# Does an non-moving electrically conductive metal become magnetic in a full-DC offset AC field?

Let's assume an electrically conductive object made of metal is static and does not move.

Then let's apply an outer electric field. Clearly, a constant DC field would not induce any currents and magnetic field in the conductive object. However, a changing AC field would induce a current and therefore a changing magnetic field.

So let's apply an AC field with full-DC offset: Will there a current and a magnetic field be induced to the conductive object?

And what's the official term to call AC fields with full-DC offset anyway? (Because they aren't AC, are they?)

• What do you mean by full-DC offset? Is that a current opossing the current generated by the changing externalmagnetic field? Oct 1 '20 at 21:32
• No, actually the current that creates the external field is a 100% DC-offset AC current, hence the field created by the current should be a 100% DC-offset AC field as well: The resulting field is not a strict AC field with changing poles, but its field strenght is fluctuating. Oct 1 '20 at 21:39

As an example, consider that the conductive object is simply a wire loop. If we let the external field be a magnetic field, we may apply Faraday's law directly, such that the induced electro-motive force (emf) $$\mathcal{E}$$ is
$$\mathcal{E} = -\frac{\partial \phi}{\partial t}$$ where $$\phi$$ is the magnetic flux penetrating the loop. We let the external field be a sum of an AC and a DC component, such that the flux through the loop is the sum of the two contributions, i.e. $$\phi(t) = \phi_{dc} + \phi_{ac}(t)$$. The DC component does not vary with time and hence does not contribute to the emf according to the above equation - only the time-varying AC component does. Thus, the induced emf is the same whether the DC offset is present or not.