Question

The number density of free electrons in a copper conductor estimated in Example 3.1 is \(8.5 × 10^{28} m^{−3}\). 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 \(2.0 × 10^{−6} m^2\) and it is carrying a current of 3.0 A.

Answer 1

Number density of free electrons in a copper conductor, \(n = 8.5 × 10^{28} m^{−3}\) Length of the copper wire, \(l = 3.0 m\)
Area of cross-section of the wire,\( A = 2.0 × 10^{−6} m^2\)
Current carried by the wire,\( I = 3.0 A\), which is given by the relation,
\(I = nAeV_d\)
Where, 
e = Electric charge = \(1.6 × 10^{−19} C\)
\(V_d = Drift\space  velocity =\frac{ Length \space of \space the \space wire (I)}{Time \space taken\space  to\space  cover l (t)}\)
\(I = nAe\frac{l}{t}\)
\(t = \frac{nAel}{I}\)
\(t = \frac{3 \times 8.5 \times 10^{28} \times 2 \times10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(t = 2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is \(2.7 \times 10^{4} s.\)