Faraday's law and induced emf in a loop

In summary, Faraday's law states that the induced emf in a conducting loop by a changing flux is given by dx/dt. In a circuit with the loop in series with wires, the actual conducting loop receives the voltage "across" it, which can be calculated by dividing the voltage of the loop by its resistance. However, in the case of time-changing magnetic fields, the voltage between any two points cannot be defined unless a path is specified. The net voltage once around the loop depends on the geometry of the loop, the external inducing flux, and the resistance of the loop. When a current exists in the loop, a magnetic flux is generated with polarity so as to oppose the original external field, known as the law of L
  • #1
kobulingam
10
0
(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?
 
Physics news on Phys.org
  • #2
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
 
Last edited:
  • #3
cabraham said:
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?
 
Last edited by a moderator:
  • #4
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.
 
  • #5
kobulingam said:
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
 
Last edited by a moderator:

Related to Faraday's law and induced emf in a loop

What is Faraday's Law?

Faraday's Law, also known as Faraday's Induction Law, states that when there is a change in magnetic flux through a conducting loop, an electromotive force (emf) is induced in the loop. This emf is proportional to the rate of change of magnetic flux.

What is magnetic flux?

Magnetic flux is a measure of the amount of magnetic field passing through a given area. It is represented by the symbol Φ and is measured in units of webers (Wb).

How is emf induced in a loop?

Emf is induced in a loop when there is a change in magnetic flux through the loop. This can occur through various means such as moving a magnet near the loop, changing the strength of the magnetic field, or changing the orientation of the loop in the magnetic field.

What is the formula for calculating induced emf?

The formula for calculating induced emf is emf = -N(dΦ/dt), where N is the number of turns in the loop and (dΦ/dt) is the rate of change of magnetic flux.

What are some applications of Faraday's Law?

Faraday's Law has numerous applications in everyday life, such as in generators and transformers used in power plants. It is also used in devices like microphones and speakers, and in technologies like magnetic levitation trains and induction cooktops.

Similar threads

Replies
27
Views
1K
Replies
4
Views
1K
Replies
25
Views
1K
Replies
2
Views
729
  • Electromagnetism
Replies
6
Views
771
Replies
11
Views
2K
Replies
2
Views
6K
Replies
1
Views
1K
Replies
10
Views
3K
  • Electromagnetism
Replies
20
Views
1K
Back
Top