Question

Calculate the Hall voltage induced on a patient’s heart while being scanned by an MRI unit. Approximate the conducting path on the heart wall by a wire $\mathbf{7}\mathbf{.}\mathbf{50}$cm long that moves at $\mathbf{10}\mathbf{.}\mathbf{0}$ perpendicular to a $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{‐}\mathbf{T}$magnetic field.

Short answer

The induced hall voltage is $1.13\times {10}^{‐2}\mathrm{V}$.

Step-by-step solution

Given information

The diameter of the aorta is $L=7.50cm\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=7.50\times {10}^{‐2}m$The magnetic field strength of the earth is $\mathrm{B}=1.50\mathrm{T}$

The speed of the aluminum rod is $\mathrm{v}=10.0\mathrm{cm}/\mathrm{s}\left(\frac{1\mathrm{m}/\mathrm{s}}{100\mathrm{cm}/\mathrm{s}}\right)=0.10\mathrm{m}/\mathrm{s}$

Concept Introduction

The Hall effect is defined as the generation of a voltage differential (the Hall voltage) across an electrical conductor that is transverse to the current and perpendicular to the applied magnetic field, and this voltage can be estimated using the equation,

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

Calculation of the Hall Voltage

Substitute the given data in equation (1) will give the value of induced hall voltage such that,

$$\begin{gathered} \mathrm{E}=\left(1.50\mathrm{T}\right)\times \left(7.50\times {10}^{‐2}\mathrm{m}\right)\times 0.10\mathrm{m}\backslash \mathrm{s} \\ =1.13\times {10}^{‐2}\mathrm{V} \end{gathered}$$

Therefore, the induced hall voltage is $1.13\times {10}^{‐2}\mathrm{V}$.