Solenoid Passing through square coil-Finding electric field

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sweetdion
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Solenoid Passing through square coil--Finding electric field

Homework Statement


A very long solenoid with n turns of wire per unit length and radius b carries a current I(t) which decays with time as I(t)=I0e-t/T. The direction of current in the solenoid is as shown below. The solenoid passes through a single-turn square coil of wire with side length slightly larger than 2b and resistance R.

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a) Determine the electric field both inside and outside the solenoid as the current is decaying. Provide both a magnitude and direction.

b) Determine the current through the square coil as a function of time, Ic(t) as the solenoid current decays.

c) Determine the total energy dissipated in the resistance of the square coil from over the time period t=0 to ∞. Where does the energy come from?

Homework Equations


The induced emf can be obtained from Faraday’s law as
∫ E dr = -d/dt ∫B dA for an open surface S

The Attempt at a Solution



I'm going to start with part a
If we let c=dB/dt, the rate of change of the magnetic field
We choose that surface to be a circle of radius r, and we note by symmetry that the electric field must be tangential to it so for r<=R,
Ein2PiR=cPir2
So Ein=1/2cr tangential to the surface

Eout=2Pir=cPiR2
So Eout=cR2/2r tangential to the surface

I will go onto the next parts next but I'm unsure if this is the right way to find the electric field...

thanks in advance,
sweetdion
 
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although I am a student like yourself, I arrived at the same answers

note that B is obtained by ampere's law and that B =
mu x n x I(t)

and that outside the (ideal) solenoid
B is zero
 
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so how would we relate the Ic with the square to find part b? This must be Faraday's law...
 


well I don't think the question is too clear, but I'm assuming the wire somehow jumps onto the square coil then jumps off again onto the solenoid proper..

according to faraday's law, emf = d/dt (flux)

but instead of parsing the electromotive force into its definition as the line integral of electric field, we will now use its other definition as applied to circuitry, namely
I = emf / R

I = dflux/dt / R