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Chapter 19: Problem 97

An electromagnetic flowmeter is to be used to measure blood speed. A magnetic field of \(0.115 \mathrm{T}\) is applied across an artery of inner diameter \(3.80 \mathrm{mm}\) The Hall voltage is measured to be \(88.0 \mu \mathrm{V} .\) What is the average speed of the blood flowing in the artery?

Short Answer

Expert verified
Answer: The average speed of the blood flowing in the artery is approximately 2.092 m/s.

Step by step solution

01

Identify given values

In this problem, we are given the following values: - The magnetic field strength \((B) = 0.115\:T\) - The inner diameter of the artery \((d) = 3.80\:mm\) - The Hall voltage \((V) = 88.0\:\mu V\)
02

Convert the units

Since the units in the problem are not consistent, we need to convert them before we can make use of the formula. Let's convert \(d\) from mm to meters (m) and \(V\) from \(\mu V\) to volts (V). - \(d = 3.80\:mm \cdot \frac{1\:m}{1000\:mm} = 3.80 \times 10^{-3}\:m\) - \(V = 88.0\:\mu V \cdot \frac{1\:V}{10^6\:\mu V} = 88.0 \times 10^{-6}\:V\)
03

Calculate the average speed

Now we can use the formula to calculate the average speed of the blood \((v)\) in the artery. $$ v = \frac{V}{Bd} $$ Plug in the values for \(V\), \(B\), and \(d\): $$ v = \frac{88.0 \times 10^{-6}\:V}{0.115\:T \times 3.80 \times 10^{-3}\:m} $$
04

Solve for the average speed

Now let's solve for the average speed \((v)\): $$ v = \frac{88.0 \times 10^{-6}\:V}{0.115\:T \times 3.80 \times 10^{-3}\:m} = 2.092\:m/s $$ Hence, the average speed of the blood flowing in the artery is approximately \(2.092\:m/s\).

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Most popular questions from this chapter

A square loop of wire of side \(3.0 \mathrm{cm}\) carries \(3.0 \mathrm{A}\) of current. A uniform magnetic field of magnitude \(0.67 \mathrm{T}\) makes an angle of \(37^{\circ}\) with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?
You want to build a cyclotron to accelerate protons to a speed of $3.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ The largest magnetic field strength you can attain is 1.5 T. What must be the minimum radius of the dees in your cyclotron? Show how your answer comes from Newton's second law.
A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?
A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)
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