Answer:
a. 50.9 mΩ b. 9.32 × 10¹⁸ electrons/s c. 4.55 × 10⁵ A/m² d. i. resistance and current density ii. They would decrease.
Step-by-step explanation:
a. The resistance of the copper wire is given by
R = ρl/A where ρ = resistivity of copper wire = 1.68 × 10⁻⁸ Ωm, l = length of copper wire = 10 m and A = cross-sectional area of coper wire = πd²/4 where d = diameter of copper wire = 2.05 mm = 2.05 × 10⁻³ m
A = πd²/4
= π(2.05 × 10⁻³ m)²/4
= 13.2025/4 × 10⁻⁶ m²
= 3.3 × 10⁻⁶ m².
So R = ρl/A
= 1.68 × 10⁻⁸ Ωm × 10 m/3.3 × 10⁻⁶ m²
= 0.0509 Ω
= 50.9 mΩ
b. Since a current of 1.5 A flows through the wire, it means that 1.5 C/s, that is 1.5 Coulombs of charge flows through it per second.
Since 1 electron = 1.609 × 10⁻¹⁹ C, the number of electrons in 1.5 A is 1.5 C/s ÷ 1.609 × 10⁻¹⁹ C per electron = 1.5/1.609 × 10⁻¹⁹ C = 9.32 × 10¹⁸ electrons/s
c. The current density J = I/A where I = current = 1.5 A and A = cross-sectional area of copper wire = 3.3 × 10⁻⁶ m²
J = 1.5 A/3.3 × 10⁻⁶ m²
= 4.55 × 10⁵ A/m²
d. i. If the diameter were twice the initial diameter, d' = 2d, then the resistance and current density would change since they are dependent of the cross-sectional area of the wire which is then dependent on the diameter of the wire.
ii. If the diameter were twice the initial diameter, d' = 2d, then since the cross-sectional area A' = πd'²/4 = π(2d)²/4 = 4πd²/4 = 4A. So, the cross-sectional area increases by a factor of four.
The new resistance R' = ρl/4A = R/4
The new current density J' = I/4A = J/4.
So the resistance and current density would decrease.