JEE Challenger
More from Current Electricity
PhyMedAdv 2026 • Paper 2

Drift Velocity of Electrons in Metal Wire

A metal wire of cross-sectional area 0.5 mm20.5\text{ mm}^2 and length 100 m100\text{ m} is connected across a battery of e.m.f. 2 V2\text{ V} and internal resistance 1 Ω1\ \Omega. The density, atomic mass and electrical conductivity of the metal are 6.35×103 kg m−36.35 \times 10^3\text{ kg m}^{-3}, 63.5 gm/mole63.5\text{ gm/mole} and 2×108 mho m−12 \times 10^8\text{ mho m}^{-1}, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s−1\text{mm s}^{-1}) of the electrons in the wire is: [Take Avogadro's number as 6×10236 \times 10^{23} and charge of the electron as 1.6×10−19 C1.6 \times 10^{-19}\text{ C}.]

Options

A

0.052

B

0.104

C

0.208

Correct
D

0.156

Topics & Concepts

Current ElectricityOhm's Law

Step-by-Step Solution

To find the drift velocity of the electrons in the wire, we use the fundamental relationships governing electric current, material properties, and electron mobility.

Step 1: Resistance of the Wire

The electrical resistance RR of a wire of length LL, cross-sectional area AA, and conductivity σ\sigma is given by: R=LσAR = \frac{L}{\sigma A}

Given:

  • Length, L=100 mL = 100\text{ m}
  • Cross-sectional area, A=0.5 mm2=0.5×10−6 m2A = 0.5\text{ mm}^2 = 0.5 \times 10^{-6}\text{ m}^2
  • Electrical conductivity, σ=2×108 mho m−1\sigma = 2 \times 10^8\text{ mho m}^{-1}

Substituting the values: R=100(2×108)×(0.5×10−6)=100100=1 ΩR = \frac{100}{(2 \times 10^8) \times (0.5 \times 10^{-6})} = \frac{100}{100} = 1\ \Omega

Step 2: Total Current in the Circuit

The wire is connected across a battery of e.m.f. E=2 VE = 2\text{ V} and internal resistance r=1 Ωr = 1\ \Omega.

The total resistance of the circuit is: Rtotal=R+r=1 Ω+1 Ω=2 ΩR_{\text{total}} = R + r = 1\ \Omega + 1\ \Omega = 2\ \Omega

Thus, the electric current II flowing through the wire is: I=ERtotal=2 V2 Ω=1 AI = \frac{E}{R_{\text{total}}} = \frac{2\text{ V}}{2\ \Omega} = 1\text{ A}

Step 3: Number Density of Conduction Electrons

The number density nn (number of free electrons per unit volume) of the metal is related to its mass density dd, Avogadro's number NAN_A, and molar mass MM: n=d⋅NAMn = \frac{d \cdot N_A}{M}

Given:

  • Density, d=6.35×103 kg m−3d = 6.35 \times 10^3\text{ kg m}^{-3}
  • Atomic mass, M=63.5 g mol−1=63.5×10−3 kg mol−1M = 63.5\text{ g mol}^{-1} = 63.5 \times 10^{-3}\text{ kg mol}^{-1}
  • Avogadro's number, NA=6×1023 mol−1N_A = 6 \times 10^{23}\text{ mol}^{-1}
  • 11 conduction electron per atom

Substituting the values: n=(6.35×103 kg m−3)×(6×1023 mol−1)63.5×10−3 kg mol−1n = \frac{(6.35 \times 10^3\text{ kg m}^{-3}) \times (6 \times 10^{23}\text{ mol}^{-1})}{63.5 \times 10^{-3}\text{ kg mol}^{-1}} n=6.35×10363.5×10−3×6×1023=105×6×1023=6×1028 m−3n = \frac{6.35 \times 10^3}{63.5 \times 10^{-3}} \times 6 \times 10^{23} = 10^5 \times 6 \times 10^{23} = 6 \times 10^{28}\text{ m}^{-3}

Step 4: Drift Velocity Calculation

The electric current II is related to the drift velocity vdv_d by: I=nAevdI = n A e v_d

Rearranging for drift velocity vdv_d: vd=InAev_d = \frac{I}{n A e}

Where e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C} is the charge of an electron.

Substituting the known quantities: vd=1(6×1028)×(0.5×10−6)×(1.6×10−19)v_d = \frac{1}{(6 \times 10^{28}) \times (0.5 \times 10^{-6}) \times (1.6 \times 10^{-19})} vd=13×1022×1.6×10−19=14.8×103 m s−1v_d = \frac{1}{3 \times 10^{22} \times 1.6 \times 10^{-19}} = \frac{1}{4.8 \times 10^3}\text{ m s}^{-1}

Converting vdv_d from m s−1\text{m s}^{-1} to mm s−1\text{mm s}^{-1}: vd=1034.8×103 mm s−1=14.8 mm s−1≈0.208 mm s−1v_d = \frac{10^3}{4.8 \times 10^3}\text{ mm s}^{-1} = \frac{1}{4.8}\text{ mm s}^{-1} \approx 0.208\text{ mm s}^{-1}

Conclusion

The drift velocity of the electrons in the wire is 0.208 mm s−10.208\text{ mm s}^{-1}, which corresponds to Option (C).

Drift Velocity of Electrons in Metal Wire | Physics PYQ Solution - JEE Challenger