Question

A silver wire that is 2.5 mm in diameter and 37 cm long carries a current of 169 mA. How many electrons per second pass a given cross-section of the wire?

Answer 1

By the definition of the current we have:

$$I=\dfrac{Q}{t},$$ $$Q=It.$$

From the other hand,

$$Q=ne.$$

Therefore, we can write:

$$It=ne,$$ $$\dfrac{n}{t}=\dfrac{I}{e},$$ $$\dfrac{n}{t}=\dfrac{169\cdot10^{-3}\ A}{1.6\cdot10^{-19}\ C}=1.06\cdot10^{18}\ \dfrac{electrons}{s}.$$