## Materials Science and Engineering: An Introduction

$n = 4.12 \times 10^{28} m^{-3}$
Given: A hypothetical metal is known to have an electrical resistivity of $3.3 \times 10^{-8 }(Ω-m)$. A current of 25 A passes through a specimen of this metal 15 mm thick. When a magnetic field of 0.95 tesla is simultaneously imposed in a direction perpendicular to that of the current, a Hall voltage of $-2.4 \times 10^{-7} V$ is measured. Required: number of free electrons per cubic meter Solution: Using the rearranged form of Equation 18.8: $n = \frac{σ}{|e|μ_{e}} = \frac{( 3.03 \times 10^{7} (Ω-m)^{-1})}{(1.602 \times 10^{-19} C)(4.59 \times 10^{-3} m^{2}/V-s)} = 4.12 \times 10^{28} m^{-3}$