Question

A metal wire of cross-sectional area \(0.5\,\text{mm}^2\) and length \(100\,\text{m}\) is connected across a battery of e.m.f. \(2\,\text{V}\) and internal resistance \(1\,\Omega\). The density, atomic mass and electrical conductivity of the metal are \(6.35 \times 10^3\,\text{kg m}^{-3}\), \(63.5\,\text{gm/mole}\) and \(2 \times 10^8\,\text{mho m}^{-1}\), respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in \(\text{mm s}^{-1}\)) of the electrons in the wire is:
(Take Avogadro’s number as \(6 \times 10^{23}\) and charge of the electron as \(1.6 \times 10^{-19}\,\text{C}\).)

• A. $0.052$
• B. $0.104$
• C. $0.208$
• D. $0.156$

Hint:

Remember to convert molar mass to kg/mole ($M = 0.0635\text{ kg/mole}$) to maintain SI unit consistency when using density in $\text{kg/m}^3$. This is a common source of decimal errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the drift velocity of electrons in a metal wire given its physical dimensions, material properties (conductivity, density, molar mass), and the electrical circuit parameters it is connected to. We assume each atom contributes one conduction electron.

Step 2: Key Formula or Approach:

• Resistance of wire: $R = \frac{L}{\sigma A}$

• Current in the circuit: $I = \frac{E}{R + r}$

• Drift velocity: $v_d = \frac{I}{neA}$

• Electron number density: $n = \frac{\rho \times N_A}{M_{molar}}$

Step 3: Detailed Explanation:

• Calculate the resistance of the wire ($R$): \[ A = 0.5 \text{ mm}^2 = 0.5 \times 10^{-6} \text{ m}^2, \quad L = 100 \text{ m}, \quad \sigma = 2 \times 10^8 \text{ mho m}^{-1} \] \[ R = \frac{100}{(2 \times 10^8)(0.5 \times 10^{-6})} = \frac{100}{100} = 1\text{ }\Omega \]
• Calculate the current ($I$) flowing through the wire: \[ E = 2\text{ V}, \quad r = 1\text{ }\Omega \] \[ I = \frac{2}{1 + 1} = 1\text{ A} \]
• Calculate the electron number density ($n$): \[ \rho = 6.35 \times 10^3 \text{ kg m}^{-3}, \quad M = 63.5 \text{ g/mole} = 0.0635 \text{ kg/mole} \] \[ n = \frac{(6.35 \times 10^3) \times (6 \times 10^{23})}{0.0635} = 6 \times 10^{28} \text{ m}^{-3} \]
• Calculate the drift velocity ($v_d$): \[ v_d = \frac{1}{(6 \times 10^{28})(1.6 \times 10^{-19})(0.5 \times 10^{-6})} \] \[ v_d = \frac{1}{4.8 \times 10^3} \approx 0.0002083 \text{ m s}^{-1} \]
• Convert to $\text{mm s}^{-1}$: \[ v_d \approx 0.208 \text{ mm s}^{-1} \]

Step 4: Final Answer:
The drift velocity is $0.208\text{ mm s}^{-1}$.