Consider a solenoid of length l, no. of turns N, n = N/l, area A, and negligible resistance. In its middle is placed a single-turn loop of same area (well, slightly smaller) in the solenoid's middle. The loop has, for the moment, a very high resistance so there is no appreciable loop current. The solenoid is sinusoidally excited with voltage v1. ø11 is the flux in the solenoid due to the current i1 in the solenoid, and ø21 is the flux in the loop generated by the current i1 in the solenoid. Obviously, ø21 = ø11.
So this would seem a tightly coupled ideal transformer since all the flux generated by the solenoid also cuts the area of the loop, so there is no leakage flux nor turns resistances (the loop resistance forms the load for the transformer & so is external to the transformer).
Easily shown:
L1 = self-inductance of the solenoid = Nø11/i1 = μ0n2Al,
and mutual inductance M = ø21/i1 = μ0nA.
Also, M = k√(L1 L2) where L2 is the self-inductance of the loop. Since the system is tightly coupled, k ~ 1.
This then says L2 = M2/L1 = μ0A/l
but this cannot be since it says L2 is a function of a property of the solenoid (namely l). In fact, computing the self-inductance of the loop is known to be exceedingly difficult, requiring elliptical integrals and what-not.
Now, an objection could be raised: since there is negligible current in the loop, the loop's self-inductance is meaningless. So let's now impose finite resistance R to the loop:
This is then a transformer with a secondary load R and secondary (loop) current i2. The equations for this transformer, assuming zero solenoid resistance, are
v1 = L1 di1/dt + M di2/dt
v2 = -i2*R = L2 di2/dt + M di1/dt.
Invoking phasors and using the derived expression for M gives
V2/V1 = μ0nAR/{μ0 n2lAR - jω(μ0 n2 lA L2 - μ02 n2 A2)}
= 1/{N -(jω/R)(NL2 - μ0nA)}
and invoking the knowledge as before that v2/v1 must = V2/V1 = 1/N (ideal transformer again) we see that
NL2 = μ0nA
which again says L2 = μ0A/l as before when i2 was = 0.
So that's my dilemma: how can L2 simply depend on the length l of the solenoid and its area when we all know L2 is a very difficult compute.
So this would seem a tightly coupled ideal transformer since all the flux generated by the solenoid also cuts the area of the loop, so there is no leakage flux nor turns resistances (the loop resistance forms the load for the transformer & so is external to the transformer).
Easily shown:
L1 = self-inductance of the solenoid = Nø11/i1 = μ0n2Al,
and mutual inductance M = ø21/i1 = μ0nA.
Also, M = k√(L1 L2) where L2 is the self-inductance of the loop. Since the system is tightly coupled, k ~ 1.
This then says L2 = M2/L1 = μ0A/l
but this cannot be since it says L2 is a function of a property of the solenoid (namely l). In fact, computing the self-inductance of the loop is known to be exceedingly difficult, requiring elliptical integrals and what-not.
Now, an objection could be raised: since there is negligible current in the loop, the loop's self-inductance is meaningless. So let's now impose finite resistance R to the loop:
This is then a transformer with a secondary load R and secondary (loop) current i2. The equations for this transformer, assuming zero solenoid resistance, are
v1 = L1 di1/dt + M di2/dt
v2 = -i2*R = L2 di2/dt + M di1/dt.
Invoking phasors and using the derived expression for M gives
V2/V1 = μ0nAR/{μ0 n2lAR - jω(μ0 n2 lA L2 - μ02 n2 A2)}
= 1/{N -(jω/R)(NL2 - μ0nA)}
and invoking the knowledge as before that v2/v1 must = V2/V1 = 1/N (ideal transformer again) we see that
NL2 = μ0nA
which again says L2 = μ0A/l as before when i2 was = 0.
So that's my dilemma: how can L2 simply depend on the length l of the solenoid and its area when we all know L2 is a very difficult compute.
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