A long wire connects a switch to a bulb. The free electrons in the wire drift with a speed of \(1.5\ \mathrm{mm\,s^{-1}}\). The electric field travels along the wire with a speed of \(3\times10^{8}\ \mathrm{m\,s^{-1}}\). The ratio of the time taken by an electron to drift from the switch to the bulb to the time taken by the field to reach the bulb is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\times10^{11}\)
When the switch is closed, the electric field is set up along the wire with the speed of an electromagnetic wave. The electrons only drift slowly. Both cover the same length \( l \), and time is length divided by speed. So the ratio of the times is the inverse ratio of the speeds.
\( v_d=1.5\ \mathrm{mm\,s^{-1}}=1.5\times10^{-3}\ \mathrm{m\,s^{-1}} \)
\( \begin{aligned} \frac{t_e}{t_f} &= \frac{l/v_d}{l/c} \\ &= \frac{c}{v_d} \\ &= \frac{3\times10^{8}}{1.5\times10^{-3}} \\ &= 2\times10^{11} \end{aligned} \)
Therefore, \( t_e/t_f=2\times10^{11} \). Hence, the bulb glows almost at once, even though each electron moves very slowly.
NCERT (2026-27):
Example 3.1 – “(ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 × 108 m s–1 … The drift speed is, in comparison, extremely small; smaller by a factor of 10–11.”