph is P = 5.0 × 10−4 W. Using this in the P = V2/R relation leads
to V = 0.10 Vs.
39. Eq. 2626 gives the rate of thermal energy production:
P = iV = (10.0A)(120V) = 1.20 kW. Dividing this into the 180 kJ necessary to cook the three hotdogs leads to the result
t = 150 s.
40. The resistance is R = P/i2 = (100 W)/(3.00 A)2 = 11.1 Ω.
41. (a)...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.
 Fall '08
 schuller
 Magnetism, Work

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