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: Identify the given values
- Length of the wire (L) = 50 cm = 0.50 m (convert cm to m by dividing by 100)
- Cross-sectional area (A) = 1 mm² = 1 × 10⁻⁶ m² (convert mm² to m² by multiplying by 10⁻⁶)
- 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 given values:
\[ R = \frac{2 \, \text{V}}{4 \, \text{A}} = 0.5 \, \Omega \]

### Step 3: Use the formula for resistance in terms of resistivity
The resistance (R) of a wire can also be expressed as:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 4: Rearranging the formula to find resistivity
To find the resistivity \( \rho \), we rearrange the formula:
\[ \rho = \frac{R \cdot A}{L} \]

### Step 5: Substitute the known values into the equation
Now we can substitute the values we have:
- \( R = 0.5 \, \Omega \)
- \( A = 1 \times 10^{-6} \, \text{m}^2 \)
- \( L = 0.50 \, \text{m} \)

So,
\[ \rho = \frac{0.5 \, \Omega \cdot (1 \times 10^{-6} \, \text{m}^2)}{0.50 \, \text{m}} \]

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

### Final Answer
The resistivity of the wire is \( 1 \times 10^{-6} \, \Omega \cdot \text{m} \).

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