A cylindrical conductor of length 2 m and area of cross-section carries an electric current of when its ends are connected to a battery. Mobility of electrons in the conductor is The value of is

Question

A cylindrical conductor of length 2 m and area of cross-section $0.2 {mm}^{2}$ carries an electric current of $1.6 A$ when its ends are connected to a $2 V$ battery. Mobility of electrons in the conductor is $α \times 10^{- 3} m^{2} / V . s .$ The value of $α$ is :

$(electron concentration = 5 \times 10^{28} / m^{3} and electron charge = 1.6 \times 10^{- 19} C)$ [2026]

Smart Achievers · Accepted Answer

(1)

$V_{d} = μ E = μ \frac{V}{ℓ}$

$I = n e A V_{d}$

$V_{d} = \frac{I}{n e A}$

$μ = \frac{I l}{N n e A}$

$μ = \frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{- 19} \times 2 \times 10^{- 7}}$

$μ = 1 \times 10^{- 3} m^{2} / V s$

$α = 1$

Solution

Ans.
(1)

$V_{d} = μ E = μ \frac{V}{ℓ}$

$I = n e A V_{d}$

$V_{d} = \frac{I}{n e A}$

$μ = \frac{I l}{N n e A}$

$μ = \frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{- 19} \times 2 \times 10^{- 7}}$

$μ = 1 \times 10^{- 3} m^{2} / V s$

$α = 1$

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Solution

Ans. (1) Vd=μE=μVℓ I=neAVd Vd=IneA μ=IlNneA μ=1.6×32×5×1026×1.6×10-19×2×10-7 μ=1×10-3 m2/V s α=1

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Ans.
(1)

$V_{d} = μ E = μ \frac{V}{ℓ}$

$I = n e A V_{d}$

$V_{d} = \frac{I}{n e A}$

$μ = \frac{I l}{N n e A}$

$μ = \frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{- 19} \times 2 \times 10^{- 7}}$

$μ = 1 \times 10^{- 3} m^{2} / V s$

$α = 1$