Question

A slice of copper is 1.0mm thick and carries a current of 10A. A magnetic field of flux density 1.0T, correctly applied, produces a maximum Hall voltage of 0.6x10^-6V between the edges of the slice. Calculate the number of free charge carriers per unit volume.

Answer 1

The Hall voltage on the opposite faces:

$$V_H=\frac{IB}{nte},\\\space\\ n=\frac{IB}{V_Hte},\\\space\\ n=\frac{10·1}{(0.6·10^{-6})(0.001)(1.602·10^{-19})}=1.04·10^{29}\text{ m}^{-3}$$