Discussion of Faradays and Lenz law.

[yungman]

This started in the MIT professor's thread. But this part is concentrated on Faradays and Lenz law in constant magnetic flux loop and transformer that is very interesting and not quite relate to the MIT professor discussion. I want to bring this out and continue discussion about the Faraday's Law:

$$Emf = -\frac { d \Phi}{d t}$$

The next two posts are just copies of Claude "Cabraham" post on his explanations. He is a long time transformer designer and on his last leg of his PHD program, I take his opinion very seriously and from reading the two posts, I do have some question. I hope others can join in.

 

[yungman]

This is his first post unedited:

Faraday: v = -N*d(phi)/dt.
Ohm: J = sigma*E.
Lorentz: F = q*(E + uXB).

The induced voltage is described in Faraday's Law, FL. But we must be careful. A time changing mag flux is related to the emf (voltage) induced in the loop per FL. But the flux "phi" is a net flux, not the external flux exciting the loop.

When the time varying, herein "ac", mag flux phi, excited the loop, there is charge motion per Lorentz force law, LFL. Free electrons in the wire are moved in a direction determined by E & B. E acts tangentially, B acts normally. Hence a rotational field condition is present & the electrons circulate around the loop. But hold on. The moving e- constitute a current, which produces another mag field. If the external mag flux density is called "Be", & internal is "Bi", then Bnet = B = Be + Bi. Of course, the law of Lenz, LL, tells us that the induced or internal B field opposes the external B in polarity.

In addition, we have another thing going on. As e- transit through the wire, they crash into lattice particles & lose some energy. This energy is radiated in the form of photon emission. It is around 5 to 7 micron in wavelength, & is felt as heat in the infrared region. Charges accumulate due to said collisions, & these charges have their own associated E field, since e- carry E fields due to their own charge.

This accumulation of e- charge gives rise to an ir-rotational (conservative) E field. This E field has no curl & is not an emf, but a drop. It is a polarizing type of force, its curl is zero, as it can not drive electrons around the closed loop.

So what is the voltage? How is it determined? The voltage V, is simply the line integral along a particular path of E*dl. But E has 3 components, the induced E field due to external B, the E field due to the current in the wire & its ac B field, Bi, & the static ir-rotational E field due to charge accumulation incurred via electron collisions w/ the lattice structure.

The voltage from a to b, is the line integral of the composite E fields along a chosen path. Inside the Cu wire, what is the voltage value, Vcu? We are measuring voltage from terminals a & b, w/ the path as the inside of the Cu winding.

The external ac mag flux density Be, gives rise to an induced E field, Ee, such that electrons in the Cu wire are moved along the wire, w/ said Be acting normal. This Ee is rotational.

But the induced current, Iloop, has its own B field, ac, we will call Bi. Bi is an ac field, & it has the opposite polarity of Be. It induces an E field & voltage or emf in the loop. This is the self-inductance of the loop. This is Ei, also rotational. For a very low loop resistance, the Bi cancels the Be almost entirely. What remains is the E field due to charges accumulating due to collisions between electrons & lattice structure. Call this one Ec. as it is due to charged particles.

Hence Enet = E = Ee + Ei + Ec. But Ee & Ei nearly cancel perfectly inside a low resistance conductor. If the loop were open, Ei tends toward zero, & the loop voltage is maximum due to no cancelling between Ee & Ei. Closing the loop via a resistive load results in current & a counter-balancing E & B fields, Ei & Bi. So inside the copper we still have Ec. If the entire loop was very low resistance so that Ee & Ei nearly cancel entirely, we still have Ec.

The Cu sec winding in my example is 0.1 ohm. The 10A current times the 0.1 ohm results in the voltage drop of 1.0V. The line integral of the composite E field along the Cu path results in Ee & Ei almost cancelling, & Ec*dl giving us around 1.0V. In other words, as soon as the ext mag field, Be, enters the Cu wire, current is induced. The induced current has a strong B field, Bi, that cancels the external, & equilibrium is reached.

You seem to be looking for the 120V induced emf inside the copper in distributed form. But don't forget that there is a counter-emf generated as well. The distributed emf sources are nearly perfectly canceled by the distributed counter-emf sources. But the Ec component does not get cancelled. It accounts for the 1.0V drop inside the Cu.

To better visualize this, consider a low resistance loop of wire, 0.010 ohm, closed & immersed in an ac mag field, Be. The Lorentz force moves the free electrons in the wire. This is current. But the current generates its own mag field, Bi. The law of Lenz tells us that they oppose each other. If the current is 1.0A, w/ a 0.010 ohm loop resistance, the voltage around the loop is 0.010 volt.

But we now open the loop, keeping the flux & area the same. The current plummets to near zero, but the voltage increases to 10V. Here, the ac mag flux produces an induced emf of 10V in the open state. When the loop closes, the net loop voltage is a mere 0.010V.

Why the difference? Of course, it is the cancellation. The external & internal parts of B & E account for the drastic difference in voltage between high & low impedance conditions. With a high-Z loop, the external B field is unopposed. Without induced current, the E field is due to the external B field, & the full voltage is realized since there is no loop current to cancel it.

When the loop resistance is low, the current generates cancellation of the external fields. The voltage is the line integral of all 3 phenomena. How can the Cu wire have just 1.0V, when the heater load has 119V, when their paths start & end at the same 2 points, a & b? They are in parallel, yet differing voltages are found.

Why? Answer is cancellation, & charge accumulation due to differing resistance values. I believe I've made my case, but someone other than me should affirm.

Claude

 

[yungman]

This is his second post unedited:

You're on the right track. I type slowly & long posts take a lot of time. I gave a xfmr example, but my explanation was eventually focused on a loop immersed in an ac B field, like that of an antenna receiving rf. A xfmr has one more thing going on.

A loop immersed in an rf B field in free space is subjected to induction. But said loop has an area which receives a specific amount of radiated power. This is induction w/ constant power. In the open circuit state, v = -N*d(phi)/dt. Also, phi = Ac*B, where Ac is the cross-sectional area of the loop, & B is mag flux density.

When the loop is open, Be, the external mag field, is related to the loop voltage Vloop, as follows. Vrms = Bpk/(4.443*f*Ac*N), where f is frequency, N is turns, per Faraday's law, FL. But if we close the loop in a high value of resistance R, we get current.

This current generates a field which opposes Be, so we call it Bi. I covered the rest previously. An equilibrium is reached when the loop resistance R is low enough so that the Bi cancels Be. Lowering R further does not increase the current. The voltage reduces as R is lowered, i.e. loop voltage decreases w/ decreasing R & increasing current.

It has to be this way per conservation of energy law, CEL. This is a constant power condition. The loop receives a limited amount of rf power, & the loop power cannot exceed the incident power per CEL. Hence Bi cancels Bi when R is low enough.

Now the xfmr is examined. When open secondary is measured, 120V appears at the terminals. Let's use these parameters for the xfmr including core. Vrms = 120V, Ac = (5cm X 5cm) 25 cm^2, f = 60 Hz, N = 120 turns both pri & sec, lc = core path length = 50 cm, Rsec = 0.1 ohm, & mu_r = relative permeability of core including incidental gap = 1000.

The B field in the core computes to 1.501 tesla per FL (15,010 gauss, a typical value for a grain oriented silicon steel material). To get H, we divide by mu, where mu = mu0*mu_r. Since NI = integral H*dlc, we get a magnetizing current of 0.498 amp, or 0.5A rounding off.

So we have a xfmr w/ sec open, 120V rms, & 0.5A magnetizing current, Imag. What happens when we connect the 11.9 ohm heater load across the secondary. The 0.1 ohm sec winding resistance is in series w/ the 11.9 ohm heater, for a total of 12.0 ohm, & the sec current, Isec = 10A. The terminal voltage drops by 1.0V to 119V rms.

The 1.5 tesla is the core when open is Be in this case. It requires an H to sustain it, w/ Imag of 0.5A. If the sec is loaded, that load current, induced by Be/Ee, tends to produce a mag flux, or "mmf", opposite in polarity to Be. This is Bi. Only 0.5A of counter-mmf will cancel the 1.5 tesla of Be. So where does the 120V come from, as you just asked?

A xfmr is not operating under a constant power condition like a loop in free space. A xfmr operates w/ constant voltage. The primary is connected across a good strong well-regulated constant voltage source, CVS. The power company goes through great effort to insure the voltage at our outlet is 120V rms.

As soon as load current is drawn at the sec, the counter-mmf produces Bi cancelling Be, resulting in a drop in terminal voltage. But the xfme primary is connected to a CVS. Said CVS then outputs an increase in current which counters the counter-mmf. The additional primary current provides "counter-counter-mmf". Just as the counter-mmf (or "Bi" if you prefer) resulted in counter-emf & a drop in voltage, the counter-counter-mmf produces a counter-counter-emf & an increase in voltage.

As long as the primary is excited by a good solid CVS which has the power capability to meet the load demands, said CVS will offer any current necessary to keep Vpri at 120V rms. Thus the cancellation of Be by Bi, is countered by increased Ipri.

But the mag flux still cannot enter the Cu sec winding to any large degree. Since the sec Cu resistance & the heater load are in series, their current is identical. Hence the 120V is divided between the 0.1 ohm & 11.9 ohm. When current exists in 2 different resistances in series, the higher resistance material incurs more electron to latiice collisions, & more accumulated charges. The charges provide their E field, Ec. When all 4 components of E are evaluated, we get 1.0V in the Cu sec, & 119V in the heater load.

In a nutshell, the CVS sets up a core flux, which sets up a sec E field & voltage. When loaded, the sec current produces Bi/Ei which cancels the Be/Ee. The CVS then forces an equilibrium condition by providing just enough primary current so thet Vloop = 120V. It's a CVS, that is what it does. Lattice collisions take place more frequently in the higher resistance medium. But being in series the current in each of the 2 media must be equal. Hence charges build up & the E field due to charges, Ec, adds or subtracts to/from Ee, Ei, & counter-Ee.

Dr. Lewin has a good paper on Ec which I'll dig up & post. Did this help?

Claude

Thanks for the detail reply, this is very informative. My question from this post is it is almost like the Faraday's Law should be written as:

$$P_{(induce)} = -k \frac{d\Phi}{dt}$$

The reason is because of the first part where the $\Phi$ is constant given the constant area loop. So due to Bi and Ei, that tend to cancel the Be and Ee so the lower the loop resistance R, the more Bi and Ei, this lower the induced emf. So the net result is really a constant power equation rather than constant voltage like the original equation:

$$Emf = -\frac{d\Phi}{dt}$$




In the transformer case, the primary is hooked to the wall power where it is true constant voltage. Be will increase as the load draw more current. This is true constant voltage until the current drawn is so high that both the resistance of the primary kick in and lower the net primary volt on the coil. This together with the resistance of the secondary further lower the secondary voltage that result the drop of output voltage as load current increases.

Please comment on this.