Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A uniform copper wire of length 1 m and cross-sectional area $5 \times {10^{ - 7}}{m^2}$ carries a current of 1 A. Assuming that there are $8 \times {10^{28}}$ free electrons per m3 in copper, how long will an electron take to drift from one end of the wire to the other?

Option 1)
$0.8 \times {10^3}$ sec

Option 2)
$1.6\times {10^3}$ sec

Option 3)
$3.2\times {10^3}$ sec

Option 4)
$6.4\times {10^3}$ sec

Answers (1)

As we learnt

Translatory motion of charge -

$i=nqvA$

- wherein

If there are $n$ particles per unit volume each having charge $q$ and moving with velocity $v$

The drift velocity of electrons is given by

${V_d} = \frac{I}{{enA}}$$

If l is the length of the wire, the time taken is

$t = \frac{\ell }{{{V_d}}} = \frac{{\ell \,e\,n\,A}}{I} = \frac{{1 \times 1.6 \times {{10}^{ - 19}} \times 8 \times {{10}^{28}} \times 5 \times {{10}^{ - 7}}}}{1}$$

= $6.4 \times {10^3}$$ sec

Option 1)

$0.8 \times {10^3}$$ sec

Option 2)

$1.6\times {10^3}$$ sec

Option 3)

$3.2\times {10^3}$$ sec

Option 4)

$6.4\times {10^3}$$ sec

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question