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When calculating the flux integral $$-\iint_{\Sigma}{\frac{d\overrightarrow{B}}{dt} \cdot \overrightarrow{dA}} $$ in Faraday's law for a flat surface, which direction does $\overrightarrow{dA}$ point?

Is the rule to curl my right hand in the direction of the current then the thumb points in the direction of $\overrightarrow{dA}$? So, say the surface is in the x-y plane and we look at it from the top, the boundary goes counterclockwise, does $\overrightarrow{dA}$ point up?

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    $\begingroup$ $d\vec A$ is better than $\vec{dA}$, and $d\vec \Sigma$ is even better, in case the vector potential or Poynting vector show up. $\endgroup$
    – JEB
    Commented Jun 5, 2023 at 16:09

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Unless the surface through which you're calculating the flux is closed (in which case the canonical choice is to choose the area vector outward, you choose the direction of the area vector $\hat{A}$. There's no rule, here; you just need to be consistent throughout the whole calculation. You'll find that the choice doesn't affect any physical consequences. A standard choice is to choose $\vec{A}$ so that the flux is positive. This makes reasoning with Lenz's law easier.

Note that the right-hand rule just doesn't apply for this calculation: you're computing a dot product, not a cross-product. Later on, you might need to use the right-hand rule to figure out the direction of the induced current, but that's not a part of the calculation of the flux.

To be a little more careful about this. Faraday's Law states that a changing magnetic flux induces an emf, mathematically given by $$ \textrm{emf} = \oint_C \vec{E}\cdot d\vec{s}=-\frac{d}{dt}\int_S\vec{B}\cdot d\vec{A} = -\frac{d\Phi_{\textrm{mag}}}{dt}\,, $$ where the curve $C$ is the boundary of the surface $S$, oriented according to the right-hand rule. Which is to say, you choose an orientation for the surface $S$, point your thumb along this direction, and your fingers then curl in the direction along which we're integrating along $C$.

You can choose any orientation for the surface $S$ as long as you consistently choose the integration direction along $C$.

Now, often in practice, we compute the absolute value of the right-hand side of Faraday's Law to get the magnitude of the induced emf, and we use Lenz's Law to figure out the direction of the induced electric field, induced current, and/or induced magnetic field. This separates the calculation into two pieces which makes it easier.

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  • $\begingroup$ I know how to find out which way the current goes but what about the induced emf? Is it always negative? $\endgroup$
    – Ray
    Commented Jun 5, 2023 at 15:53
  • $\begingroup$ @Ray What do you mean by "negative"? emf is the line integral of the electric field: emf = $\int \vec{E}\cdot d\vec{l}$. In electrostatics, this is just the the electric potential difference between the initial and final points of the path along which we're integrating. For Faraday's Law, we know that $\oint \vec{E}\cdot d\vec{l} = -\frac{d}{dt}\int_A\vec{B}\cdot d\vec{A}$. Then, the right-hand rule (used properly) tells you the direction of the electric field induced by the changing magnetic field, and this electric field can drive a current. $\endgroup$
    – march
    Commented Jun 5, 2023 at 16:00
  • $\begingroup$ @Ray I've added a little to the post to make some of these connections clear. The right-hand rule does show up directly in Faraday's Law in the sense of matching the orientation of the surface to the direction of the path along which we're integrating the induced electric field. $\endgroup$
    – march
    Commented Jun 5, 2023 at 16:21

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