A wire of 50 cm long, `1mm^(2)` in cross-section carries a current of 4 A, when connected to a 2 V battery, the resistivity of wire is

Options

A. `2 xx 10^(-7) Omega m`
B. `5 xx 10^(-7) Omega m`
C. `4 xx 10^(-6) Omega` m
D. `1 xx 10^(-6) Omega m`

Explanation

To find the resistivity of the wire, we will follow these steps:

### Step 1: Write down the given data
- Length of the wire (L) = 50 cm = 0.5 m
- Cross-sectional area (A) = 1 mm² = 1 x 10⁻⁶ m²
- Current (I) = 4 A
- Voltage (V) = 2 V

### Step 2: Calculate the resistance (R) using Ohm's Law
Ohm's Law states that:
\[ R = \frac{V}{I} \]
Substituting the values:
\[ R = \frac{2 \, \text{V}}{4 \, \text{A}} = 0.5 \, \Omega \]

### Step 3: Use the formula relating resistivity (ρ), resistance (R), length (L), and area (A)
The formula for resistance in terms of resistivity is:
\[ R = \frac{\rho L}{A} \]
We can rearrange this to find the resistivity:
\[ \rho = \frac{R \cdot A}{L} \]

### Step 4: Substitute the values into the formula
Now substituting the values we have:
- R = 0.5 Ω
- A = 1 x 10⁻⁶ m²
- L = 0.5 m

So,
\[ \rho = \frac{0.5 \, \Omega \cdot (1 \times 10^{-6} \, \text{m}^2)}{0.5 \, \text{m}} \]

### Step 5: Simplify the expression
\[ \rho = \frac{0.5 \times 10^{-6}}{0.5} \]
\[ \rho = 1 \times 10^{-6} \, \Omega \cdot \text{m} \]

### Final Answer
The resistivity of the wire is:
\[ \rho = 1 \times 10^{-6} \, \Omega \cdot \text{m} \]

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