Answer to Question #280983 in Electric Circuits for Anonymous005

Question

The cross-sectional area of a wire containing 0.67 A is 4.9 × 10^–7 m^2 . If there are 8.5 × 10^28 charge carriers per m^3 in this wire, what is the drift speed of the charge carriers?  

Answer 1

We can find the drift speed of the charge carriers as follows:

$$v_d=\dfrac{I}{nAq},$$ $$v_d=\dfrac{0.67\ A}{8.5\times10^{28}\ \dfrac{el}{m^3}\times4.9\times10^{-7}\ m^2\times1.6\times10^{-19}\ C},$$ $$v_d=1.0\times10^{-4}\ \dfrac{m}{s}.$$