Faraday's law and induced emf in a loop

[kobulingam]

(As magnitudes)
Faraday's law states that induced emf in a conducting loop by a changing flux, say x, is:

emf = dx/dt


Now if we have a circuit with the loop in series with wires (which are then tied together), how do I picture the voltages/currents in this circuit?

I mean, the actual conducting loop taking the changing flux gets the voltage "across" it, correct?

So if I wanted to calculate the current in this circuit, I just divide the voltage of the conducting loop by it's resistance (ignoring the extra stubs of wires which are in series with this), correct?

BUT, if above is correct, what if the wires have a lot of resistance, say 10 times the resistance of the wire loop?

 

[cabraham]

But the voltage around the loop is not merely the time derivative of the external flux, but rather the external minus internal flux. When a current exists in the loop a magnetic flux is generated with polarity so as to *oppose* the original external field. This is known as the law of Lenz.

The voltage once around the loop equals the current times the total resistance per Ohm's law, and also must equal -d@/dt, where @ = external flux minus internal flux. As far as the "voltage across it" goes, that's a little ambiguous. When not dealing w/ time-changing mag fields, the voltage "across" two points is well-defined. But in the time-varying mag field case, i.e. induction, the voltage between any two points cannot be defined unless a path is specified. In the first case path does not matter, but with induction it matters.

When you ask about the voltage "across" the loop, I presume that you mean "once around the loop starting and stopping at the same point". That makes more sense than "across". The net voltage once around the loop depends on the geometry of the loop, the external inducing flux, and the resistance of the loop.

Did this help?

Claude

 

[kobulingam]

But the voltage around the loop is not merely the time derivative of the external flux, but rather the external minus internal flux. When a current exists in the loop a magnetic flux is generated with polarity so as to *oppose* the original external field. This is known as the law of Lenz.

The voltage once around the loop equals the current times the total resistance per Ohm's law, and also must equal -d@/dt, where @ = external flux minus internal flux. As far as the "voltage across it" goes, that's a little ambiguous. When not dealing w/ time-changing mag fields, the voltage "across" two points is well-defined. But in the time-varying mag field case, i.e. induction, the voltage between any two points cannot be defined unless a path is specified. In the first case path does not matter, but with induction it matters.

When you ask about the voltage "across" the loop, I presume that you mean "once around the loop starting and stopping at the same point". That makes more snese than "across". The net voltage once around the loop depends on the geometry of the loop, the external inducing flux, and the resistance of the loop.

Did this help?

Claude

I have something like this in mind http://img299.imageshack.us/img299/3151/loopandstubky1.jpg

The whole thing is made of wire, but only the circular part is OPENED, while the other extended part is insulated wire that is closed (no area to receive flux).

So if we apply a changing flux to the circular loop, an EMF appears on that loop. Now how much current will occur in that circuit (circular loop plus that stub of wire)? Is it the EMF from Faraday's law divided by resistance of the circular part or divided by resistance of whole circuit?

 

[jtbell]

In any closed circuit (without branches) the current in the circuit is the total emf divided by the total resistance. In your example, you must include the resistance of the "extension" even though it doesn't contribute to the emf.

 

[cabraham]

I have something like this in mind http://img299.imageshack.us/img299/3151/loopandstubky1.jpg

The whole thing is made of wire, but only the circular part is OPENED, while the other extended part is insulated wire that is closed (no area to receive flux).

So if we apply a changing flux to the circular loop, an EMF appears on that loop. Now how much current will occur in that circuit (circular loop plus that stub of wire)? Is it the EMF from Faraday's law divided by resistance of the circular part or divided by resistance of whole circuit?

It's not a simple problem. The changing flux applied to the loop is the *external* flux. When current exists, another flux exists, namely *internal* flux. The net emf around the loop is V = -d@/dt, where "@" is the net flux obtained by subtracting the internal from the external. The external is a fixed value, but the internal flux depends on the current which depends on the resistance R as well.

Ohm's law will always be upheld. If the loop resistance is R, then the loop voltage divided by the loop current will always equal R. But the loop voltage, or emf if you prefer depends on the external flux value, the rate of change (freq), the geometry of the loop, and R, the resistance of the loop. Computing the emf based on these quantities is not that hard, but not trivial. If I get time in the next week or so, I might take a crack at it. In a nutshell, for very large values of R, the current is small as is the internal magnetic flux. In that case, the net flux is virtually identical to the external flux since the internal flux is tiny. Thus with high R value, the emf is determined by external flux, frequency, and loop geometry. Varying R will change the current in accordance w/ Ohm's law, but the voltage stays about the same as long as R is very large.

But as R decreases, the current I increases as does the internal magnetic flux. Subtracting the internal from the external results in smaller net flux @. Thus the emf is reduced as well. Then at a low enough R value, the loop enters constant current mode of operation. Reducing R further has little effect on I, but V decreases accordingly per Ohm.

Have I helped?

Claude