Question

A conductor of length 1.5 m and area of cross-section \( 3 \times 10^{-5} \, \text{m}^2 \) has electrical resistance of 15 \( \Omega \). The current density in the conductor for an electric field of 21 \( \text{V/m} \) is.

• A. \( 0.7 \times 10^6 \, \text{A/m}^2 \)
• B. \( 0.7 \times 10^{-6} \, \text{A/m}^2 \)
• C. \( 0.7 \times 10^{-5} \, \text{A/m}^2 \)
• D. \( 0.7 \times 10^5 \, \text{A/m}^2 \) \bigskip

Hint:

To find the current density, first calculate the resistivity using Ohm's law and then use it to find \( J \) with the formula \( J = \frac{E}{\rho} \).

Answer 1

The Correct Option is B

Step 1: Using Ohm's law \( J = \frac{E}{\rho} \), where \( J \) is the current density, \( E \) is the electric field, and \( \rho \) is the resistivity. The resistivity \( \rho \) can be calculated using: \[ R = \rho \cdot \frac{L}{A} \] where \( R = 15 \, \Omega \), \( L = 1.5 \, \text{m} \), and \( A = 3 \times 10^{-7} \, \text{m}^2 \). \[ \rho = \frac{R \cdot A}{L} = \frac{15 \times 3 \times 10^{-7}}{1.5} = 3 \times 10^{-6} \, \Omega \cdot \text{m} \] Now, substituting the values: \[ J = \frac{21}{3 \times 10^{-6}} = 0.7 \times 10^7 \, \text{A/m}^{-6} \] Thus, the current density is \( \boxed{0.7 \times 10^7 \, \text{A/m}^{-6}} \). \bigskip