Pacemaker and Wire with Current

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Homework Help Overview

The discussion revolves around the effects of magnetic fields on cardiac pacemakers and the calculations related to magnetic fields generated by current-carrying wires. The original poster presents a problem involving a pacemaker's sensitivity to a magnetic field and seeks to determine the distance from a wire carrying a specific current.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the force equation and the magnetic field equation, questioning the need for unit conversions and the accuracy of their calculations. There is an exploration of how to properly set up the equations for the magnetic field and force per unit length between parallel conductors.

Discussion Status

Some participants have provided guidance on correcting equations and maintaining precision in calculations. There is an ongoing exploration of different interpretations of the problem, particularly regarding the calculations for magnetic fields and forces, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of maintaining significant figures in their calculations and the potential for rounding errors affecting their results. There is also mention of specific homework constraints related to the problem setup.

myersb05
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A cardiac pacemaker can be affected by a static magnetic field as small as 1.7 mT. How close can a pacemaker wearer come to a long, straight wire carrying 27 A?

I know that F=ILBsin(theta). I attempted to solve using I=27, L as unknown, F=1.7, and Bwire=mupi*I/2pir with an unknown r that would be equal to the L.

I don't think there is a need to convert the 1.7mT to Teslas because my answer should be in mm.
 
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Hi myersb05,

myersb05 said:
A cardiac pacemaker can be affected by a static magnetic field as small as 1.7 mT. How close can a pacemaker wearer come to a long, straight wire carrying 27 A?

I know that F=ILBsin(theta). I attempted to solve using I=27, L as unknown, F=1.7, and Bwire=mupi*I/2pir with an unknown r that would be equal to the L.

I don't think there is a need to convert the 1.7mT to Teslas because my answer should be in mm.

There's no need to deal with the force equation; the question just asks at what distance is the field of a long straight wire equal to 1.7mT.

I think you will need to convert to T, since B is in the numerator and r is in the denominator.

Also, it might just be a typo, but your B equation does not look right. It should be:

<br /> B = \frac{\mu_0 I}{2\pi r}<br />
 
Worked perfectly. Thanks again Al. That was the same problem I had last night. My equation was written improperly in my notes. I am stuck on another one if you have time.

Two long parallel conductors separated by 14.0 cm carry currents in the same direction. The first wire carries a current I1 = 3.00 A, and the second carries I2 = 8.00 A. (Assume the conductors lie in the plane of the page.)

(a) What is the magnitude of the magnetic field created by I1 and at the location of I2?
(b) What is the force per unit length exerted by I1 on I2?
C and D are the opposites of those.

I found the F/L by using uI1I2/2pid. So that came out as .000034 which is correct and applies to both wires.
I attempted to use Bwire=u8.0/2pi.14=.000011 and that was incorrect. I tried the same for the 3 amp wire and got .000004 which is also incorrect.
 
myersb05 said:
Worked perfectly. Thanks again Al. That was the same problem I had last night. My equation was written improperly in my notes. I am stuck on another one if you have time.

Two long parallel conductors separated by 14.0 cm carry currents in the same direction. The first wire carries a current I1 = 3.00 A, and the second carries I2 = 8.00 A. (Assume the conductors lie in the plane of the page.)

(a) What is the magnitude of the magnetic field created by I1 and at the location of I2?
(b) What is the force per unit length exerted by I1 on I2?
C and D are the opposites of those.

I found the F/L by using uI1I2/2pid. So that came out as .000034 which is correct and applies to both wires.
I attempted to use Bwire=u8.0/2pi.14=.000011 and that was incorrect. I tried the same for the 3 amp wire and got .000004 which is also incorrect.

Are you saying that 0.000004T is not the correct answer for part a? That's what I am getting; maybe you need to keep more digits (that results is roughly 6% to 7% or so off from the non-rounded answer), or maybe they want different units?
 
WebAssign is telling me that .000004 is close but that I should keep more digits. I can't come up with an answer that is any less rounded than .000004.
 
You found that result by calculating,

<br /> \frac{4\pi \times 10^{-7} (3)}{2 \pi (0.14)}<br />

right? The factors of \pi will cancel; do you mean your calculator only gives you one digit?

I was thinking you rounded some answers to get your result, and to get the right answer you should not round at all.
 
Yea, that is exactly what I typed into my TI-89 and I got exactly .000004
 
myersb05 said:
Yea, that is exactly what I typed into my TI-89 and I got exactly .000004

Well let's try to factor it out some:

<br /> \frac{4\pi \times 10^{-7} (3)}{2 \pi (0.14)}\to \frac{4 \times 10^{-7} (3)}{2 (0.14)} \to \frac{6}{.14} \times 10^{-7}<br />

So do 6/.14, and then add on the exponential part after you have more than one digit. Does that give the right answer?
 
I didn't even notice that you had a TI-89. Here's what you can do:

As soon as you get the answer of 0.000004, hit the up arrow, and then hit enter. That will bring the value down to the input line, with all the digits you need.

(You can also go into the mode menu and set it to show differing numbers of digits, but the above is what I usually do.)
 
  • #10
Got it, thanks again Al
 
  • #11
Glad to help!

It wasn't until after I posted the algebra in post #8 that I finally thought about the fact that I had a TI-89 and have the same issue quite often, so I end up doing the little key dance from post #9 all the time without even thinking about it.
 
  • #12
Yea, I'll have to remember that
 

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