Answer
$μ_{e} = 4.59 \times 10^{-3} m^{2}/V-s $
Work Step by Step
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:
electron mobility for this metal
Solution:
Using Equation 18.4, compute for the conductivity:
$σ =\frac{1}{ρ} = \frac{1}{3.3 \times 10^{-8 }(Ω-m)} = 3.03 \times 10^{7} (Ω-m)^{-1}$
Using a combination of Equations 18.20b and 18.18:
$μ_{e} = |R_{H}|σ = \frac{|V_{H}|dσ}{I_{x}B_{z}}$
$μ_{e} = \frac{(|-2.4 \times 10^{-7}|)(15 \times 10^{-3} m)(3.03 \times 10^{7} (Ω-m)^{-1})}{(25 A)(0.95 tesla)} = 4.59 \times 10^{-3} m^{2}/V-s $
