Question

What Hall voltage is produced by a$\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{‐}\mathbf{T}$field applied across a role="math" localid="1654155783323" $\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{‐}\mathit{c}\mathit{m}$diameter aorta when blood velocity is$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{/}\mathbf{s}$?

Short answer

The induced hall voltage is $\mathbf{3}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{3}}\mathbf{}\mathbf{V}$

Step-by-step solution

Given information

The diameter of the aorta is $\mathrm{L}=2.60\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=2.60\times {10}^{‐2}\mathrm{m}$

The magnetic field strength of the earth is $\mathrm{B}=0.200\mathrm{T}$

The speed of the aluminum rod is$\mathrm{v}=60.00\mathrm{cm}/\mathrm{s}\left(\frac{1\mathrm{m}/\mathrm{s}}{100\mathrm{cm}/\mathrm{s}}\right)=0.6\mathrm{m}/\mathrm{s}$

Define the Hall effect

The principle of the Hall effect is used to generate an induced voltage when a charge carrier moves in an applied magnetic field perpendicular to the direction of motion and can be defined as,

$\mathbf{E}\mathbf{=}\mathbf{BLv}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluating the hall voltage

Substitute the given data in equation (1), which will result,

$$\begin{gathered} \mathrm{E}=\mathrm{BLv} \\ =(0.200\mathrm{T})\times (2.60\times {10}^{‐2}\mathrm{m})\times (0.6\mathrm{m}/\mathrm{s}) \\ =3.12\times {10}^{‐3}\mathrm{V} \end{gathered}$$

Therefore, the induced hall voltage is$3.12\times {10}^{‐3}\mathrm{V}$