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Magnetic field change for induced current

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant 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]

    3. 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?
     
  2. jcsd
  3. May 7, 2016 #2

    TSny

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    Does your ##L_2## represent the correct length?
     
  4. May 7, 2016 #3
    You are right, so L2 should be be done with r1? That gives me 0.3 meters
     
  5. May 7, 2016 #4

    TSny

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    Yes. Make sure you keep enough significant figures in the intermediate calculations so that you don't produce too much "round off error".
     
  6. May 7, 2016 #5
    Gotcha, always keep sig figs. I'm now getting for the final answer 6.08 T/s Is that right?
     
  7. May 7, 2016 #6

    TSny

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    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: May 7, 2016
  8. May 7, 2016 #7
    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
     
  9. May 7, 2016 #8

    TSny

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    I added an additional comment at the end of my last post. See if it helps.
     
  10. May 7, 2016 #9
    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!
     
  11. May 7, 2016 #10

    TSny

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    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.
     
  12. May 7, 2016 #11
    The diameter of 10 cm has 1 sig fig right, so shouldn't it have 1 sig fig?
     
  13. May 7, 2016 #12

    TSny

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    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.
     
  14. May 7, 2016 #13
    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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