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Potential difference between the ends of a current carrying conductor of length 50 cm is 10 volt. If the drift velocity of free electrons in the conductor is $8 \times 10^{- 5} m / s$ , find the mobility of electrons in $m^{2} V^{- 1} S^{- 1}$ .

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$8.45 \times 10^{- 6}$

$2.15 \times 10^{- 6}$

$4 \times 10^{- 6}$

$6 \times 10^{- 5}$

answer is C.

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Detailed Solution

Electric field inside the conductor

$E = \frac{V}{l} = \frac{10}{0.5} \frac{V}{m} = 20 \frac{V}{m}$

$∴$ Mobility $μ = \frac{V_{d}}{E}$

$∴ μ = \frac{8 \times 10^{- 5}}{20} m^{2} V^{- 1} S^{- 1} = 4 \times 10^{- 6} m^{2} V^{- 1} S^{- 1}$

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