Understanding how a conductor's properties affect its resistance is fundamental in electrical physics. The resistance of a wire depends on its material, length, and cross-sectional area. In this problem, we explore the effect of changing the wire's diameter while keeping its length and material constant.

Here are the details provided for the conductor wire:

• Initial diameter of the wire ($\text{d}_1$) = 0.5 mm
• Initial resistance of the wire ($\text{R}_1$) = 400 Ω
• The length of the wire remains the same.
• The new diameter ($\text{d}_2$) is double the initial diameter, so $\text{d}_2 = 2 \times \text{d}_1$.

Resistance of a Conductor: Basic Principles

The resistance ($\text{R}$) of a conductor is directly proportional to its length ($\text{L}$) and inversely proportional to its cross-sectional area ($\text{A}$). It also depends on the material's resistivity ($\rho$). The formula for resistance is given by:

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

Where:

• $\rho$ (rho) is the resistivity of the material (constant for a given material).
• $\text{L}$ is the length of the conductor.
• $\text{A}$ is the cross-sectional area of the conductor.

Cross-sectional Area and Diameter Relationship

For a cylindrical wire, the cross-sectional area ($\text{A}$) is that of a circle, which is given by:

$$\text{A} = \pi \text{r}^2$$

Where 'r' is the radius of the wire. Since the diameter ($\text{d}$) is twice the radius ($\text{d} = 2\text{r}$), we can express the radius as $\text{r} = \frac{\text{d}}{2}$. Substituting this into the area formula:

$$\text{A} = \pi \left(\frac{\text{d}}{2}\right)^2 = \pi \frac{\text{d}^2}{4}$$

Impact of Diameter on Wire Resistance

From the resistance formula $\text{R} = \rho \frac{\text{L}}{\text{A}}$ and the area formula $\text{A} = \pi \frac{\text{d}^2}{4}$, we can substitute $\text{A}$ into the resistance formula:

$$\text{R} = \rho \frac{\text{L}}{\pi \frac{\text{d}^2}{4}} = \frac{4\rho \text{L}}{\pi \text{d}^2}$$

Since $\rho$, $\text{L}$, and $\pi$ are constants for the same wire and length, we can see that the resistance ($\text{R}$) is inversely proportional to the square of the diameter ($\text{d}^2$). This means:

$$\text{R} \propto \frac{1}{\text{d}^2}$$

This relationship is crucial for solving this problem involving the conductor's resistance.

Step-by-Step Resistance Calculation

We are given an initial resistance ($\text{R}_1$) for an initial diameter ($\text{d}_1$), and we need to find the new resistance ($\text{R}_2$) when the diameter is doubled ($\text{d}_2 = 2\text{d}_1$).

Applying the Inverse Square Relationship:

Given the inverse square relationship $\text{R} \propto \frac{1}{\text{d}^2}$, we can set up a ratio comparing the initial and final states of the wire's resistance:

$$\frac{\text{R}_1}{\text{R}_2} = \frac{1/\text{d}_1^2}{1/\text{d}_2^2} = \frac{\text{d}_2^2}{\text{d}_1^2}$$

Now, substitute the relationship $\text{d}_2 = 2\text{d}_1$ into the ratio:

$$\frac{\text{R}_1}{\text{R}_2} = \frac{(2\text{d}_1)^2}{\text{d}_1^2}$$

$$\frac{\text{R}_1}{\text{R}_2} = \frac{4\text{d}_1^2}{\text{d}_1^2}$$

$$\frac{\text{R}_1}{\text{R}_2} = 4$$

Solving for the New Resistance:

We know $\text{R}_1 = 400 \, \Omega$. Substitute this value into the equation:

$$\frac{400 \, \Omega}{\text{R}_2} = 4$$

To find $\text{R}_2$, rearrange the equation:

$$\text{R}_2 = \frac{400 \, \Omega}{4}$$

$$\text{R}_2 = 100 \, \Omega$$

Therefore, when the diameter of the wire conductor is doubled, its resistance becomes one-fourth of its original value, which is 100 Ω.