Question

Using information in Example$\mathbf{20}\mathbf{.}\mathbf{6}$, what would the Hall voltage be if a $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{T}$field is applied across a gauge copper wire ($\mathbf{2}\mathbf{.}\mathbf{588}$mm in diameter) carrying a $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{‐}\mathbf{A}$current?

Short answer

The induced hall voltage in the copper wire is $1.47\times {10}^{‐6}\mathrm{V}$

Step-by-step solution

Given information

The diameter of the aorta is$\mathrm{L}=2.588\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=2.588\times {10}^{‐3}\mathrm{m}$

The magnetic field strength of the earth is$\mathrm{B}=2.00\mathrm{T}$

The current flowing through the copper wire is$\mathrm{I}=20.0\mathrm{A}$

The charge on the electron is$\mathrm{e}=1.6\times {10}^{‐19}\mathrm{C}$

Concept Introduction

The Hall effect is defined as the generation of an induced voltage across a current-carrying conductor that is transverse to the current and perpendicular to the applied magnetic field, which can be expressed as,

$\mathbf{E}\mathbf{}\mathbf{=}{\mathbf{BLv}}_{\mathbf{D}}$…………………(1)

Calculation for Hall Voltage

To calculate the drift velocity in the copper wire we use the following equation,

$\mathrm{I}={\mathrm{neAv}}_{\mathrm{D}}$…………………(2)

where n is the number of electrons per cubic meter, is the charge of an electron, $\mathrm{A}$is the wire's cross-sectional area, and ${\mathrm{v}}_{\mathrm{D}}$is the drift velocity.

The cross-sectional area $\mathrm{A}$of the copper wire can be calculated as,

$\mathrm{A}={\mathrm{\pi r}}^2$

Where r is the radius of the copper wire and $\mathrm{r}=\frac{\mathrm{L}}{2}$,

Therefore the cross-sectional area will be,

$$\begin{gathered} \mathrm{A}=3.14\times {\left(\frac{2.588\times {10}^{‐3}\mathrm{m}}{2}\right)}^2 \\ =5.26\times {10}^{‐6}\mathrm{m} \end{gathered}$$

Therefore using equation (2), we can calculate the drift velocity such that,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{D}}=\frac{1}{\mathrm{neA}} \\ =\frac{20.0\mathrm{A}}{\left(8.34\times {10}^{28}{\mathrm{m}}^{‐3}\right)\times \left(1.6\times {10}^{‐19}\mathrm{C}\right)\times \left(5.26\times {10}^{‐6}{\mathrm{m}}^2\right)} \\ =2.85\times {10}^{‐4}\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, we can calculate the value of induced hall voltage using the equation (1), such that

$$\begin{gathered} \mathrm{E}=\left(2.00\mathrm{T}\right)\times \left(2.588\times {10}^{‐3}\mathrm{m}\right)\times \left(2.85\times {10}^{‐4}\mathrm{m}/\mathrm{s}\right) \\ =1.47\times {10}^{‐6}\mathrm{V} \end{gathered}$$

Therefore, the induced hall voltage in copper wire is $1.47\times {10}^{‐6}\mathrm{V}$.