Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 41

An electromagnetic flowmeter is used to measure blood flow rates during surgery. Blood containing Na" ions flows due south through an artery with a diameter of \(0.40 \mathrm{cm} .\) The artery is in a downward magnetic field of \(0.25 \mathrm{T}\) and develops a Hall voltage of \(0.35 \mathrm{mV}\) across its diameter. (a) What is the blood specd (in \(\mathrm{m} / \mathrm{s}\) )? (b) What is the flow rate (in \(\mathrm{m}^{3} / \mathrm{s}\) )? (c) The leads of a voltmeter are attached to diametrically opposed points on the artery to measure the Hall voltage. Which of the two leads is at the higher potential?

Short Answer

Expert verified
Based on the given information, calculate the blood velocity, flow rate, and identify the higher potential lead in the artery. 1. Calculate the blood velocity: \(v = \frac{0.35*10^{-3}}{0.25 * 0.40*10^{-2}}\) \(v ≈ 0.035\,m/s\) 2. Calculate the blood flow rate: Radius: \(r = 0.20*10^{-2}\,m\) Cross-sectional area: \(A = \pi * (0.20*10^{-2})^2\) Flow rate: \(Q = A * v\) \(Q ≈ 4.42\times10^{-7} \,m^3/s\) 3. Higher potential lead: Using the right-hand rule, the higher potential lead is on the right side. Blood velocity: \(0.035\,m/s\) Blood flow rate: \(4.42\times10^{-7} \,m^3/s\) Higher potential lead: Right side

Step by step solution

01

Calculate the blood velocity

To find the blood velocity, we can use the formula for the Hall voltage produced in an electromagnetic flowmeter: \(V_H = B * D * v\), where \(V_H\) is the Hall voltage, \(B\) is the magnetic field, \(D\) is the diameter of the artery, and \(v\) is the blood velocity. We are given \(V_H=0.35\,mV=0.35*10^{-3}\,V\), \(B=0.25\,T\), and \(D=0.40\,cm=0.40*10^{-2}\,m\). To calculate the blood velocity \(v\), we can rearrange the formula: \(v = \frac{V_H}{B * D} = \frac{0.35*10^{-3}}{0.25 * 0.40*10^{-2}}\). Now, plug in the given values to find the blood velocity.
02

Calculate the blood flow rate

To find the blood flow rate, we need to multiply the blood velocity by the cross-sectional area of the artery. The cross-sectional area of a cylinder (in this case, the artery) is given by: \(A = \pi * r^2\), where \(A\) is the area, and \(r\) is the radius of the artery. We have the diameter \(D=0.40*10^{-2}\,m\), so the radius is \(r = \frac{D}{2} = 0.20*10^{-2}\,m\). Now, we can calculate the area: \(A = \pi * (0.20*10^{-2})^2\). To find the flow rate \(Q\), we multiply the area by the blood velocity: \(Q = A * v\). Plug in the calculated values for \(A\) and \(v\) to find the flow rate.
03

Determine the higher potential lead

We can use the right-hand rule to determine which of the two leads is at a higher potential. Since blood is flowing south (downwards), and the magnetic field is also pointing downward, the cross-product of the blood velocity and magnetic field will give the direction of the Hall voltage: \(V_H = v \times B\). Following the right-hand rule, if the thumb points south (downwards) and the index finger points towards the magnetic field (downwards), then the middle finger points to the right, which is the higher potential lead.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?
The magnetic field in a cyclotron is \(0.360 \mathrm{T}\). The dees have radius \(82.0 \mathrm{cm} .\) What maximum speed can a proton achicve in this cyclotron?
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish $^{12} \mathrm{C}^{+}\( and \)^{14} \mathrm{C}^{+}$ ions that have the same charge. The \(^{12} \mathrm{C}^{+}\) ions move in a circle of diameter \(25 \mathrm{cm} .\) (a) What is the diameter of the orbit of \(^{14} \mathrm{C}^{+}\) ions? (b) What is the ratio of the frequencies of revolution for the two types of ion?
Point \(P\) is midway between two long, straight, parallel wires that run north- south in a horizontal plane. The distance between the wires is $1.0 \mathrm{cm} .\( Each wire carrics a current of \)1.0 \mathrm{A}$ toward the north. Find the magnitude and direction of the magnetic field at point \(P\)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Rotational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Electricity

Read Explanation

Measurements

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free