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. `2xx10^(-7) Omega-m`
B. `5xx10^(-7) Omega-m`
C. `4xx10^(-6) Omega-m`
D. `1xx10^(-6) Omega-m`

Answer

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

### Step 1: Understand the given data
- Length of the wire (L) = 50 cm = 0.5 m (since 1 cm = 0.01 m)
- Cross-sectional area (A) = 1 mm² = \(1 \times 10^{-6}\) m² (since 1 mm = 0.001 m)
- Current (I) = 4 A
- Voltage (V) = 2 V

### Step 2: Use Ohm's Law to find the resistance (R)
Ohm's Law states that:
\[ V = I \cdot R \]
From this, we can rearrange to find R:
\[ R = \frac{V}{I} \]

Substituting the values:
\[ R = \frac{2 \, \text{V}}{4 \, \text{A}} = 0.5 \, \Omega \]

### Step 3: Use the formula for resistivity (ρ)
The resistivity (ρ) of a material can be calculated using the formula:
\[ \rho = R \cdot \frac{A}{L} \]

### Step 4: Substitute the values into the resistivity formula
We have:
- R = 0.5 Ω
- A = \(1 \times 10^{-6}\) m²
- L = 0.5 m

Now substituting these values:
\[ \rho = 0.5 \, \Omega \cdot \frac{1 \times 10^{-6} \, \text{m}^2}{0.5 \, \text{m}} \]

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

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

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