electric-circuitselectric-currentelectricityelectromagnetic-inductionelectromagnetism
For this above question, how is even a back emf induced in the circuit because of the coil. Doesn't Faraday's Law say the a change in flux threading an external coil will induce an emf. So then how can the coil induce a back emf in its own circuit?
Below is the solution. Which looks right if a back-emf could be produced in the first place.
is the magnetic field produced inside the soft iron core aligned with the magnetic field produced in the primary coil, or does it oppose it?
A conductive (iron) transformer core has both diamagnetic (excludes magnetic field) and ferromagnetic (enforces magnetic field) properties.
For a microsecond or ten after the primary current is established,
the B field in the core is low, because eddy currents in the iron oppose the applied H
field. After 1000 microseconds, those eddy currents have died
out, and the iron magnetization reinforces the magnetic induction
provided by the primary winding.
$$B \simeq \mu * H$$
Many thin laminations, rather than a solid iron block, makes a core that magnetizes more quickly (and helps energy efficiency), because it minimizes that changeover time. Lenz's law applies both to the eddy currents, and to the secondary currents, but does not relate directly to "the" magnetic field. The B field comes from the sum of those currents and the iron magnetic polarization, and isn't a simple superposition of parts, because magnetization of iron is not linear. That constant "mu" in the equation is ... not exactly a constant.
My text book says that 'when the secondary current is on, the magnetic field it creates is in the opposite direction to the magnetic field of the primary current.
There is some truth in that statement, because secondary current has the effect of reducing the magnetization of the iron, and at low magnetization, the system IS linear. Still, 'the magnetic field' ought to mean the B field, inside the core, and not some theoretical H field due to part of the nearby currents. That B field is composed of induction by two (or more) transformer windings, and by eddy currents, and by the ferromagnetic polarization.
While you can add the H field sources, it is not valid (because of polarization) to confuse that sum with the B field in the core. The polarization is not linear, and that fact will cause nonideal behavior of the transformer, which should NOT be ignored.
Magnetic flux and Faraday's law are defined purely mathematically. Because mathematical definitions typically deal with idealized concepts, subtle issues start to show up when we try to apply such definitions to more complicated physical situations. In particular, we usually need some physical principles to go with the mathematics; the physics tells us how to apply the mathematics.
Magnetic flux and Faraday's Law provide good examples of such subtleties and complications. I'll walk through the definitions carefully, pointing out the stumbling blocks, and then come back to Faraday's wheel. Most importantly, when talking about Faraday's wheel, there's one subtlety that requires additional input from physics, which is how to define a circuit.
We start with the mathematical definition of flux: \begin{equation} \Phi_B = \iint_S \vec{B} \cdot d\vec{A}. \end{equation} You probably know that $\vec{B}$ is the magnetic field vector, $d\vec{A}$ is the differential area element (the direction of which is normal to the surface with some continuous choice of either "outward" or "inward" direction), and $S$ is the surface over which the integral is performed. There are numerous idealizations built into this integral. Among them, we are assuming that we know exactly how to delimit this surface with exactly zero thickness and perfectly distinct boundaries.
This can raise subtle issues when we try to talk about real physical situations. For example, we might want to talk about the flux through a wire loop. But a physical wire has nonzero thickness. Should the surface of integration end at the center of the wire? At its outer edge? How do you specify the outer edge if the wire is not in a plane? You'll frequently hear people saying that Faraday's law only applies to "infinitely thin" wires. I would suggest that it's more correct to say that Faraday's law only applies naively to thin wires; it can be used in more general situations with careful application.
Next, we come to Faraday's Law, which relates the electromotive force $\mathcal{E}$ and the the boundary of your surface to the time-derivative of the magnetic flux as \begin{equation} \mathcal{E} = \int_{\partial S} \vec{E} \cdot d\vec{l} = -\frac{\mathrm{d}\Phi_B} {\mathrm{d}t}, \end{equation} where $\partial S$ is the boundary of the surface we used to define the flux. There are some more subtleties here. First, and quite simply, you have to be very specific about the surface and the boundary that you are dealing with. Second, the derivative is the total derivative, rather than the partial derivative. This can be important when the surface of integration is changing in time, and when the magnetic field is changing in space and time.
That last part is a really subtle issue, even though it's about something so major as how we define a circuit. I'll quote Jackson to explain it: "The electric field $\vec{E}$ is the electric field at $d\vec{l}$ in the coordinate system or medium in which $d\vec{l}$ is at rest, since it is that field that causes current to flow if a circuit is actually present." [Classical Electrodynamics, third edition, section 5.15.] So for a circuit in which the path of electrons is changing relative to the conductor through which they're moving, you have to use a changing circuit path.
Now, applied to Faraday's wheel (the homopolar generator), the first point is that you have to specifically define the surface that you're talking about — and hence the boundary of that surface. Students frequently rush past this point, and take the disc itself as the surface. But this is wrong because the boundary of that surface is the edge of the wheel, so technically you would be calculating the electromotive force around the wheel. But you're measuring the potential between the center and the edge, so that surface can't work. Instead, you need to take a surface whose boundary passes between the center contact and the contact on the edge, then goes out from each of them, and over to some galvanometer or something. A standard choice for the surface is a square that it is perpendicular to the wheel, with one edge along the wheel's axis and the next edge along a radius of the wheel. Now the paradox comes up when you realize that $\vec{B}$ is actually parallel to this surface, so $\vec{B} \cdot d\vec{A}$ is zero everywhere and always. So it would be reasonable to think that \begin{align} \frac{d \Phi_B}{dt} &= \frac{d}{dt} \iint_S \vec{B} \cdot d\vec{A} \\ &= \iint_S \frac{d}{dt} \left( \vec{B} \cdot d\vec{A} \right) \\ &= \iint_S \frac{d}{dt} \left( 0 \right) \\ &= \iint_S 0 \\ &= 0, \end{align} which would imply $\mathcal{E} = 0$. But that's not what's measured. It turns out that the derivation I just gave is wrong.
The (wrong) derivation above used the partial derivative, and ignored movement of the surface. You can resolve this paradox if you remember that the derivative in question is the total derivative, and that the surface might be moving. Now, Jackson told us that we need to use a path that's stationary with respect to the medium. So the surface we're integrating over is moving with respect to stationary magnetic field we've been told about, and we need to differentiate the stationary field with respect to the moving surface. The standard way of describing the total derivative in a material that's moving with velocity $\vec{v}$ at a given point is to use the "material derivative" (also commonly known as the "convective derivative"): \begin{equation} \frac{d}{dt} = \frac{\partial}{\partial t} + \vec{v} \cdot \vec{\nabla}. \end{equation} To be a little more specific, you use this material derivative to get the derivative with respect to moving coordinates when you're given a field in static coordinates. Now we get \begin{align} \frac{d \Phi_B}{dt} &= \frac{d}{dt} \iint_S \vec{B} \cdot d\vec{A} \\ &= \iint_S \left[ \left( \vec{v} \cdot \vec{\nabla} \right) \vec{B} \right] \cdot d\vec{A} \\ &= \iint_S \left[ \vec{\nabla} \times \left( \vec{B} \times \vec{v} \right) + \vec{v} (\vec{\nabla} \cdot \vec{B}) \right] \cdot d\vec{A} \\ &= \iint_S \left[ \vec{\nabla} \times \left( \vec{B} \times \vec{v} \right) \right] \cdot d\vec{A} \\ &= \int_{\partial S} \left( \vec{B} \times \vec{v} \right) \cdot d\vec{l} \end{align} which is nonzero. [To go from the second to third line, I used a standard vector calculus identity with assumptions about $\vec{v}$ for our case. To go from the third to fourth, I used the fact that $\vec{\nabla} \cdot \vec{B} = 0$. To get to the last line, I used the Stokes theoreom. Remember that $\partial S$ is the boundary of $S$.]
Alternatively, we could differentiate with respect to stationary coordinates, but note that the surface is changing in those coordinates. If we look at a little segment of the boundary $\Delta \vec{l}$, and suppose it's moving with velocity $\vec{v}$, then after time $\Delta t$ has passed, this has introduced a new amount of area into the surface given by \begin{equation} \Delta \vec{A} = (\vec{v} \Delta t) \times \Delta \vec{l}. \end{equation} This changes the flux integral by \begin{equation} \Delta \Phi_B = \vec{B} \cdot \Delta \vec{A} = \vec{B} \cdot (\vec{v} \times \Delta \vec{l}) \Delta t, \end{equation} which implies \begin{equation} \frac{\Delta \Phi_B}{\Delta t} = \vec{B} \cdot (\vec{v} \times \Delta \vec{l}). \end{equation} Now we can take the limits of small $\Delta t$ and small $\Delta \vec{l}$, then integrate up the contributions from all those little $\Delta \vec{l}$ segments around $\partial S$ to get the total derivative of the flux: \begin{align} \frac{d \Phi_B}{dt} &= \int_{\partial S} \vec{B} \cdot (\vec{v} \times d\vec{l}) \\ &= \int_{\partial S} \left( \vec{B} \times \vec{v} \right) \cdot d\vec{l}. \end{align} To get from the first to second line, I just used the usual scalar triple product. And this is the same result as the one we got using the material derivative — again, it's nonzero.
I'm not saying that any of this is obvious at all; it's certainly not obvious from the simple mathematical definitions I've given above. But that's all just mathematics. The key point to remember when applying it to physics is Jackson's point: You need to define your circuit with respect to the medium through which the electrons are actually moving.
There's really nothing that tells us the answers a priori — that's why humans had to do the experiments to figure out how physics works. There's a pretty good discussion of subtleties regarding Faraday's Law on Wikipedia's Faraday paradox page, but the better discussion of the wheel comes on the Faraday induction page. And of course, we should all read Jackson more carefully.
Best Answer
The circuit has an EMF $\mathcal E_0$ in the form of the battery of $12~\mathrm V\,.$
The current $I$ through the circuit is not constant right from the beginning.
It was zero when the circuit was open.
After a sufficient amount of time-interval, $I$ would attain a steady value $I_0\,.$
Prior to that $\dot I \ne 0\,.$
It can't go from $0$ to $I_0$ at an instant.
So, as the current $I$ changes at the rate $\dot I(t),$ there then arises the induced electromotive force which would tend to run the current in such a direction so as to oppose the flux change.
So, applying the law of conservation of energy, we get $$\mathcal E_0 + \underbrace{\left(-~ \mathrm L~\dot I(t)\right)}_\textrm{Back EMF} = RI(t);\tag I $$ assuming the direction of the current driven by the battery as positive.