/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Using an electromagnetic flowmet... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Using an electromagnetic flowmeter (Fig. P19.69), a heart surgeon monitors the flow rate of blood through an artery. Electrodes \(A\) and \(B\) make contact with the outer surface of the blood vessel, which has interior diameter \(3.00 \mathrm{~mm} .\) (a) For a magnetic field magnitude of \(0.0400 \mathrm{~T}\), a potential difference of \(160 \mu \mathrm{V}\) appears between the electrodes. Calculate the speed of the blood. (b) Verify that electrode \(A\) is positive, as shown. Does the sign of the \(\mathrm{emf}\) depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain.

Short Answer

Expert verified
The speed of the blood is approximately \(1.33 \, m/s\). Electrode \(A\) is indeed the positive electrode. The sign of the emf depends on the prevalent type of ion charge - it would be reverse for negative charges and same for positive charges.

Step by step solution

01

Calculate the speed of blood

To solve for the speed (v) of the blood in the artery, use the formula for electromagnetic induction given by \(v = \frac{E}{Bd}\), where E is the induced potential difference, B is the magnetic field strength, and d is the diameter of the artery. Inputs are \(E = 160 \times 10^{-6} V\), \(B = 0.04 T\) and \(d = 3.00 \times 10^{-3} m.\)
02

Substitute the values into the equation

Plugging in the given values into the formula, we get \(v = \frac{160 \times 10^{-6}}{0.04 \times 3.00 \times 10^{-3}}\), which calculates to \(v \approx 1.33 \, m/s\). This is the speed at which the blood is moving through the artery.
03

Identify the positive electrode

Looking at the direction of blood flow and the given magnetic field in the figure, we can apply the right-hand rule. Pointing our right thumb in the direction of blood flow and our fingers in the direction of the magnetic field, our palm will face the positive electrode, which verifies electrode A's positivity.
04

Determine dependence of emf on ion charges

The emf produced does not depend on whether the ions in the blood are predominantly positively or negatively charged. It only depends on the movement of charges. As long as charges (positive or negative) are moving, they will induce emf. The polarity of the emf would, however, be affected by the type of the charge predominantly present - reverse direction for predominantly negative charges, same direction for predominantly positive charges.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Induction
Electromagnetic induction is a fundamental principle of physics explaining how a changing magnetic field can produce an electric current in a conductor. This fascinating phenomenon is described by Faraday's law of induction, which essentially states that when a conductor moves through a magnetic field (or vice versa), an electromotive force (emf) is induced in the conductor.

For instance, in the case of an electromagnetic flowmeter, which is used to measure the speed of blood in an artery, the conductor is the blood itself—a salty fluid containing charged ions. As the blood flows through a magnetic field, these charges experience a force called the 'Lorentz force,' which leads to the separation of charges and results in the creation of an emf across electrodes placed on the outer surface of the blood vessel.

In simpler terms, as the blood moves, it carries charged particles with it through the magnetic field. The movement of these charged particles generates an electric potential, which can be measured and used to calculate the velocity of the flowing blood.
Blood Flow Rate
In medical applications, it's crucial to monitor the rate at which blood travels through the body's circulatory system. The flow rate can provide insights into a patient's health and help diagnose various conditions. An electromagnetic flowmeter is a device that helps achieve this by measuring the speed of flow without obstructing the blood vessel.

The flow rate is directly proportional to the cross-sectional area of the blood vessel and the velocity of fluid flow. Using principles of electromagnetism, the speed of the blood is determined, and by knowing the diameter of the artery, one can calculate the flow rate using the formula:

Flow Rate (Q) = Cross-sectional Area (A) × Velocity (v)

where the area is \( \pi r^2 \), with r being the radius of the artery. In our problem, once we've calculated the speed of blood using the induced emf, we could also find the volume of blood flowing per second, if needed, to assess a patient's circulatory health.
Right-hand Rule
The right-hand rule is a useful mnemonic for understanding the direction of various vectors in electromagnetism, such as the magnetic field, current, and force. In the context of an electromagnetic flowmeter, the right-hand rule helps us figure out the direction of the induced emf and, consequently, which electrode will be positive.

To apply the right-hand rule, you extend your right hand with the thumb pointing in the direction of the moving charged particles (or current), which in this case is the blood flow. Your fingers should point in the direction of the magnetic field. Then, your palm will indicate the direction of the force on a positive charge. Since the force direction and induced emf are related, this will help in understanding that the lead A will have a positive polarity in the setup given the blood movement direction and magnetic field orientation. Confirming the polarity can assist in ensuring the accurate functioning of the flowmeter.
Induced emf
The concept of induced emf is integral to the operation of an electromagnetic flowmeter. Electromotive force, or emf, refers to the voltage developed by any source of electrical energy such as a battery or dynamo. In this context, it's the voltage across the electrodes in the flowmeter created by the process of electromagnetic induction.

As the ions in the blood move through the artery and encounter the perpendicular magnetic field, they create an electric field due to their separation. This separation of charge leads to a potential difference (induced emf) across the electrodes, which can be measured. It's this measured potential difference that we use in the equation \( v = \frac{E}{Bd} \) to find the speed of blood flow.

In the exercise, the induced emf doesn't depend on whether the ions in the blood are primarily positive or negative, as emf is generated as long as there is movement of charge. However, the predominant type of charge would affect the direction the emf is induced. If the charges moving through the magnetic field were primarily negative, the polarity would reverse compared to that induced by positive charges moving in the same direction.

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 in a circular path perpendicular to a uniform magnetic ficld with a magnitude of \(2.0 \mathrm{mT}\). If the speed of the electron is \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\), determine (a) the radius of the circular path and (b) the time it takes to complete one revolution.

A proton is at rest at the plane vertical boundary of a region containing a uniform vertical magnetic field \(B\). An alpha particle moving horizontally makes a head-on elastic collision with the proton. Immediately after the collision, both particles enter the magnetic field, moving perpendicular to the direction of the field. The radius of the proton's trajectory is \(R\). Find the radius of the alpha particle's trajectory. The mass of the alpha particle is four times that of the proton, and its charge is twice that of the proton.

A straight wire carrying a \(3.0-\mathrm{A}\) current is placed in a uniform magnetic field of magnitude \(0.28 \mathrm{~T}\) directed perpendicular to the wire. (a) Find the magnitude of the magnetic force on a section of the wire having a length of \(14 \mathrm{~cm}\). (b) Explain why you can't determine the direction of the magnetic force from the information given in the problem.

The magnetic field \(40.0 \mathrm{~cm}\) away from a long, straight wire carrying current \(2.00 \mathrm{~A}\) is \(1.00 \mu \mathrm{T} .\) (a) At what dis- tance is it \(0.100 \mu \mathrm{T} ?\) (b) At one instant, the two conductors in a long household extension cord carry equal 2.00-A currents in opposite directions. The two wires are \(3.00 \mathrm{~mm}\) apart. Find the magnetic ficld \(40.0 \mathrm{~cm}\) away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one-tenth as large? (d) The center wire in a coaxial cable carries current \(2.00 \mathrm{~A}\) in one direction, and the sheath around it carries current \(2.00 \mathrm{~A}\) in the opposite direction. What magnetic field does the cable create at points outside?

At the cquator, near the surface of Earth, the magnetic field is approximately \(50.0 \mu \mathrm{T}\) northward and the electric field is about \(100 \mathrm{~N} / \mathrm{C}\) downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron with an instantaneous velocity of \(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) directed to the east in this enyironment.

See all solutions

Recommended explanations on Physics Textbooks

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.