Question:

A wire of length 2.0 m and resistance 4.0 Ω is connected to a 12-volt battery. If the wire has a uniform cross-sectional area of 0.001 m^2, then calculate:

a) The current flowing through the wire. b) The potential difference across the wire. c) The resistivity of the wire material.

Assume the wire is made of a homogeneous material.

[Provide step-by-step explanation and solution]

Answer:

a) To find the current flowing through the wire, we can use Ohm's Law, which states that the current (I) flowing through a conductor is equal to the potential difference (V) across the conductor divided by its resistance (R).

Given: Resistance (R) = 4.0 Ω Potential difference (V) = 12 V

Using Ohm's Law:

I = V/R I = 12 V / 4.0 Ω I = 3.0 A

Therefore, the current flowing through the wire is 3.0 A.

b) The potential difference across the wire is the same as the potential difference provided by the battery, which is 12 V.

Therefore, the potential difference across the wire is 12 V.

c) The resistivity (ρ) of the wire material can be determined using the formula:

R = (ρ * L) / A

Where: R is the resistance of the wire, L is the length of the wire, A is the cross-sectional area of the wire, and ρ is the resistivity of the wire material.

Given: Length of the wire (L) = 2.0 m Resistance of the wire (R) = 4.0 Ω Cross-sectional area of the wire (A) = 0.001 m^2

Rearranging the formula, we have:

ρ = (R * A) / L ρ = (4.0 Ω * 0.001 m^2) / 2.0 m ρ = 0.002 Ω * m^2 / 2.0 m ρ = 0.001 Ω * m

Therefore, the resistivity of the wire material is 0.001 Ω * m.

In conclusion: a) The current flowing through the wire is 3.0 A. b) The potential difference across the wire is 12 V. c) The resistivity of the wire material is 0.001 Ω * m.