In a Hall-effect experiment, a current of $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{A}}$sent lengthwise through a conductor$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$wide,$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$ long, and$\mathbf{10}\mathbf{}\mathbf{\text{μm}}$ thick produces a transverse (across the width) Hall potential difference of$\mathbf{10}\mathbf{}\mathbf{\text{μV}}$ when a magnetic field of$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{T}}$is passed perpendicularly through the thickness of the conductor. From these data, find (a) the drift velocity of the charge carriers and (b) the number density of charge carriers. (c) Show on a diagram the polarity of the Hall potential difference with assumed current and magnetic field directions, assuming also that the charge carriers are electrons.

The current is, $i=3\text{A}$

The wide of conductor is, $d=1 \text{cm}=0.01 \text{m}$

The length of conductor is, $l=4 \text{cm}=0.04 \text{m}$

The potential difference is, $V=10{\text{μV=10×10}}^{-6}\text{V}$

The magnetic field is,$B=1.5\text{T}$

According totheconcept of electric field,it is the ratio of potential difference applied to the distance between those plates. According totherelation between electric field, magnetic field, and velocity, we can find the drift velocity. From equation 28-12, we can find the number density. From 3-D diagram, we can show the polarity of the hall potential.

Formula:

$\mathit{E}\mathbf{=}{\mathit{v}}_{\mathbf{d}}\mathit{B}\mathbf{}$

$\mathit{n}\mathbf{=}\frac{\mathbf{B}\mathbf{i}}{\mathbf{V}\mathbf{l}\mathbf{e}}$

As we know, the electric field is the ratio of potential difference to the width of the conductor.

$E=\frac{V}{d}$ ......(1)

Substitute all the value in the above equation.

$$\begin{gathered} E=\frac{10\times {10}^{-6}\text{V}}{0.01\text{m}} \\ E=10\times {10}^{-4}\raisebox{1ex}{$\text{V}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right. \end{gathered}$$

Now by using the relation between drift velocity and electric field,

$E=v_{d}B$ .......(2)

$v_d=\frac{E}{B}$

By substituting the value, we can get

$$\begin{gathered} v_d=\frac{10\times {10}^{-4}{\displaystyle \raisebox{1ex}{$\text{V}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.}}{1.5\text{T}} \\ {v}_d=6.67\times {10}^{-4}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right. \\ {v}_d\approx 6.7\times {10}^{-4}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right. \end{gathered}$$

Hence the drift velocity of charge carriers is$v_d=6.7\times {10}^{-4}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

According to equation 28-12, the number density, i.e., the number of charges per unit volume is written as

$n=\frac{Bi}{Vle}$

From equation, we can write the potential in terms of electric field as

$V=Ed$

By substituting this equation in number density, we can write

$n=\frac{Bi}{Edle}$

But according to equation (2), $v_d=\frac{E}{B}$

Therefore,

$n=\frac{i}{v_{d}dle}$

Substitute all the value in the above equation.

$$\begin{gathered} n=\frac{3\text{A}}{6.67\times {10}^{-4}{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\times 0.01\times 10\times {10}^{-6}\times 1.6\times {10}^{-19}} \\ n=2.8\times {10}^{29}/\text{m} \end{gathered}$$

Hence the number density of charge carriers$n=2.8\times {10}^{29}/\text{m}$

By using the 3-D diagram, we assume the magnetic field along up, and width is along east west line, and we assume the current going along north direction. The flow of electron is opposite to the direction of the current. So by using the right hand rule, we conclude that the west side of the conductor is negative and the east side is positive.