How Does \(\oint E.dl = -\int dB/dt .dS\) Illustrate Electromagnetic Induction?

[darkwhite87]

Maxwell's Equation HELP!

This is one of the Maxwell's Equations that I can't understand:

$$\oint$$ E.dl = -$$\int$$ dB/dt .dS

From what I understand, $$\int$$E.dl gives the potential difference along the line. Therefore $$\oint$$E.dl should always equal to zero because it is the potential difference of a point compared with itself..is my logic right?

So how can Maxwell's equation $$\oint$$ E.dl = -$$\int$$ dB/dt .dS
be valid when the surface integral of dB/dt is not zero??

Can someple please explain this to me

 

[vanesch]

This is probably one of the most confusing issues to learn, and there are different, incompatible but equivalent, views on the issue.

The first point is: E is a conservative vectorfield ONLY in electrostatics. That means: E can be derived from a scalar potential (and hence has rotation 0) only in the case of electrostatics. When magnetic fields change, then E is NOT a conservative field anymore, and E is NOT the gradient of a scalar potential. But, the scalar potential remains... but gets a different meaning now.

So if there are changing magnetic fields, the rotation of E is not 0 anymore, and we cannot write anymore E = - grad V. In fact, the relation becomes now: E = - grad V - dA/dt where A is the so-called electromagnetic vector potential. What the equation you wrote down represents, is the value of the rotation of the E-field in the case of changing B fields.

Now, people (especially electrical engineers) like so much to use the scalar potential (the thing that is read out by volt meters), that they find a trick: they introduce a so-called "electromotive force" (emf), which is the compensation of the difference between true scalar potentials and the actual path integral of the E-field around a circuit. But "emf" is not a potential in the sense that it is a function of (x,y,z). It is a function of the circuit under consideration. Moreover, often the changes in magnetic field are confined to certain lumped circuit components: selfs and transformers. As such, one can often associate the "emf" to these lumped circuit components, and pretend that the scalar potential is still in relation to the E-field according to the E = -grad V relationship outside of these components. So you just "add these emf" in when setting up your circuit equations, and most of the time things are fine like that. It doesn't work, however, when the changing B-fields are spread over the entire circuit.

 

[darkwhite87]

thanks, this helped me to understand more.

 

[rbj]

Now, people (especially electrical engineers) like so much to use the scalar potential (the thing that is read out by volt meters), that they find a trick: they introduce a so-called "electromotive force" (emf), which is the compensation of the difference between true scalar potentials and the actual path integral of the E-field around a circuit.

the salient bottom line that we electrical engineers like to think is this: suppose you have a static electric field somewhere of maybe 10 zillion volts/meter. if you have a high impedance voltmeter and hold the two voltmeter probes apart by 10 cm (1/10 meter), you would measure a voltage of 1 zillion volts. so that's a nasty E field, i wouldn't want to be there. but even in the static field of 10 zillion volts/meter, if the two probes are right next to each other, virtually the same place but not quite touching, the voltage from point A to point A is zero because the potential difference is zero.

now, suppose you have a copper wire (conducts electricity nicely) bent in a loop with the two ends virtually at the same place (but not quite touching), the voltage between the two points is zero in that big static E-field. but now if the B-field that is perpendicular to the loop is changing in time, there is a voltage measured between those two ends of the wire at virtually the same point A. the voltage between point A and point A is not zero if you take a closed path around that has some changing magnetic field inside the closed path.

that's my not-pretty, but practical way of looking at it.

 

[vanesch]

now, suppose you have a copper wire (conducts electricity nicely) bent in a loop with the two ends virtually at the same place (but not quite touching), the voltage between the two points is zero in that big static E-field. but now if the B-field that is perpendicular to the loop is changing in time, there is a voltage measured between those two ends of the wire at virtually the same point A. the voltage between point A and point A is not zero if you take a closed path around that has some changing magnetic field inside the closed path.

Right. You can illustrate this with the following. Replace your copper wire by a tungsten wire with some resistance, in a changing magnetic field. Close the loop. Now ground that loop in some point, and put your black rod of the volt meter there.
A current is going to flow in the loop because of the changing B-field.
Now, we're going to scan the loop with our red test rod. First we put it next to the black one, on the reference ground. We find normally 0 V. Next, we slide it slowly around the loop. Because of Ohm's law, we will find an increasing voltage, the further we get around the loop... up to a point, where suddenly, the voltage reading becomes negative!
What happened ? We moved the wire of the red test rod through the magnetic field to go and test on the other side ! The wires of the test rod are also part of the test loop!
There's no way to measure the scalar potential difference between two points when there is a varying magnetic field without taking into account the loop formed by the testrods...