Question

Answer (Detailed Solution Below)

Option 2 : $1.47 \times 10^{-4}\ \mathrm{m/s}$ / $1.47 \times 10^{-4}\ \mathrm{m/s}$

Detailed Solution

सही उत्तर $1.47 \times 10^{-4}\ \mathrm{m/s}$ / $1.47 \times 10^{-4}\ \mathrm{m/s}$ है।

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Question:

A current of $2\ \mathrm{A}$ flows through a copper wire of cross-sectional area $1\ \mathrm{mm^2}$. If the free-electron density is $8.5 \times 10^{28}\ \mathrm{m^{-3}}$, the drift velocity is approximately: / $1\ \mathrm{mm^2}$ अनुप्रस्थ-काट क्षेत्रफल के ताँबे के तार में $2\ \mathrm{A}$ धारा बहती है। यदि मुक्त-इलेक्ट्रॉन घनत्व $8.5 \times 10^{28}\ \mathrm{m^{-3}}$ हो, तो अनुगमन वेग लगभग है:

$7.4 \times 10^{-5}\ \mathrm{m/s}$ / $7.4 \times 10^{-5}\ \mathrm{m/s}$

$1.47 \times 10^{-4}\ \mathrm{m/s}$ / $1.47 \times 10^{-4}\ \mathrm{m/s}$

$2.9 \times 10^{-4}\ \mathrm{m/s}$ / $2.9 \times 10^{-4}\ \mathrm{m/s}$

$5.9 \times 10^{-4}\ \mathrm{m/s}$ / $5.9 \times 10^{-4}\ \mathrm{m/s}$

Correct Answer: $1.47 \times 10^{-4}\ \mathrm{m/s}$ / $1.47 \times 10^{-4}\ \mathrm{m/s}$

$v_d = I/(neA) = 2/(8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}) \approx 1.47 \times 10^{-4}\ \mathrm{m/s}$. / $v_d = I/(neA) = 2/(8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}) \approx 1.47 \times 10^{-4}\ \mathrm{m/s}$।

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