Magnetic field change for induced current

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Homework Help Overview

The problem involves a circular loop of wire in a uniform magnetic field, where the goal is to determine the rate of change of the magnetic field required to induce a specific current. The context includes concepts from electromagnetism, specifically Faraday's law of induction and resistance calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the length of the wire and its implications on the resistance. There are questions about the accuracy of the values used for the radius and area in the calculations. Some participants suggest recalculating with more significant figures to reduce round-off error.

Discussion Status

The discussion is ongoing with participants sharing their calculations and results. There is a focus on ensuring accuracy in the calculations and significant figures. Some participants have arrived at different values for the rate of change of the magnetic field, indicating a productive exploration of the problem.

Contextual Notes

There is ambiguity regarding the significant figures of the given values, particularly the diameter of the loop and the current. Participants are considering how these ambiguities affect their calculations and final answers.

Sho Kano
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Homework Statement


A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 10 cm formed from wire of diameter 2.5 mm and resistivity 8.5×10-8 Ω·m. At what rate must the magnitude of the magnetic field change to induce a 10 A current in the loop?

Homework Equations


[itex]\phi \quad =\quad BA,\quad B\bot A\\ { V }_{ induced }\quad =\quad \frac { d\phi }{ dt } \quad =\quad IR\\ R\quad =\quad \frac { \rho { L }_{ 2 } }{ { A }_{ 2 } }[/itex]

The Attempt at a Solution


[itex]Givens:\\ \\ { d }_{ 1 }\quad =\quad 10/100\quad =\quad 0.1\quad meters\\ { r }_{ 1 }\quad =\quad \frac { { d }_{ 1 } }{ 2 } \quad =\quad 0.05\quad meters\\ { d }_{ 2 }\quad =\quad 2.5/1000\quad =\quad 0.0025\quad meters\\ { r }_{ 2 }\quad =\quad \frac { { d }_{ 2 } }{ 2 } \quad =\quad 0.0013\quad meters\\ { L }_{ 2 }\quad =\quad 2\pi { r }_{ 2 }\quad =\quad 0.0082\quad meters\\ { A }_{ 1 }\quad =\quad \pi { { r }_{ 1 } }^{ 2 }\quad =\quad 0.0079\quad sqr\quad meters\\ { A }_{ 2 }\quad =\quad { \pi { { r }_{ 2 } } }^{ 2 }\quad =\quad 5.31\quad x\quad { 10 }^{ -6 }\quad sqr\quad meters\\ I\quad =\quad 10A\\ \rho \quad =\quad 8.5\quad x\quad { 10 }^{ -8 }\quad \Omega \quad meters[/itex]

[itex]Calculations:\\ \frac { d\phi }{ dt } \quad =\quad { A }_{ 1 }\frac { dB }{ dt } \quad =\quad IR\quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 } } \\ { A }_{ 1 }\frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 2 } }{ { A }_{ 2 }{ A }_{ 1 } } \quad =\quad 0.1665\quad T/s[/itex]

Which is wrong?
 
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Does your ##L_2## represent the correct length?
 
TSny said:
Does your ##L_2## represent the correct length?
You are right, so L2 should be be done with r1? That gives me 0.3 meters
 
Yes. Make sure you keep enough significant figures in the intermediate calculations so that you don't produce too much "round off error".
 
TSny said:
Yes. Make sure you keep enough significant figures in the intermediate calculations so that you don't produce too much "round off error".
Gotcha, always keep sig figs. I'm now getting for the final answer 6.08 T/s Is that right?
 
Sho Kano said:
I'm now getting for the final answer 6.08 T/s Is that right?
I'm getting closer to 7 T/s. I suspect you have too much round off error. Go back and recalculate ##r_1##, ##r_2##, ##A_1##, ##A_2##, and ##L## to at least 3 significant figures. Or, better, express your final answer symbolically in terms of the given quantities and then plug in the numbers only at the end of the calculation.
 
Last edited:
TSny said:
I'm getting closer to 7 T/s.
Using the same numbers?
i = 10
p = 8.5 10-8
L = 0.3
A1 = 0.0079
A2 = 5.31 10-6
dB/dt = (i*p*L)/(A1*A2) = 6.0788 T/s
 
I added an additional comment at the end of my last post. See if it helps.
 
TSny said:
I added an additional comment at the end of my last post. See if it helps.
Okay I got it now,
[itex]\frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 1 } }{ { A }_{ 1 }{ A }_{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { \rho 2\pi \frac { { d }_{ 1 } }{ 2 } }{ { \pi }^{ 2 }{ (\frac { { d }_{ 1 } }{ 2 } ) }^{ 2 }{ (\frac { { d }_{ 2 } }{ 2 } ) }^{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { 16\rho \pi { d }_{ 1 } }{ { \pi }^{ 2 }{ { d }_{ 1 } }^{ 2 }{ d }_{ 2 }^{ 2 } } \quad =\quad 6.9264\quad =\quad 7\quad \frac { T }{ s }[/itex]

Thanks!
 
  • #10
OK. Since the numbers given in the problem have two significant figures, your answer should also be expressed to 2 significant figures: 6.9 T/s.
 
  • #11
TSny said:
OK. Since the numbers given in the problem have two significant figures, your answer should also be expressed to 2 significant figures: 6.9 T/s.
The diameter of 10 cm has 1 sig fig right, so shouldn't it have 1 sig fig?
 
  • #12
Well, 10 cm is ambiguous in terms of whether the 0 is counted as significant, unless your instructor has his or her own rules for that. Same for the 10 A. Since the other numbers are given to 2 sig figs, it suggests to me that the 10's also are to 2 significant figures. But, use your own judgment.
 
  • #13
TSny said:
Well, 10 cm is ambiguous in terms of whether the 0 is counted as significant, unless your instructor has his or her own rules for that. Same for the 10 A. Since the other numbers are given to 2 sig figs, it suggests to me that the 10's also are to 2 significant figures. But, use your own judgment.
Yea, it's ambiguous. But it does seem likely that 10 is actually 10. Thanks, the round off error was very real.
 

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