Question

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 µV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?

Answer 1

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When we are dealing with Hall voltage, it is necessary to have the values of the current, the magnetic field, the length, the area and the number of carriers at hand. The Hall voltage equation is given by,$V_h = \frac{iB}{neL}$

Where,

i= current

B= Magnetic field

L = Length

n = number of charge carriers

e= charge of a electron

We need replace and solve for n,

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

Therefore the density of charge carrier is $6.6*10^{27}e/m^3$