4
$\begingroup$

In my textbook, as a preliminary to Faraday's law of induction, magnetic flux is defined over a closed loop as

$$\Phi_B = \oint \vec{B}\cdot d\vec{A}$$

Then it draws a parallel with electric flux and says: "as in electric flux, $d\vec{A}$ is a vector of magnitude $dA$ that is perpendicular to a differential area $dA$". But the electric flux is through a closed surface, and the direction of $d\vec{A}$ is defined as coming out of the surface. The book does not mention anything about how direction is the direction of area vector is defined in case of magnetic flux. Also, there is not necessarily a current in the loop so we can't assign a direction using that?

I wasn't able to find an explicit clarification anywhere else on the web.

$\endgroup$
1
  • $\begingroup$ The direction of the surface vector just determines the sign of your flux. This is pretty much arbitrary, and it is easy to adjust such that you get the signs that you'd like your outcome to have. $\endgroup$
    – Danu
    Commented Jun 10, 2014 at 13:40

3 Answers 3

3
$\begingroup$

Indeed you have two possibilities. However it does not matter here, in the sense that you can fix the direction you want in absence of further requirements, provided obviously, you are coherent with your choice at each step when you solve a particular problem. The "right" prescription has to be fixed as soon as you state Faraday's law of induction. In that case you have to fix a direction along the loop used to compute the integral of the electric field and the direction of $d\vec{A}$. They are linked by the "law of the right hand": The preferred direction $d\vec{\ell}$ along the loop is that from the palm to fingertips of your right hand when it surrounds the loop. Then, the associated preferred direction of $d\vec{A}$ is indicated by the thumb. With these choices, Faraday's law of induction is stated: $$\oint_{+\partial \Sigma} \vec{E} \cdot d\vec{\ell} = - \frac{d}{dt} \int_{+\Sigma} \vec{B}\cdot d\vec{A}$$

$\endgroup$
2
  • $\begingroup$ The key phrase here is "it does not matter". Perhaps it goes without saying, but once an orientation has been chosen for a particular problem it cannot be changed. $\endgroup$
    – garyp
    Commented Jun 10, 2014 at 13:46
  • $\begingroup$ Yes you are right, I was a bit too sloppy. I corrected my answer taking your remark into account. $\endgroup$ Commented Jun 10, 2014 at 14:02
1
$\begingroup$

First, note that "perpendicular to a differential area $dA$" limits the direction to two directions: "in" and "out." This should be obvious for a flat surface, but is also true for curved surfaces. In that case, imagine a flat plane tangent to the surface at a given point; that will limit the direction of $d\vec{A}$ to two directions at that point (the direction of $d\vec{A}$ will vary with position on your surface).

Second, note that reversing the direction of $d\vec{A}$ will change the algebraic sign of $\Phi_B$. This is okay, because you're not directly measuring the flux; you're measuring the eletcric and magnetic fields, the currents, and the voltages. What is physical about the magnetic flux and Faraday's law of induction is the magnitude and direction of the induced voltage. The convention about the direction doesn't affect the magnitude; it only affects the direction. The direction of the induced voltage in Faraday's law is given by Lenz's law; it is the negative sign in the mathematical form of Faraday's law. This is an expression of the conservation of energy. As long as the direction that you use to calculate the magnetic flux is consistent with the direction you choose when you use Faraday's law.

But, in order to apply Faraday's law, you need a current. That current will define a direction. So for physicists, it's conventional to use the direction of the current to define which direction is "out" of the surface. The right hand rule convention in this case has you curl your fingers in the direction of the current around the surface. Your thumb points "out" of the surface. In this manner, the direction of the path around the edge of the surface determines which direction is "into" the surface and which direction is "out of" the surface.

Mathematicians still use the same right hand rule to relate the direction around the edge of the surface with the direction "into" and "out of" the surface, but there is not an inherent reason to pick one or the other direction for the path around the edge of the surface.

$\endgroup$
0
$\begingroup$

It is said to be perpendicular to the surface in the outward direction.

$\endgroup$

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.