Answer to Question #170958 in Electric Circuits for Shaine

Question

A current of 350 mA flows through a gold wire with a diameter of 1.0 mm and a length of 25 cm. There are 5.90 x 1028 free electrons/m3in a gold. 

A) What is the current density in the wire?

B) What is the magnitude of the drift velocity of the free electrons in the wire?

Answer 1

(a) By the definition of the current density, we have:

$$J=\dfrac{I}{A}=\dfrac{4I}{\pi d^2}=\dfrac{4\cdot350\cdot10^{-3}\ A}{\pi\cdot(1.0\cdot10^{-3}\ m^2)^2}=4.46\cdot10^{-5}\ \dfrac{A}{m^2}.$$

(b) The magnitude of the drift velocity of the free electrons in the wire can be found as follows:

$$v=\dfrac{J}{n|e|}=\dfrac{4.46\cdot10^{-5}\ \dfrac{A}{m^2}}{5.9\cdot10^{28}\ \dfrac{el}{m^3}\cdot|-1.6\cdot10^{-19}\ C|}=4.72\cdot10^{-15}\ \dfrac{m}{s}.$$