Which lead has a higher potential in this electromagnetic flowmeter calculation?

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Homework Help Overview

The discussion revolves around an electromagnetic flowmeter used to measure blood flow rates, specifically focusing on the calculation of blood speed and flow rate, as well as determining which lead of a voltmeter is at a higher potential based on the Hall voltage developed across an artery in a magnetic field.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of the right hand rule to determine the direction of force and potential difference between the leads. Questions arise regarding the correct application of the rule and the location of positive ions in relation to the leads.

Discussion Status

The discussion is ongoing, with participants questioning the application of the right hand rule and clarifying the direction of force. There is a focus on identifying which lead is at a higher potential, but no consensus has been reached yet.

Contextual Notes

Participants are working with specific values for blood speed and flow rate, but there is uncertainty regarding the correct interpretation of the right hand rule and its implications for the potential difference between the leads.

matt85
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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 cm. The artery is in a downward magnetic field of 0.25 T and develops a Hall voltage of 0.35 mV across its diameter. (a) What is the blood speed (in m/s)? (b) What is the flow rate (in m3/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?

1. (a) 0.35 m/s; (b) 4.4 times 10-6 m3/s; (c) the east lead
2. (a) 0.43 m/s; (b) 4.4 times 10-6 m3/s; (c) the east lead
3. (a) 0.43 m/s; (b) 5.7 times 10-6 m3/s; (c) the south lead
4. (a) 0.35 m/s; (b) 4.4 times 10-6 m3/s; (c) the south lead
5. (a) 0.43 m/s; (b) 4.4 times 10-6 m3/s; (c) the south lead

I have worked through the problem, and figured out that it is .43 m/s and the flow rate is 4.4E-6 m3/s. I am not sure how to figure out which lead is at the higher potential though. When I apply the right hand rule, it seems like I get a force to the west? So, which is at a higher potential, the east or south lead?

Thanks,
Matt
 
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Oops, I meant to say 0.35 m/s.
 
matt85 said:
When I apply the right hand rule, it seems like I get a force to the west?
Try again. Are you using the correct right hand?
So, which is at a higher potential, the east or south lead?
On which side will you find the positive ions?
 
To the south?
 
Which way is the force?
 
Force is to the east?
 
Well I guess that's the last choice, but yes. RHR says fingers down, thumb south, palm faces east, that's where the positive ions are, that's where the voltage is high.
 

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