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Chapter 19: Problem 42

An electromagnetic flowmeter is used to measure blood flow rates during surgery. Blood containing ions (primarily Na") flows through an artery with a diameter of \(0.50 \mathrm{cm} .\) The artery is in a magnetic field of $0.35 \mathrm{T}\( and develops a Hall voltage of \)0.60 \mathrm{mV}$ across its diameter. (a) What is the blood speed (in \(\mathrm{m} / \mathrm{s}\) )? (b) What is the flow rate (in \(\mathrm{m}^{3} / \mathrm{s}\) ) \(?\) (c) If the magnetic field points west and the blood flow is north, is the top or bottom of the artery at the higher potential?

Short Answer

Expert verified
Question: Given an artery's diameter, a magnetic field, and the Hall voltage developed across the diameter, calculate the blood speed, flow rate, and determine the direction of higher potential in the artery. Answer: (a) The blood speed is \(0.343\, \mathrm{m}/\mathrm{s}\). (b) The flow rate is \(4.250 \times 10^{-6}\, \mathrm{m}^3/\mathrm{s}\). (c) The bottom of the artery is at the higher potential.

Step by step solution

01

Write the equation for Hall voltage related to blood speed and artery diameter.

The formula for Hall voltage \(V_H\) in terms of blood speed \(v\), artery diameter \(D\), and magnetic field \(B\) is: \(V_H = v D B\)
02

Solve for blood speed \(v\) using the provided diameter, magnetic field, and Hall voltage values.

We are given the diameter \(D = 0.50\, \mathrm{cm} = 0.005\, \mathrm{m}\), the magnetic field \(B=0.35\, \mathrm{T}\), and the Hall voltage \(V_H = 0.60\, \mathrm{mV} = 6.0\times 10^{-4} \mathrm{V}\). Substitute these values into the formula and solve for \(v\): \(v = \frac{V_H}{D B} = \frac{6.0\times 10^{-4}}{0.005 \times 0.35} \mathrm{m}/\mathrm{s}\)
03

Calculate the blood speed \(v\).

Now, compute the value for blood speed \(v\): \(v = \frac{6.0\times 10^{-4}}{0.005 \times 0.35} = 0.343\, \mathrm{m}/\mathrm{s}\)
04

Write the formula for flow rate, and express the flow rate in terms of the blood speed, artery diameter, and artery cross-sectional area.

The formula for flow rate \(Q\) is given by: \(Q = v \times A\) where \(A\) is the cross-sectional area of the artery. Since the artery has a circular cross-section, the area can be expressed as \(A = \pi (\frac{D}{2})^2\)
05

Calculate the flow rate for the blood.

Substitute the blood speed and artery diameter values into the flow rate formula. Remember to change the units to m^3/s: \(Q = 0.343 \times \pi (\frac{0.005}{2})^2 = 4.250 \times 10^{-6}\, \mathrm{m}^3/\mathrm{s}\)
06

Determine the direction of higher potential in the artery using the right-hand rule.

The magnetic field points west, and the blood flow is north. Use the right-hand rule: Point your right-hand fingers in the direction of blood flow (north), then curl them in the direction of the magnetic field (west). Your right-hand thumb now points in the direction of the induced Hall voltage (downwards). So, the bottom of the artery is at the higher potential. The final answers are: (a) The blood speed is \(0.343\, \mathrm{m}/\mathrm{s}\). (b) The flow rate is \(4.250 \times 10^{-6}\, \mathrm{m}^3/\mathrm{s}\). (c) The bottom of the artery is at the higher potential.

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