L₁ D₁ L₂ D2 Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length. a) What is the resistance of this wire? R = Hint A potential difference of AV = 1.40 V is applied across the wire. b) What is the magnitude of the current density in the thin part of the wire? Hint J1 = c) What is the magnitude of the current density in the thick part of the wire? J₂ = d) What is the magnitude of the electric field in the thin part of the wire? E1 = Hint e) What is the magnitude of the electric field in the thick part of the wire? E2 =

University Physics Volume 2
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Chapter9: Current And Resistance
Section: Chapter Questions
Problem 28P: An aluminum wire 1.628 mm in diameter (14-gauge) carries a current of 3.00 amps, (a) What is the...
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L₁
D₁
L₂
D2
Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a
circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of
L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length.
a) What is the resistance of this wire?
R =
Hint
A potential difference of AV = 1.40 V is applied across the wire.
b) What is the magnitude of the current density in the thin part of the wire?
Hint
J1
=
c) What is the magnitude of the current density in the thick part of the wire?
J₂ =
d) What is the magnitude of the electric field in the thin part of the wire?
E1
=
Hint
e) What is the magnitude of the electric field in the thick part of the wire?
E2
=
Transcribed Image Text:L₁ D₁ L₂ D2 Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length. a) What is the resistance of this wire? R = Hint A potential difference of AV = 1.40 V is applied across the wire. b) What is the magnitude of the current density in the thin part of the wire? Hint J1 = c) What is the magnitude of the current density in the thick part of the wire? J₂ = d) What is the magnitude of the electric field in the thin part of the wire? E1 = Hint e) What is the magnitude of the electric field in the thick part of the wire? E2 =
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