Respuesta :
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
[tex]R=\frac{L}{A\sigma }[/tex]
[tex]R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }[/tex]
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
[tex]i = \frac{V}{R}[/tex]
[tex]i = \frac{0.060}{0.0178}[/tex]
i = 3.4 A
(c)
Current density is given as
[tex]J = \frac{i}{A}[/tex]
[tex]J = \frac{3.4}{5.3\times10^{-6}}[/tex]
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
[tex]E = \frac{J}{\sigma }[/tex]
[tex]E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}[/tex]
E = 9.01 x 10⁻³ V/m
