Question

(a) In Fig. 28-8, show that the ratio of the Hall electric field magnitude E to the magnitude E_cof the electric field responsible for moving charge (the current) along the length of the strip is

$\frac{\mathbf{E}}{{\mathbf{E}}_{\mathbf{C}}}\mathbf{=}\frac{\mathbf{B}}{\mathbf{n}\mathbf{e}\mathbf{\rho }}$

where $\rho$ is the resistivity of the material and nis the number density of the charge carriers. (b) Compute this ratio numerically for Problem 13. (See Table 26-1.)

Short answer

a) The ratio of the hall electric field magnitude E to the magnitude E_c of the electric field responsible for moving charge along the length of the strip is

$\frac{\mathrm{E}}{{\mathrm{E}}_{\mathrm{C}}}=\frac{\mathrm{B}}{\mathrm{ne\rho }}$

b) For copper,

$\frac{E}{E_c}=2.84\times {10}^{-3}$

Step-by-step solution

Identification of given data

$$\begin{gathered} \rho =1.69\times {10}^{-8}\Omega .m \\ n=8.47\times {10}^{28}/m^3 \end{gathered}$$

Significance of Hall effect

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:

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

(a) Showing that the ratio of the Hall electric field magnitude E to the magnitude EC of the electric field responsible for moving charge (the current) along the length of the strip is  EEc=Bneρ

Use the equation 26-11 $E_C=\rho J,$

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

$J=ne{v}_d$ …(i)

By substituting this equation in theelectric field, we can get

$E_C=\rho ne{v}_d$

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

$E=v_{d}B$ …(ii)

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

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

$\frac{E}{E_c}=\frac{\left(B\right)}{\left(\rho ne\right)}$ .....(iii)

(b) Compute the ratio numerically for Problem 13

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

$n=8.47\times {10}^{28}/m^{3}and$ $\rho =1.69\times {10}^{-8}\Omega .m$$E/E_c=\frac{0.65T}{(1.69\times {10}^{-8}\Omega .m\times 8.47\times {10}^{28}/m^3\times 1.6\times {10}^{-19}C)}$

$$\begin{gathered} E/E_c=\frac{0.65}{22.9\times {10}^1} \\ E/E_c=2.84\times {10}^{-3} \end{gathered}$$