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

At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?
The strip in the diagram is used as a Hall probe to measure magnetic fields. (a) What happens if the strip is not perpendicular to the field? Does the Hall probe still read the correct field strength? Explain. (b) What happens if the ficld is in the plane of the strip?
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
A solenoid of length \(0.256 \mathrm{m}\) and radius \(2.0 \mathrm{cm}\) has 244 turns of wire. What is the magnitude of the magnetic field well inside the solenoid when there is a current of \(4.5 \mathrm{A}\) in the wire?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing carbon (atomic mass 12 u), oxygen \((16 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering the magnetic field. The carbon and oxygen lines are separated by \(2.250 \mathrm{cm}\) on the photographic plate, and the unknown element makes a line between them that is \(1.160 \mathrm{cm}\) from the carbon line. (a) What is the mass of the unknown clement? (b) Identify the element.
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Solid State Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Particle Model of Matter

Read Explanation

Oscillations

Read Explanation

Further Mechanics and Thermal 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