Find the induced magnetic current on the inner of two rings

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SUMMARY

The induced current in the inner loop of a circular configuration with a diameter of 1.60 mm and a resistance of 1.10×10-2 Ω is calculated based on the changing current in the outer loop, which transitions from +1A to -1A over 0.08 seconds. The initial calculations yielded an induced current of 3590 nA, but the correct induced current is 4.79 nA. The discrepancy arises from the incorrect application of Faraday's law of electromagnetic induction and the calculation of magnetic flux.

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  • Familiarity with Ohm's law (I=V/R)
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  • Review the application of Faraday's law in calculating induced electromotive force (emf)
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Homework Statement


A small, 1.60-mm-diameter circular loop with R = 1.10×10−2Ω is at the center of a large 120-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from + 1A to -1A in 8.00×10^−2s .

What is the induced current in the inner loop? in nA

Homework Equations



I=ε/R
ε=Phi/t
Phi=A*B

The Attempt at a Solution



I=ε/R

Phi=(Area)*(Magnitude of field)
Phi = (PI*(8x10^-4^2 m^2)*??)

I am completely unsure of what to do after this. Please help! Thanks.
 
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Ok, I tried this:

B=mu*I/2R

B1=4PIx10^-7 * +1 / 2*(1.1x10^-2)= 0.000785 T

B2=4PIx10^-7 * -1 / 2*(1.1x10^-2)= -0.000785 T

I=V/R

V=PHI1-PHI2/t

PHI1=A*B1

PHI1=PI*(8x10^-4)^2 * 0.000785 =1.58x10^-9

PHI2=A*B2=PI*(8x10^-4)^2 * - 0.000785=-1.58x10^-9

V=(1.58x10^-9)*2/(8x10^-2)
V=3.95x10^-8

I=V/R

I=(3.95x10^-8)/(1.1x10^-2)
I=3.59x10^-6 A =3590 nA

Yes?I submitted, but this was not correct. The correct answer was 4.79 nA. Can anyone please explain how?
 
Last edited:

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