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Faraday's law and gauss's law for magnetism apparent contradiction

by fishingspree2
Tags: apparent, contradiction, faraday, gauss, magnetism
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fishingspree2
#1
Apr17-13, 04:08 PM
P: 141
I was reading here http://www.physicsforums.com/library...tem&itemid=294

gauss law's for magnetism says
∮B⋅dA = 0

but then faraday's law has d/dt ∮B⋅dA in it. Well if its 0 then d/dt of 0 is 0.
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jtbell
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Apr17-13, 04:39 PM
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In Gauss's Law, the integral is over a closed surface, e.g. a complete sphere, cylinder, etc.

In Faraday's Law, the integral is over an open surface whose edge is the path used in the line integral of E on the other side of the equation. For example, the surface might be a hemisphere, whose edge is a circular path. The integral of B would be over the hemisphere, and the integral of E would be around the circular path.
fishingspree2
#3
Apr17-13, 04:57 PM
P: 141
Quote Quote by jtbell View Post
In Gauss's Law, the integral is over a closed surface, e.g. a complete sphere, cylinder, etc.

In Faraday's Law, the integral is over an open surface whose edge is the path used in the line integral of E on the other side of the equation. For example, the surface might be a hemisphere, whose edge is a circular path. The integral of B would be over the hemisphere, and the integral of E would be around the circular path.
If I understand correctly, Gauss's law is for a surface off which someone walking could never fall. Like the Earth for example.

Whereas Faraday's Law is for surfaces like when we thought the earth was flat.

Is that correct?

vanhees71
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Apr17-13, 05:53 PM
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Faraday's law and gauss's law for magnetism apparent contradiction

There is, of course no contradiction, because a closed surface has no boundary, and thus for a closed surface [itex]A[/itex], you have
[tex]\int_A \mathrm{d}^2 \vec{F} \cdot \vec{B}=0.[/tex]
On the other hand there is Faraday's Law that reads for any surface [itex]A'[/itex]
[tex]\int_{A'} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}=-\int_{\partial A'} \mathrm{d} \vec{x} \cdot \vec{E}.[/tex]
You can take out the time derivative of the integral, but you have to take into account that there is a piece from the change of the area with time (if it's moving). In this form Faraday's Law reads
[tex]\frac{\mathrm{d}}{\mathrm{d} t} \int_{A'} \mathrm{d}^2 \vec{F} \cdot \vec{B}=-\int_{\partial A'} \mathrm{d} \vec{x} \cdot (\vec{E}+\vec{v} \times \vec{B}),[/tex]
where [itex]\vec{v}=\vec{v}(t,\vec{x})[/itex] is the velocity field of the area (including its boundary).

If [itex]A'=A[/itex] is closed, you have [itex]\partial A'=0[/itex], and thus the right-hand side vanishes. This means that if the magnetic flux through a surface vanishes at some time [itex]t_0[/itex], it must vanish at any time. This means Gauß's Law for the magnetic field (which says that there are no magnetic charges) is consistent with the dynamics of the magnetic field.

Of course, all this is much more simply stated for the local form of Maxwell's equations. Faraday's Law reads
[tex]\partial_t \vec{B}=-\vec{\nabla} \times \vec{E}.[/tex]
Taking the divergence of the whole equation leads to
[tex]\partial_t (\vec{\nabla} \cdot \vec{B})=0,[/tex]
which means
[tex]\vec{\nabla} \cdot \vec{B}=f(\vec{x})=\text{const in time}.[/tex]
According to Gauß's Law, which states that
[tex]\vec{\nabla} \cdot \vec{B}=0[/tex]
demands that [itex]f(\vec{x})=0[/itex]. So again we see that the two homogeneous Maxwell equations are consistent.

There is also a consistency condition for the inhomogeneous Maxwell equations. This gives the conservation law of the electric charge, i.e., the continuity equation
[tex]\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0.[/tex]


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