The 16-gauge copper wire with radius 5.1 x 10-4 m and its area of cross section is 8.20 x 10-7 m 2 . The amount of electric current is 1.67 ampere passing through in it. Calculate (a) The magnitude of electric field in the wire (b) The electric potential difference between two points 50 x 103 m apart. (c) What is the resistance of 50 m length of copper wire? Whereas the resistivity of copper wire is 1.72 x 10-8 Ω
Answers
Answer: The 16-gauge copper wire with radius 5.1 x 10-4 m and its area of cross section is 8.20 x 10-7 m 2 . The amount of electric current is 1.67 ampere passing through in it. Calculate (a) The magnitude of electric field in the wire (b) The electric potential difference between two points 50 x 103 m apart. (c) What is the resistance of 50 m length of copper wire? Whereas the resistivity of copper wire is 1.72 x 10-8 Ω
Explanation:
Given:
The 16-gauge copper wire with radius 5.1 x 10-4 m and its area of cross section is 8.20 x 10-7 m 2 . The amount of electric current is 1.67 ampere passing through in it.
To find:
Calculate (a) The magnitude of electric field in the wire
(b) The electric potential difference between two points 50 x 103 m apart.
(c) What is the resistance of 50 m length of copper wire? Whereas the resistivity of copper wire is 1.72 x 10-8 Ω
Solution:
From given, we have,
(a)
We use the formula, I = AE/ρ
E = Iρ/A
E = (1.67 × 1.72 × 10^{-8}) / (8.20 × 10^{-7})
∴ E = 0.035 V/m
The magnitude of electric field in the wire is 0.035 V/m
(b)
The electric potential difference between two points is calculated using the formula,
V = El
V = 0.035 × 50 × 10^3
∴ V = 1750 V
The electric potential difference between two points 50 x 103 m apart is 1750 V
(c)
The resistance of 50 m length of copper wire given the resistivity of copper wire is 1.72 x 10-8 Ω is calculated using the formula,
R = ρ l/A
R = 1.72 × 10^{-8} × 50/(8.2 × 10^{-7})
∴ R = 1.048 Ω
The resistance of 50 m length of copper wire is 1.048 Ω