# Magnetic field change for induced current

## 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

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

## The Attempt at a Solution

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

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

Which is wrong?

TSny
Homework Helper
Gold Member
Does your ##L_2## represent the correct length?

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

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".

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?

TSny
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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:
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

TSny
Homework Helper
Gold Member
I added an additional comment at the end of my last post. See if it helps.

I added an additional comment at the end of my last post. See if it helps.
Okay I got it now,
$\frac { dB }{ dt } \quad =\quad I\frac { \rho { L }_{ 1 } }{ { A }_{ 1 }{ A }_{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { \rho 2\pi \frac { { d }_{ 1 } }{ 2 } }{ { \pi }^{ 2 }{ (\frac { { d }_{ 1 } }{ 2 } ) }^{ 2 }{ (\frac { { d }_{ 2 } }{ 2 } ) }^{ 2 } } \\ \frac { dB }{ dt } \quad =\quad I\frac { 16\rho \pi { d }_{ 1 } }{ { \pi }^{ 2 }{ { d }_{ 1 } }^{ 2 }{ d }_{ 2 }^{ 2 } } \quad =\quad 6.9264\quad =\quad 7\quad \frac { T }{ s }$

Thanks!

TSny
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Gold Member
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.

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?

TSny
Homework Helper
Gold Member
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.

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.