0
$\begingroup$

The Faraday-Maxwell law says that $$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$ Which is always explained as a time-varying magnetic field producing an electric field. But can we go the other way and say that a rotational electric field produces a time-varying magnetic field?

$\endgroup$

1 Answer 1

2
$\begingroup$

An equation simply says that two quantities have to be the same. It's not strictly accurate to use a word like "produces" which implies a logical directionality; the left hand side and right hand side are on equal footing, logically speaking.

In practice, often one side is "given" and the other side is "something to solve for." If you are solving for a vector field $\vec{V}$ as a function of space, then given its divergence and curl in terms of known functions, and assuming that the vector field dies off asymptotically fast enough, you can solve for $\vec{V}$ using the Helmholtz decomposition. Applying this logic to Faraday's law, we can solve for an unknown electric field using the known "source terms", which include $\partial \vec{B}/\partial t$ (as well as the charge density). This leads to the language that $\partial \vec{B}/\partial t$ "producing" $\vec{E}$. Additionally, experimentally, it is possible to produce a time varying magnetic field; the fact that it is relatively easy to control the magnetic field experimentally in some situations also makes it natural to think of the equation in this direction (at least sometimes).

You are right, that if you thought of $\vec{E}$ as a known quantity as a function of time and space and wanted to solve for the unknown magnetic field, you could use Faraday's law to integrate $\nabla \times \vec{E}$ with respect to time and recover $\vec{B}$. This situation is not very common in practice, but there is nothing logically wrong with it.

In a more general situation, we don't know $\vec{E}$, $\vec{B}$, or the charges and currents as a function of time. If we know their value and time derivatives at some initial time, we can use Maxwell's equations (plus additional equations telling us how the charges and currents evolve) to solve for all of the fields, charges, and currents in terms of the initial conditions. Then, it's not really correct to say that either $\nabla \times \vec{E}$ produces $\vec{B}$, or that $\partial\vec{B}$ produces $\vec{E}$, but rather that the evolution of all the quantities is mixed together in a complicated way.

$\endgroup$
9
  • 1
    $\begingroup$ @GRANZER The law you are investigating is defined at a point in space and time, it does not apply to a circuit. If you use the integral form of the equation it is that $\oint \vec{E}\cdot d\vec{l} = - \int d\Phi/dt$ and both the left and right hand sides are zero in the situation you describe (the sum of the EMF around a closed circuit with a steady current is zero). $\endgroup$
    – ProfRob
    Mar 19, 2022 at 8:53
  • 1
    $\begingroup$ If the wire is neutral and the current is constant in time, then $\vec{E}=0$ and $\vec{B}$ is constant so $\partial\vec{B}/\partial t=0$. $\endgroup$
    – Andrew
    Mar 19, 2022 at 13:45
  • 1
    $\begingroup$ The SUM of EMFs around a circuit is zero. Kirchoff's voltage law. en.m.wikipedia.org/wiki/Kirchhoff's_circuit_laws @GRANZER $\endgroup$
    – ProfRob
    Mar 19, 2022 at 21:43
  • 1
    $\begingroup$ @GRANZER I should amend my comment. If the wire is neutral and the current is constant in time, then $\vec{E}=0$ (or almost zero) almost everywhere, except in the circuit elements where there is a voltage drop. For example, there will be a non-zero electric field across a battery, or across a resistor. Where $\vec{E}\neq 0$, it is constant in time. Since the current is constant, $\vec{B}=0$ so $\partial \vec{B}\partial t=0$. $\endgroup$
    – Andrew
    Mar 19, 2022 at 22:56
  • 1
    $\begingroup$ @GRANZER we're back to where you started. If there is a changing magnetic field at a position then there is a curling electric field at that position. The two are inseparable. $\endgroup$
    – ProfRob
    Mar 20, 2022 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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