Question

A copper wire of length 10 m and radius \((\frac{10^{-2}} { √π})\) has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of \((\frac{V}{m})\) is:

• A. \(10^4 \)\(\frac{A}{m^2}\)
• B. \(10^6\) \(\frac{A}{m^2}\)
• C. \(10^{-5}\) \(\frac{A}{m^2}\)
• D. \(10^5\) \(\frac{A}{m^2}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Current density (\(J\)) is defined as the current per unit cross-sectional area (\(J = I/A\)).
According to the microscopic form of Ohm's Law, current density is also proportional to the electric field (\(E\)) applied across the conductor: \(J = \sigma E\), where \(\sigma\) is the electrical conductivity.
Key Formula or Approach:
1. Conductivity (\(\sigma\)): The reciprocal of resistivity (\(\rho\)). Since \(R = \rho \frac{L}{A} = \frac{1}{\sigma} \frac{L}{A}\), we have \(\sigma = \frac{L}{R A}\).
2. Area (\(A\)): For a wire with circular cross-section, \(A = \pi r^2\).
3. Current Density (\(J\)): \(J = \sigma E\).
Step 2: Detailed Explanation:
1. Calculate the Cross-sectional Area (\(A\)):
Radius \(r = \frac{10^{-2}}{\sqrt{\pi}} \text{ m}\).
\[ A = \pi r^2 = \pi \left( \frac{10^{-2}}{\sqrt{\pi}} \right)^2 \]
\[ A = \pi \times \frac{10^{-4}}{\pi} = 10^{-4} \text{ m}^2 \]
2. Calculate the Conductivity (\(\sigma\)):
Length \(L = 10 \text{ m}\).
Resistance \(R = 10 \text{ \(\Omega\)}\).
\[ \sigma = \frac{L}{R A} = \frac{10}{10 \times 10^{-4}} \]
The 10s cancel out:
\[ \sigma = \frac{1}{10^{-4}} = 10^4 \text{ S/m} \]
3. Calculate the Current Density (\(J\)):
Electric field \(E = 10 \text{ V/m}\).
\[ J = \sigma E = (10^4 \text{ S/m}) \times (10 \text{ V/m}) \]
\[ J = 10^5 \text{ A/m}^2 \]
Step 3: Final Answer:
The current density in the wire is \(10^5 \text{ A/m}^2\).