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Unformatted text preview: L I
F 1I
R = RG J G J = RG J G J = Rb2g G J = 2 R.
H 2K
H A KH L K HD K H L K
2 1 1 2 2 2 2 1 2 2 1 21. The resistance at operating temperature T is R = V/i = 2.9 V/0.30 A = 9.67 Ω. Thus,
from R – R0 = R0α (T – T0), we find
T = T0 + ⎛
⎞ ⎛ 9.67 Ω ⎞
1⎛R ⎞
1
− 1⎟ = 1.8 ×103 °C .
⎜ − 1⎟ = 20°C + ⎜
⎟⎜
−3
α ⎝ R0 ⎠
⎝ 4.5 ×10 K ⎠ ⎝ 1.1Ω
⎠ Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the
value of α used in this calculation is not inconsistent with the other units involved. Table
261 has been used.
22. Let r = 2.00 mm be the radius of the kite string and t = 0.50 mm be the thickness of
the water layer. The crosssectional area of the layer of water is
A = π ⎡(r + t ) 2 − r 2 ⎤ = π [(2.50 × 10−3 m) 2 − (2.00 × 10−3 m) 2 ] = 7.07 × 10−6 m 2 .
⎣
⎦
Using Eq. 2616, the resistance of the wet string is
R= ρL
A = (150 Ω ⋅ m ) ( 800 m ) = 1.698 ×1010 Ω.
7.07 ×10−6 m 2 The current through the water layer is
i= V
1.60 ×108 V
=
= 9.42 × 10−3 A .
R 1.698 × 1010 Ω 23. We use J = E/ρ, where E is the magnitude of the (uniform) electric field in the wire, J
is the magnitude of the current density, and ρ is the resistivity of the material. The
electric field is given by E = V/L, where V is the potential difference along the wire and L
is the length of the wire. Thus J = V/Lρ and 1045 ρ= V
115 V
=
= 8.2 × 10−4 Ω ⋅ m.
4
2
LJ
10 m 14 × 10 A m
. b gc h 24. (a) Since the material is the same, the resistivity ρ is the same, which implies (by Eq.
2611) that the electric fields (in the various rods) are directly proportional to their
currentdensities. Thus, J1: J2: J3 are in the ratio 2.5/4/1.5 (see Fig. 2624). Now the
currents in the rods must be the same (they are “in series”) so
J1 A1 = J3 A3 , J2 A2 = J3 A3 . Since A = πr2, this leads (in view of the aforementioned ratios) to
4r22 = 1.5r32 , 2.5r12 = 1.5r32 . Thus, with r3 = 2 mm, the latter relatio...
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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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