Question

When $5\,V$ potential difference is applied across a wire of length $0.1\,m$, the drift speed of electrons is $2.5\times10^{-4}$ $ms^{-1}$. If the electron density in the wire is $8\times10^{28}\, m^{-3}$ the resistivity of the material is close to

• A. $1.6\times10^{-8}\Omega\,m$
• B. $1.6\times10^{-7}\Omega\,m$
• C. $1.6\times10^{-5}\Omega\,m$
• D. $1.6\times10^{-6}\Omega\,m$

Answer 1

The Correct Option is C

To determine the resistivity of the material in the wire, we can use several physical relations. The given parameters are:

• Potential difference, V = 5\, \text{V}
• Length of the wire, L = 0.1\, \text{m}
• Drift speed of electrons, v_d = 2.5 \times 10^{-4}\, \text{ms}^{-1}
• Electron density, n = 8 \times 10^{28}\, \text{m}^{-3}

The drift speed is related to the current density J by:

J = n \times e \times v_d

where e = 1.6 \times 10^{-19}\, \text{C}\, (\text{charge of an electron}).

So, the current density J can be calculated as:

J = (8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.5 \times 10^{-4})

J = 3.2 \times 10^{6}\, \text{A/m}^2

The relation between the current density J and electric field E is given by Ohm's law in terms of resistivity:

J = \frac{E}{\rho}

where E = \frac{V}{L}, and \rho is the resistivity.

Thus,

3.2 \times 10^{6} = \frac{5}{0.1 \times \rho}

\rho = \frac{5}{0.1 \times 3.2 \times 10^{6}}

\rho = 1.6 \times 10^{-5}\, \Omega\, \text{m}

The resistivity of the material is close to 1.6 \times 10^{-5}\, \Omega\, \text{m}, which matches the correct answer.