Farday's law in two conducting loops

  • Thread starter scariari
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  • #1
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consider two conducting loops with fixed geometries and at fixed positions. Let I1 and I2 be the currents flowing in the loops 1 and 2.

how does Faraday's Law allow the currents I1 and I2 to affect each other?

farday's law says that changing the flux between loops 1 will induce an emf in loop 2. where there are 2 loops at rest, a steady current I around loop 1 will produce a magnetic field B, where some of the field lines will pass through loop 2.
but does Faraday say anything without changing the flux?

Write down a set of two equations for the currents I1(t) and I1(t) in terms of the various induction coefficients.

induction coefficients might be the biggest problem here... if self inductance is Lii then coefficients of inductance are Lij. what exactly is the difference?

under which conditions on the coefficients can the currents I1 and I2 oscillate?

i am totally lost on this one. maybe if the coefficients are opposite? no clue!
 

Answers and Replies

  • #2
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Originally posted by scariari
but does Faraday say anything without changing the flux?
Yes. He says the induced emf is zero in this case. Thus I think you got to work with time-dependent currents.

i am totally lost on this one.
I guess this will not be so hard once you have the correct formulae. Can you show your work?
 
  • #3
18
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increasing I1 for example, will have an affect on the flux of loop 2 by making it repel. This happens because the emf generated in loop 2 led to a current which was in such a direction that its field tended to cnacel this new flux. This means that the current in loop 2 is opposite to the current in loop 1, and opposite currents repel.

one equation for the current is:
I(t)=((epsilon)0/R)+ke^(-(R/L)t)

what distinguishes loop1 from 2? and i take it the induction coefficient is the epsilon?

oscillation occurs when there is a start and a stop of current, so if loop 1 and loop 2 had a gap in time of current they could oscillate, right?
 
  • #4
508
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I think you got to work out a differential equation, i.e., an equation of the form
[tex]
\frac{dI}{dt} = \dots
[/tex]
or even
[tex]
\frac{d^{2}I}{dt^2} = \dots
[/tex]
 

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