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Faraday-Neumann Law: minus sign?

  1. Jul 21, 2014 #1
    I was wondering why is there a minus sign in the equation:
    EMF= -d(flux)/dt
    If this equations is derived from a previous equation which is:
    EMF= vL•B
    Why doesent the minus sign just appear in the derivation?
  2. jcsd
  3. Jul 21, 2014 #2


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  4. Jul 21, 2014 #3
    So there has to be a constant value of the magnetic flux going threw a surface?
    But isn't that "Gauss Law" in a certain way(Only there have to be other surfaces)?
  5. Jul 21, 2014 #4


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    No; where does it say that? The magnetic flux can be constant (then the EMF is zero), or it can be changing (then the EMF is not zero).

    Gauss's law isn't the only one that can be described in terms of surfaces. Check the integral form for the Maxwell equations:

  6. Jul 22, 2014 #5
    It says in my book(translating from italian):
    "The magnetic flux of a magnetic field threw a Gaussian Surface (S) is zero"
    But anyway I understood the problem.
    Thank you for helping me.
  7. Jul 22, 2014 #6


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    Yes, that statement is correct ... that is why the divergence of the magnetic field is always zero. It means that for every field line going out, there is an equivalent one coming in.

    Just look at iron filings about a magnet: the field lines are always closed.
  8. Jul 22, 2014 #7


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    Just to make it very clear, because these issues are a constant source of confusion for beginners in electromagnetic theory.

    The most general equations are Maxwell's equations in differential form. The microscopic equations read (in a vacuum, i.e., considering all charges and currents explicitly and not making approximations to describe macroscopic electrodynamics in media)
    [tex]\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,[tex]
    [tex]\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.[/tex]
    Here [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] are the electric and magnetic components of the electromagnetic field, [itex]\rho[/itex] the electric-charge density and [itex]\vec{j}[/itex] the electric current density, and [itex]c[/itex] the speed of light in a vacuum. The equations are written in Heaviside-Lorentz units. In SI units you have other constants [itex]\epsilon_0[/itex] and [itex]\mu_0[/itex] for unit conversion, which destroy the beauty of the equations. It's easy to switch from one system of units to the other. That's why I use the more natural Heaviside-Lorentz units.

    The first equation ist Faraday's Law. You can use the integral theorem by Stokes to get an integral form. To that end let [itex]F[/itex] be any surface with boundary curve [itex]\partial F[/itex]. We assume that we have defined the surface-normal elements [itex]\mathrm{d}^2 \vec{F}[/itex] and the tangent vectors on [itex]\partial F[/itex], [itex]\mathrm{d} \vec{r}[/itex], according to the right-hand rule. Then integrating Faraday's Law over the surface and using Stokes's theorem to the first term, yields
    [tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_{F} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.[/tex]
    It is very important to write the Law in this way with the partial time derivative under the integral. One can show with some effort that you can take the time integral outside of the integral, but if the boundary is moving, there is an extra term to the naive one:
    [tex]int_{F} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B} = \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B} + \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}),[/tex]
    where [itex]\vec{v}=\vec{v}(t,\vec{r})[/itex] is the velocity of each point of the boundary curve of your surface. Thus, the correct Faraday Law in integral form reads
    [tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \Phi[/tex]
    [tex]\Phi=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B}[/tex]
    is the magnetic flux through the surface. For a detailed proof, see the Wikipedia:


    The second equation is the statement that there are no free magnetic charges, i.e., any magnet appears always with a north and a south pole. If you devide the magnet in two pieces, you'll always get two magnets both with a north and a south pole but never a separate north or south pole. The integral form is easily obtained by applying Gauss's integral theorem
    [tex]\int_{\partial V} \mathrm{d}^2 \vec{F}=0.[/tex]
    Here [itex]V[/itex] is an arbitray volume and [itex]\partial_V[/itex] it's closed boundary surface. It is very important to keep in mind that this statement holds true for closed surfaces only.

    In the same way you can find integral forms of the other two inhomogeneous Maxwell equations. The easy one is Gauss's Law (the fourth equation in my ordering of the Maxwell equations). Using Gauss's Law, using the orientation of the boundary [itex]\partial V[/itex] such that the surface-normal vectors all point out of the volume [itex]V[/itex] under consideration:
    [tex]\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{r} \rho=Q_V,[/tex]
    where [itex]Q_V[/itex] is the total charge contained in the volue [itex]V[/itex].

    Finally for the Ampere-Maxwell Law you have to take into account the additional terms when taking the time derivative of the electric field outside of the integral. Then it reads
    [tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{B} = \frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{E} + \frac{1}{c} \int_F \mathrm{d} \vec{F} \cdot (\vec{j}-\rho \vec{v}) + \frac{1}{c} \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{E}).[/tex]
    Again [itex]\vec{v}=\vec{v}(t,\vec{r})[/itex] is the velocity of each point along the boundary [itex]\partial F[/itex] of the surface, and the relative orientation of the boundary and the surface-normal elements is according to the right-hand rule.
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