Faraday's law on circular wire

[gralla55]

In my examples on Faraday's law in my book, they use a drawing of a magnet approaching a circular wire. The changing magnetic flux then induces an emf on the circle of wire, which in turn causes a current to flow.

I'm wondering if a current will flow in that wire without it having an element to provide resistance, or if it's just drawn that way for simplicity. Perhaps the internal resistance of the wire itself is good enough.

 

[cabraham]

Current will exist with or w/o R. A changing B results in a Lorentz force on the wire's free charge carriers, namely electrons. As the electrons move around the loop, if no R is present no lattice collisions take place and no photons are emitted. So the wire stays cool. If R is present, collisions result in electrons droppint from conduction band into valence band, a lower energy state. To conserve energy photons are emitted in the infrared region which we feel as heat. In a superconductor there is an induced current but no heat since no R is present.

The voltage around the loop will be related to the current per the inductive reactance value. In circuit theory lumped parameters are employed. In field theory distributed parameters must be considered because that is the real world.

 

[technician]

what would happen if there was a gap in the wire loop?

 

[cabraham]

what would happen if there was a gap in the wire loop?

You would have a small capacitance between the ends of the wire loop, resulting in a small displacement current. The induced voltage would be calculated per law of Faraday, and the current can be computed as the open circuit voltage divided by the loop total impedance. This impedance Zloop, is the wire resistance plus the inductance plus the capacitive reactance. Did I help?

Claude

 

[technician]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

 

[cabraham]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

R = ρl/A

C = ε₀A/d

L = μ₀N²A/l

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jX_L - jX_C + R

Claude

 

[technician]

R = ρl/A

C = ε₀A/d

L = μ₀N²A/l

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jX_L - jX_C + R

Claude

The formula you have given for inductance relates to a toroidal coil !
This question is about a wire loop.

 

[Andrew Mason]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

 

[cabraham]

The formula you have given for inductance relates to a toroidal coil !
This question is about a wire loop.

N=1! The wire cross section can be assumed rectangular, if round a correction factor is needed. The equation is valid for a solenoid, 1 to N turns.

Claude

 

[cabraham]

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

The current will exist longer than a "very brief time", but rather continuously. If open circuit voltage Voc is 10 volts, with R of 10 kohm, X_C of -j1.0 Mohm, and X_L of +j10 kohm, then the current I is given by

10 / (10k - j1.0M + j10k) = 10.100 microamp with angle 89.42 degree.

Claude

 

[technician]

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

 

[cabraham]

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

R = ρl/A_wire

C = ε0A_wire/d, where "d" is the space between the ends of the wire

L = μ0N²A_loop/l, where "l' is the length of the solenoid, or height of a single turn loop.

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jXL - jXC + R

I hope this helps.

Claude

 

[technician]

height of a single turn loop
do you mean diameter of the loop?

 

[cabraham]

height of a single turn loop
do you mean diameter of the loop?

Diameter of wire if round, times adjustment factor. Or, for a rectangular wire, height of wire, which is thickness in direction normal to loop plane.

Claude

 

[tech99]

I assume the magnet is on the polar axis of the loop, as the effect will vary depending on geometry. Classically, we would say that as the magnet approaches the loop, field lines cut the loop and induce an EMF. If there is a "conducting" path around the loop the electrons can move and a current will flow. Post Maxwell, we say that the field of the moving magnet creates an electric field, and this encompasses the conductor of the loop, creating an EMF round the loop. Then, if a conducting path exists around the loop, a current can flow.
When the electrons start to move they are accelerating, so they will radiate an EM wave.
The radiated wave will have E and B fields. In addition, the magnetic field associated with the movement of electrons will constitute the magnetic part of the antenna induction near field. The electric field driving the electrons will constitute the electric part of the antenna induction near field. The energy lost to radiation will be permanently taken from the kinetic energy of the magnet. The reaction on the magnet of the induction fields will constitute stored energy which can be given back to the magnet.

 

[technician]

Diameter of wire if round, times adjustment factor. Or, for a rectangular wire, height of wire, which is thickness in direction normal to loop plane.

Claude

Ok... I assumed a wire of cross sectional area 1mm² with a gap of 1mm.
This wire forms a loop of rea 1m²
I used C =ε₀A/d and got C = 8.86x10^-12F

I took the area of the loop formed by the wire to be 1m²and the wire to be square 1mm x 1mm (l = 1mm in the equation for L)
This gives me L = 4∏x10^-7x10^-3 H = 4πx10^-10H

How do you use these values of C and L to calculate X_c and X_L?

 

[tech99]

The problem you have is that the EMF induced in the wire in your case is a complex function containing many frequencies, rather than, say, a sine wave which it would be for a loop rotating in a uniform field. In the rotating case, you find the reactance at the rotation frequncy and calculate the current from Ohm's Law. The total reactance of the loop will be Xl - Xc. You find reactances as follows: Xl = 2 pi F L and Xc = 1/2 pi F C.

 

[cabraham]

Of course we should not forget to include loop resistance R as well. The capacitive & inductive reactances are per tech99's post above. R + jX_L - jX_C = Z_loop.

Claude

 

[tech99]

Hi Cabraham
I was trying to keep it simple, as resistance is small compared with reactance. But if you want to use total impedance you have to add reactance and resistance vectorially,
so Z = sqrt (R^2 + X^2)

 

[technician]

Of course we should not forget to include loop resistance R as well. The capacitive & inductive reactances are per tech99's post above. R + X_L - X_C = Z_loop.

Claude

You see the values I got for C and L...do these look reasonable?...how would you calculate X_land X_c so that I can take this further... I am OK with R
Edit. I do not see any calculation of XL or Xc !

 

[Andrew Mason]

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

A is the area enclosed by the coil and any line between the ends of the coil.

AM

 

[technician]

Ok... I assumed a wire of cross sectional area 1mm² with a gap of 1mm.
This wire forms a loop of rea 1m²
I used C =ε₀A/d and got C = 8.86x10^-12F

I took the area of the loop formed by the wire to be 1m²and the wire to be square 1mm x 1mm (l = 1mm in the equation for L)
This gives me L = 4∏x10^-7/10^-3 H = 4πx10^-4H

How do you use these values of C and L to calculate X_c and X_L?

These are the values I got for C and L...do they look reasonable?
How are these used to find X_cand X_l

 

[cabraham]

X_C = -1/Cω

X_L = Lω

ω = 2∏f

Claude

 

[technician]

X_C = -1/Cω

X_L = Lω

ω = 2∏f

Claude

What is ω in the context of this post?
I recognise ω in the context of AC theory.

 

[cabraham]

The "ω" is radian frequency in radian/second. If the induced emf is a special function like rectangular, sawtooth, etc., then it can be decomposed via Fourier series. If it is a transient event, a Fourier transform will give a continuous spectrum of frequencies.

For such a case, it may be easier to just write a 2nd order differential equation and solve in the time domain. An interesting exercise it is.

Claude

 

[technician]

The original post relates to a magnet approaching a circular coil.
If the coil is a single loop having a resistance of 1ohm and the area of the loop is 1 square metre and it is experiencing a changing flux of 1tesla/sec is the induced emf simply 1volt and the induced current simply 1Amp?
If there is a gap in the wire the emf will still be 1 volt but the current will be zero?

 

[cabraham]

The original post relates to a magnet approaching a circular coil.
If the coil is a single loop having a resistance of 1ohm and the area of the loop is 1 square metre and it is experiencing a changing flux of 1tesla/sec is the induced emf simply 1volt and the induced current simply 1Amp?
If there is a gap in the wire the emf will still be 1 volt but the current will be zero?

Not necessarily. The loop has inductance as well as resistance. We have been discussing this. Please review the whole thread. If the resistance is 1.0 ohm, but the inductive reactance is close to or greater than 1.0 ohm, the current will be less than 1.0 amp. The inductive reactance is the "X_L" mentioned above. Please review.


Claude

 

[technician]

My answer in post 26 is correct! This is an elementary Faraday's law calculation as seen in textbooks. (e = -d∅/dt)
I have followed the previous posts closely and got as far as calculating L and C for a loop of wire with a gap (quoted in post22) but then reached a dead end because of some uncertainty about ω.
Are you sure that you are not confusing the original post with AC anaysis?

 

[cabraham]

My answer in post 26 is correct! This is an elementary Faraday's law calculation as seen in textbooks. (e = -d∅/dt)
I have followed the previous posts closely and got as far as calculating L and C for a loop of wire with a gap (quoted in post22) but then reached a dead end because of some uncertainty about ω.
Are you sure that you are not confusing the original post with AC anaysis?

Even when ω is not clear, L & C enter the picture. The differential equation is:

Voc = L*di/dt + R*I + (1/C)*integral (i dt)

Solving the diff eqtn for I(t) produces the result needed. The inductive & capacitive reactances cannot be neglected simply because ω is not obvious. To simply divide Voc by "R" to get the current is wrong. Using that equation, as R went down, I would increase w/o limit.

A Voc of 1.0 V, with R = 1.0 ohm, does not always give I = 1.0 A. That would be 1.0 watt of power. If the power moving the magnet through the coil is less than 1.0 W, then the electric power cannot be 1.0 W. Energy conservation is immutable. Internal loop value of L means that as soon as current is set up, a counter-emf is present. So the loaded or closed loop voltage must always be less than open circuit value.

Claude

 

[technician]

My answer is correct. This is a standard textbook elementary example and exam question !
I think you are confused with AC circuit theory and the application of Faraday's law.
The induced emf comes about as a result of a changing magnetic flux linkage and I think you have taken this induced emf as a voltage applied to an L, C, R circuit.
I have not seen anything like your analysis in any of my textbooks and there is plenty of experimental evidence that the values in my calculation are correct.
Can you give me a reference where your explanation, especially the equation, is published so that I can check it?
I can't believe that all of my textbooks are mistaken !