91Ó°ÊÓ

$\text{d}=\text{150µ³¾}\left(\frac{{10}^{−6}\text{″¾}}{1\text{ µm }}\right)=1.50×{10}^{−4}\text{″¾}$

$\text{w}=\text{4.5mm}\left(\frac{{10}^{−3}\text{″¾}}{1\text{″¾m }}\right)=4.5×{10}^{−3}\text{″¾}$

$\text{B}=\text{0.65T}$.

The Hall Effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor, and an applied magnetic field perpendicular to the current.

Formulae are as follows:

${\text{F}}_{\text{m}}={\text{ev}}_{\text{d}}\text{B}$

${\text{v}}_{\text{d}}=\frac{\text{I}}{\text{neA}}$

${\text{F}}_{\text{e}}=\frac{{\text{V}}_{\text{H}}\text{ed}}{\text{A}}$

Where V_H is hall voltage, d is thickness, A is the area, V_dis drift velocity, F_e is electric force, I is current, e is the charge on particle, F_m is a magnetic force, and B is the magnetic field.

To find Hall voltage$\left({\text{V}}_{\text{H}}\right)$:

Here, both forces balance each other.

Hence,

${\text{F}}_{\text{m}}={\text{F}}_{\text{e}}$

${\text{ev}}_{\text{d}}\text{B}=\frac{{\text{V}}_{\text{H}}\text{ed}}{\text{A}}$

${\text{V}}_{\text{H}}=\frac{{\text{Av}}_{\text{d}}\text{B}}{\text{d}}$

Now, putting the formula of drift velocity,

$\begin{array}{c}{\text{V}}_{\text{H}}=\frac{\text{AIB}}{\text{neAd}}\\ =\frac{\text{IB}}{\text{ned}}\end{array}$

$\begin{array}{c}{V}_H=\frac{23\text{ }A×0.65\text{ }T}{8.47×{10}^{28}{\text{″¾}}^{\text{-3}}×1.50×{10}^{−6}\text{ }m\text{ }×1.6×{10}^{−19}\text{ }C}\\ =7.4×{10}^{−6}\text{ }V\end{array}$

Hence, the hall voltage is${\text{V}}_{\text{H}}=\text{7.4}×{\text{10}}^{\text{-6}}\text{V}$.

Therefore, by using the concept of hall voltage, the hall voltage through the copper strip can be determined