Question

A wire \(50 \mathrm{~cm}\) long and \(1 \mathrm{~mm}^{2}\) in cross-section carries a current of \(4 \mathrm{~A}\) when connected to a \(2 \mathrm{~V}\) battery. The resistivity of the wire is: (A) \(2 \times 10^{-7} \Omega \mathrm{m}\) (B) \(5 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4 \times 10^{-6} \Omega \mathrm{m}\) (D) \(1 \times 10^{-6} \Omega \mathrm{m}\)

Short answer

The resistivity of the wire is \(1\times10^{-6} \Omega m\) (Option D).

Step-by-step solution

Calculate the resistance using Ohm's law

Ohm's law states that the voltage is equal to the product of current (I) and resistance (R): \[V = IR\] We are given the current (I = 4 A) and the voltage (V = 2 V). We will use Ohm's law to find the resistance (R): \[R = \frac{V}{I} = \frac{2}{4}\]

Find the resistance value

By solving for R, we find its value: \[R = \frac{2}{4} = 0.5 \Omega\]

Use the resistance formula to find resistivity

The formula for the resistance in terms of resistivity (ρ), length (L), and cross-sectional area (A) is: \[R = \frac{\rho L}{A}\] We are given the length (L = 50 cm = 0.5 m) and cross-sectional area (A = \(1 mm^2 = 1\times10^{-6} m^2\)). We will use the observed resistance (R = 0.5 Ω) and the formula to find the resistivity (ρ): \[\rho = \frac{RA}{L} = \frac{0.5 \times 1\times10^{-6}}{0.5}\]

Calculate the resistivity

By solving for ρ, we find its value: \[\rho = \frac{0.5 \times 1\times10^{-6}}{0.5} = 1\times10^{-6} \Omega m\] The resistivity of the wire is therefore \(1\times10^{-6} \Omega m\), which corresponds to option (D).

Key concepts

Resistivity Calculation

Resistivity is a fundamental property that quantifies how strongly a material opposes the flow of electric current. It's crucial for determining how easy or difficult it is for current to pass through a wire. The symbol for resistivity is \( \rho \), and its units are ohm-meters (\( \Omega \text{m} \)).

To find the resistivity of a wire, we use the following formula:

• \[ R = \frac{\rho L}{A} \]
• \( R \) is the resistance in ohms (\( \Omega \))
• \( L \) is the length of the wire (in meters)
• \( A \) is the cross-sectional area (in square meters)

To solve for resistivity, rearrange the formula:

• \[ \rho = \frac{RA}{L} \]

In our example, the calculated resistivity \( \rho = 1\times10^{-6} \Omega m \) signifies that this material does not resist current flow significantly, making it a good conductor.

Electrical Resistance

Electrical resistance is the measure of an object's opposition to the passage of electric current. It's like the friction for electric circuits.
Resistance is represented by the letter \( R \) and measured in ohms (\( \Omega \)). Using Ohm's Law, which states \( V = IR \), we can determine the amount of resistance an object has, if we know the current and the voltage.

For our specific wire:

• Given: Voltage \( V = 2 \text{ V} \) and Current \( I = 4 \text{ A} \)
• Resistance \( R = \frac{V}{I} = \frac{2}{4} \Omega \)
• Calculated Resistance \( R = 0.5 \Omega \)

Therefore, a resistance of \( 0.5 \Omega \) indicates that this wire only slightly impedes the flow of electric current, which can result in efficient energy transfer through the circuit.

Cross-Sectional Area

The cross-sectional area of a wire influences how much current it can carry. It's essentially the size of the slice you would see if you cut straight through the wire. A larger cross-sectional area allows more current to pass through with less resistance, similar to how a river that is wider can carry more water.

In the formula \( R = \frac{\rho L}{A} \), the cross-sectional area \( A \) is crucial:

• A larger \( A \) means smaller \( R \), meaning less resistance and more current flow.

For the given exercise:

• The wire has an area of \( 1 \text{ mm}^2 \), which is \( 1 \times 10^{-6} \text{ m}^2 \)

Thus, with a modest cross-sectional area of \( 1 \text{ mm}^2 \), the wire in question can effectively convey 4 A of current with the given voltage.