# Induced current on a metallic ring

1. Dec 13, 2012

### DonJuvenal

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?

Best wishes,

Juvenal.

2. Dec 14, 2012

### Philip Wood

L deals with the emf due to the changing field due to the changing induced current, does it not?

3. Dec 14, 2012

### Enthalpy

You're absolutely right, the current in the ring creates a flux and its variation creates a voltage, including if the current results from an external field.

You have already taken this effect in account when using the self-inductance to compute the current. The self-inductance tells how much voltage is induced in the ring by its own current.