Final answer:
The current density in the rod is 1.59 x 10^6 A/m^2. The magnitude of the electric field applied to the rod is 75V. The resistivity of the material is 4.72 × 10^-5 Ω·m and the conductivity is 2.12 × 10^4 Ω^ -1 ·m^ -1.
Step-by-step explanation:
To find the current density, we first need to calculate the cross-sectional area of the rod. The formula for the area of a circle is A = πr^2, where r is the radius of the circle. In this case, the diameter of the rod is given as 0.2 cm, so the radius is 0.1 cm or 0.001 m. Therefore, the area is A = π(0.001^2) = 3.14 x 10^-6 m^2.
Current density (J) is defined as the current (I) divided by the cross-sectional area (A). So, J = I/A = 5A / 3.14 x 10^-6 m^2 = 1.59 x 10^6 A/m^2. Therefore, the current density in the rod is 1.59 x 10^6 A/m^2.
To calculate the magnitude of the electric field (E), we can use Ohm's Law. Ohm's Law states that the electric field (E) is equal to the current density (J) multiplied by the resistivity (ρ).
The formula for Ohm's Law is E = J × ρ. Rearranging the formula, we can solve for ρ: ρ = E / J = 75V / 1.59 × 10^6 A/m^2 = 4.72 × 10^-5 Ω·m.
The resistivity (ρ) of a material is the reciprocal of its conductivity (σ). So, the conductivity of the material can be calculated as σ = 1/ρ = 1 / 4.72 × 10^-5 Ω·m = 2.12 × 10^4 Ω^ -1 ·m^ -1.