Problem 178 - A Complete Resource Book in Physics for JEE Main

Question

When \(5 \mathrm{~V}\) potential difference is applied across a wire of length \(0.1 \mathrm{~m}\), the drift speed of electrons is \(2.5 \times\) \(10^{-4} \mathrm{~ms}^{-1}\). If the electron density in the wire is \(8 \times\) \(10^{28} \mathrm{~m}^{-3}\), the resistivity of the material is close to (A) \(1.6 \times 10^{-7} \Omega \mathrm{m}\) (B) \(1.6 \times 10^{-6} \Omega \mathrm{m}\) (C) \(1.6 \times 10^{-5} \Omega \mathrm{m}\) (D) \(1.6 \times 10^{-8} \Omega \mathrm{m}\)

Step-by-Step Solution

Answer

The resistivity of the material is close to \(1.6 \times 10^{-7} \Omega \mathrm{m}\) (Option A).

Step 1: Find the current using the drift speed formula and Ohm's law

We are given the drift speed, electron density, and potential difference. Using the drift speed formula and Ohm's law, we can solve for the current. \(I = nAvq\) Since we don't have the value for the cross-sectional area A, we will eliminate it using the resistance formula and Ohm's law. From Ohm's Law, \(I=\frac{V}{R}\), and from the resistance formula, \(R = \frac{\rho L}{A}\), \(I = \frac{V}{\rho}\frac{A}{L}\) Now, equating the two expressions for current, we get: \(nAvq = \frac{V}{\rho}\frac{A}{L}\)

Step 2: Solve for resistivity

Now, we can solve for the resistivity (\(\rho\)), as it is the only unknown variable in the equation. First, cancel out \(V\) and \(A\) from both sides of the equation: \(nqvL = \rho\) Now, plug in the given values: \(n = 8 \times 10^{28} \mathrm{ m^{-3}}\) \(v = 2.5 \times 10^{-4} \mathrm{ ms^{-1}}\) \(q = 1.6 \times 10^{-19} \mathrm{ C}\) (charge of an electron) \(L = 0.1 \mathrm{ m}\) \(\rho = \frac{nqvL}{1}\) \(\rho = \frac{(8 \times 10^{28})(2.5 \times 10^{-4})(1.6 \times 10^{-19})(0.1)}{1}\) \(\rho = 1.6 \times 10^{-7} \Omega \mathrm{m}\) The resistivity of the material is close to \(1.6 \times 10^{-7} \Omega \mathrm{m}\), which corresponds to option (A).