Answer
$B_{z} = 0.74\ T$
Work Step by Step
Given:
A metal alloy is known to have electrical conductivity and electron mobility values of $1.2
\times 10^{7} (Ω-m)^{-1}$ and $0.0050 m^{2}/V-s$, respectively. A current of 40 A passes through a specimen of this alloy that is 35 mm thick.
Required:
What magnetic field would need to be imposed to yield a Hall voltage of $-3.5 \times 10^{-7} V$?
Solution:
Using a combination of Equations 18.18 and 18.20b:
$B_{z} = \frac{|V_{H}|d}{I_{x}R_{H}} = \frac{|V_{H}|dσ}{I_{x}μ_{e}}$
$B_{z} = \frac{|(-3.5 \times 10^{-7} V)|(35 \times 10^{-3} m)(1.2
\times 10^{7} (Ω-m)^{-1})}{(40 A)(0.0050 m^{2}/V-s)} = 0.74\ Tesla $
