0
$\begingroup$

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?)

$\endgroup$
2
  • $\begingroup$ What do you mean by full-DC offset? Is that a current opossing the current generated by the changing externalmagnetic field? $\endgroup$
    – Dennis Fr
    Oct 1 '20 at 21:32
  • $\begingroup$ 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. $\endgroup$
    – Marcus
    Oct 1 '20 at 21:39
1
$\begingroup$

In short, the answer is that a current will be induced in the conductive object, as the external field is time-varying.

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.

The induced emf may manifest itself as an induced current or voltage, depending on whether the loop is closed or not.

The above is true in the steady-state case. If one instead considers suddenly turning on the external field at some instance in time, the DC component will briefly cause an emf in the loop as well. This transient response occurs until the charges within the wire has re-distributed in response to the external field, reaching the steady-state case as discussed above. For more on this transient behavior, you can read up on charge relaxation time.

If instead of a wire the conductive object has an arbitrary shape and volume, it may create a response magnetic field both due to induced currents, and magnetization if e.g. the conductive object is made of a magnetic material.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.