91Ó°ÊÓ

$$\begin{gathered} \mathrm{ÒÏ}=1.69×{10}^{-8}\mathrm{Î\copyright }.m \\ n=8.47×{10}^{28}/m^3 \end{gathered}$$

According to the Hall effect's basic tenets, a current-carrying conductor or semiconductor exposed to a perpendicular magnetic field can measure a voltage at an angle to the current direction.

By using the concept of electric field, which is, according to equation 26-11, the resistivity of material times current density. By substituting the value of current density in terms of drift velocity, we can calculate the total electric field. But according to Hall Effect, the electric field is velocity of that charges into magnetic field. By using the value of resistivity of copper, we can find the ratio for copper.

Formula:

$$\begin{gathered} E_c=\mathrm{ÒÏ}ne{v}_d \\ E=v_{d}B \end{gathered}$$

Use the equation 26-11 $E_C=\mathrm{ÒÏ}J,$

Where J is the current density and $\mathrm{ÒÏ}$is the resistivity of material. The current density is also written as the form of drift velocity as

$J=ne{v}_d$ …(¾±)

By substituting this equation in theelectric field, we can get

$E_C=\mathrm{ÒÏ}ne{v}_d$

Also according to Hall Effect, we can write the electric field as

$E=v_{d}B$ …(¾±¾±)

From equation (i) and (ii), taking ratio, we can write that

$\frac{E}{E_c}=\frac{\left(v_{d}B\right)}{\left(\mathrm{ÒÏ}ne{v}_d\right)}$

From equation (iii), we can calculate for copper as

$n=8.47×{10}^{28}/m^{3}and$ $\mathrm{ÒÏ}=1.69×{10}^{-8}\mathrm{Î\copyright }.m$$E/E_c=\frac{0.65T}{(1.69×{10}^{-8}\mathrm{Î\copyright }.m×8.47×{10}^{28}/m^3×1.6×{10}^{-19}C)}$