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DonJuvenal
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I've just got confused about how the induced current in a metallic ring is calculated.
Consider a metallic ring (radius a, resistance R and inductance L) immersed in a oscillating magnetic field, which is oriented orthogonally to the plane of the ring.
The variation in flux of magnetic field induces a FEM. Modelling the ring as a resistor in series with an inductance, give us the well know result about the induced current and its phase relative to the external field. More concretely:
if B(t) = Bo Cos(ω t) then
I_induced(t) = - (π R^2 Bo ω /L)/sqrt((ωL)^2 + R^2) * cos (ω t + δ)
with δ= atan (R/(wL)).
Thus, the external field sets the induced current.
Why the magnetic field due to the induced current is not taken into account? The total field is the sum of the external uniform field plus the inhomogeneous induced one, therefore the magnetic flux across the ring is not only given by the external field. Should the induced field be taken into account in a self-consistent procedure?
I appreciate in advance your replies.
Best wishes,
Juvenal.
Consider a metallic ring (radius a, resistance R and inductance L) immersed in a oscillating magnetic field, which is oriented orthogonally to the plane of the ring.
The variation in flux of magnetic field induces a FEM. Modelling the ring as a resistor in series with an inductance, give us the well know result about the induced current and its phase relative to the external field. More concretely:
if B(t) = Bo Cos(ω t) then
I_induced(t) = - (π R^2 Bo ω /L)/sqrt((ωL)^2 + R^2) * cos (ω t + δ)
with δ= atan (R/(wL)).
Thus, the external field sets the induced current.
Why the magnetic field due to the induced current is not taken into account? The total field is the sum of the external uniform field plus the inhomogeneous induced one, therefore the magnetic flux across the ring is not only given by the external field. Should the induced field be taken into account in a self-consistent procedure?
I appreciate in advance your replies.
Best wishes,
Juvenal.