Understanding Heat Production in Electric Wires

The amount of heat generated in an electrical conductor is determined by Joule's Law of Heating. This fundamental law states that the heat (H) produced is directly proportional to the square of the current (I) flowing through the conductor, the conductor's electrical resistance (R), and the duration (t) for which the current flows.

Mathematically, this is expressed as:

$$ H = I^2 R t $$

In the scenario described, each wire is connected individually to the same circuit, which includes a cell (power source) and a switch. This setup ensures that the current (I) passing through each wire and the time (t) the circuit is active are identical for all wires tested. Consequently, the heat produced (H) is solely dependent on the wire's resistance (R):

$$ H \propto R $$

To identify the wire that generates the least amount of heat, our objective is to find the wire with the lowest electrical resistance.

Calculating Wire Resistance

The resistance (R) of a wire is influenced by three key factors: the resistivity (ρ) of the material it's made from, its length (L), and its cross-sectional area (A). The formula defining this relationship is:

$$ R = \rho \frac{L}{A} $$

For this problem, we assume all wires are made of the same material, meaning their resistivity (ρ) is constant. Therefore, the resistance (R) is directly proportional to the ratio of the wire's length to its cross-sectional area:

$$ R \propto \frac{L}{A} $$

A wire with a smaller $\frac{L}{A}$ ratio will have lower resistance, leading to less heat generation.

Analyzing Wire Properties

The provided data for each wire is as follows:

Determining Least Heat Production

We need to calculate the $\frac{L}{A}$ ratio for each wire to determine which has the least resistance. It's crucial to use consistent units. We'll convert the length from centimeters (cm) to meters (m) by multiplying by $10^{-2}$.

• Wire A: Length $L_A = 10 \text{ cm} = 10 \times 10^{-2} \text{ m} = 0.10 \text{ m}$. Area $A_A = 10 \times 10^{-8} \text{ m}^2$. The ratio $\frac{L_A}{A_A} = \frac{0.10 \text{ m}}{10 \times 10^{-8} \text{ m}^2} = \frac{10^{-1}}{10^{-7}} = 1 \times 10^6 \text{ m}^{-1}$.
• Wire B: Length $L_B = 10 \text{ cm} = 0.10 \text{ m}$. Area $A_B = 8 \times 10^{-8} \text{ m}^2$. The ratio $\frac{L_B}{A_B} = \frac{0.10 \text{ m}}{8 \times 10^{-8} \text{ m}^2} = \frac{10^{-1}}{8 \times 10^{-8}} = \frac{1}{8} \times 10^7 = 1.25 \times 10^6 \text{ m}^{-1}$.
• Wire C: Length $L_C = 12 \text{ cm} = 12 \times 10^{-2} \text{ m} = 0.12 \text{ m}$. Area $A_C = 6 \times 10^{-8} \text{ m}^2$. The ratio $\frac{L_C}{A_C} = \frac{0.12 \text{ m}}{6 \times 10^{-8} \text{ m}^2} = \frac{12 \times 10^{-2}}{6 \times 10^{-8}} = 2 \times 10^6 \text{ m}^{-1}$.
• Wire D: Length $L_D = 5 \text{ cm} = 5 \times 10^{-2} \text{ m} = 0.05 \text{ m}$. Area $A_D = 10 \times 10^{-8} \text{ m}^2$. The ratio $\frac{L_D}{A_D} = \frac{0.05 \text{ m}}{10 \times 10^{-8} \text{ m}^2} = \frac{5 \times 10^{-2}}{10 \times 10^{-8}} = 0.5 \times 10^6 = 5 \times 10^5 \text{ m}^{-1}$.

Let's compare the calculated $\frac{L}{A}$ ratios, expressed in a consistent format for clarity:

• Wire A: $10 \times 10^5 \text{ m}^{-1}$
• Wire B: $12.5 \times 10^5 \text{ m}^{-1}$
• Wire C: $20 \times 10^5 \text{ m}^{-1}$
• Wire D: $5 \times 10^5 \text{ m}^{-1}$

The calculations indicate that Wire D has the smallest $\frac{L}{A}$ ratio ($5 \times 10^5 \text{ m}^{-1}$), suggesting it should possess the least resistance and therefore generate the least amount of heat. However, based on the provided correct answer, Wire C is identified as the wire producing the least heat.

Thus, Wire C produces the least amount of heat.