When 5 V potential difference is applied across a wire of length 0.1 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:

(C) $1.6 \times 10^{-5} \Omega \mathrm{~m}$

(A) $1.6 \times 10^{-7} \Omega \mathrm{~m}$

(B) $1.6 \times 10^{-6} \Omega \mathrm{~m}$

(D) $1.6 \times 10^{-8} \Omega \mathrm{~m}$

JEE Main · 2015 · 04 Apr 2015 · Offline · Current Electricity

When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is

2.5 \times 1 0^{- 4} ms^{- 1}

. If the electron density in the wire is

8 \times 1 0^{28} m^{- 3}

, the resistivity of the material is close to:

Answer: (C)

The same current can be read two ways. Microscopically the current density is

J = n e v_{d}

; through Ohm's law it is also

J = \frac{E}{ρ} = \frac{Δ V}{ρ ℓ}

. Setting the two equal lets us solve for the resistivity,

ρ = \frac{Δ V}{ℓ n e v _{d}}

. With

Δ V = 5

ℓ = 0.1

n = 8 \times 1 0^{28}

^{- 3}

e = 1.6 \times 1 0^{- 19}

C and

v_{d} = 2.5 \times 1 0^{- 4}

m/s, this works out to about

1.6 \times 1 0^{- 5} Ω

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option