A 4m long copper wire has a diameter of 2mm and carries a current of 10A.

Determine the ff:
a. current density in the wire
b. electric field
c. potential difference between the ends of the wire
d. resistance of the wire

To determine the current density in the wire, we need to find the cross-sectional area of the wire.

a. Current density (J) is defined as the current (I) passing through a cross-sectional area (A) of a conductor. Mathematically, it is given by J = I / A.

The diameter of the wire is given as 2 mm. We can use this to find the radius (r) of the wire, which is half the diameter. So, r = 2 mm / 2 = 1 mm = 0.001 m.

Now, we can calculate the cross-sectional area (A) using the formula A = π * r^2, where π (pi) is approximately 3.14159.

A = 3.14159 * (0.001 m)^2 = 3.14159 * 0.000001 m^2 = 0.0000031416 m^2.

Since we know that the current passing through the wire is 10 A, we can now calculate the current density.

J = I / A = 10 A / 0.0000031416 m^2 ≈ 3,185,289 A/m^2.

Therefore, the current density in the wire is approximately 3,185,289 A/m^2.

b. To determine the electric field (E), we can use Ohm's law extended to include resistivity (ρ), which is a property of the material the wire is made of. The formula is E = ρ * J, where ρ is given for copper as 1.7 * 10^-8 Ωm.

E = (1.7 * 10^-8 Ωm) * (3,185,289 A/m^2) ≈ 54.13 V/m.

Therefore, the electric field inside the copper wire is approximately 54.13 V/m.

c. The potential difference (V) across the wire can be found using Ohm's law: V = I * R, where R is the resistance of the wire.

We can find the resistance of the wire using the formula R = ρ * (L / A), where L is the length of the wire.

The length of the wire is given as 4 m, and we previously calculated the cross-sectional area as 0.0000031416 m^2.

R = (1.7 * 10^-8 Ωm) * (4 m / 0.0000031416 m^2) ≈ 0.0216 Ω.

Now we can calculate the potential difference across the wire.

V = I * R = 10 A * 0.0216 Ω = 0.216 V.

Therefore, the potential difference between the ends of the wire is approximately 0.216 V.

d. We already found the resistance of the wire as 0.0216 Ω.

Therefore, the resistance of the wire is 0.0216 Ω.