Ncert Exercises & Solutions

3. 13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m^–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^–6 m² and it is carrying a current of 3.0 A.

Number of density of free electrons in a copper conductor, n = 8.5 $\times$ 10²⁸ m^-3

Length of the copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 $\times$ 10^-6 m²

Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeV_d

Where, e = Electric charge = 1.6 $\times$ 10^-19 C

$V_{d} = Drift velocity = \frac{Length of the wire (l)}{Time taken to cover l (t)}$
$I = nAe \frac{l}{t}$
$t = \frac{nAel}{I}$
$= \frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{- 6} \times 1.6 \times 10^{- 19}}{3.0}$
$= 2.7 \times 10^{4} s$

Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7 \times 10^{4} s$

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