Suppose we have a conducting ring in a constant magnetic field $\vec{B}$. Suppose that the ring is being deformed. We know from Faraday's law that such an action will induce a current in the loop. Because of this currect a magnetic field $\vec{B}_{\text{induced}}$ would appear, which will weaken the net magnetic field (Lenz's law). So in total, the magnetic field would change ($\vec{B}+\vec{B}_{\text{induced}}$). However in every texbook there is an assumption that $B$ is always constant, that is: $\mathcal{E}=-\mathrm{d} \Phi/\mathrm{d} t=B(\mathrm{d} A/\mathrm{d} t)$, even though there are induced fields which change the magnetic field. Why do we ignore the induced fields in Faraday's law?
[Physics] Why the induced field is ignored in Faraday’s law
electromagnetism
Related Solutions
Electromagnetism – Does Induced Current’s Magnetic Field Affect the Change in Flux in Faraday’s Law?
I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
Here is one way to think about it:
When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force.
Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take a dipole magnet and move it close to the coil. As I do so, magnetic field lines cross the conductor, and generate a force on the charge carriers.
It is a convenient trick for figuring out "what goes where" to know that the induced current will flow so as to oppose the magnetic field change that generated it. In the perfect case of a superconductor, this "opposing" is perfect - this is the basis of magnetic levitation. For resistive conductors, the induced current is not quite sufficient to oppose the magnetic field, so some magnetic field is left.
The point is that the flowing of the current is instantaneous - it happens as the magnetic field tries to establish in the coil. So it's not "Apply field in coil. Coil notices, and generates an opposing field. " - instead, it is "Start to apply field in coil. Coil notices and prevents field getting to expected strength".
Not sure if this makes things any clearer...
Best Answer
The current generated in the closed wire loop of course induces a magnetic field $B_{induced}$ opposed to $B$ . The magnetic field would be $B-B_{induced}$.
But by what means you keep the constant or anyother initial magnetic field $B$ ???
By an apparatus that, trying to keep the field constant, increases or decreases the strength of its field from $B$ to $B+B_{induced}$ adding or subtracting energy to or from the wire loop depending on its motion and/or deformation. So the field remains constant :$(B+B_{induced})-B_{induced}=B$. That's the Lenz's Law : saves the energy conservation. This is a mechanism of energy exchange.
By no way the induced magnetic field $B_{induced}$ could be negleted . By principle, a so called induction motor works because of this :
"ELECTRIC MACHINERY" Fitzgerald-Kingsley-Kusko, 3rd Edition, McGraw-Hill Kogakusha.
4-2 INTRODUCTION TO POLYPHASE INDUCTION MACHINES, pages 187,189
"....the induction motor is one in which alternating current is supplied to the stator directly and to the rotor by induction or transformer action from the stator.(page 187)
.... When used as an induction motor, the rotor terminals are short-circuited......The field produced by the rotor currents therefore revolves at the same speed as the stator field, and a starting torque results, tending to turn the rotor in the direction of rotation of the stator-inducing field. (page 189)
Helmholtz Transport Theorem and Faraday's Law
Let $\:\mathbf{F}\left(\mathbf{x},t\right)\:$ be a time-varying vector field defined at points $\:\mathbf{x}=\left( x_{1},x_{2},x_{3}\right)\:$ of a region of $\:\mathbb{R}^{3}\:$ and time $\:t$. If $\:S(t)\:$ is a surface in motion-deformation described by the velocity field $\:\mathbf{w}\left(\mathbf{x},t\right)\:$ of its points, as in Figure, then for the rate of change of the flux of $\:\mathbf{F}\:$ through this surface we have
\begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{F}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}=\\ \int\limits_{S(t)} \left[\dfrac{\partial \mathbf{F}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \mathbf{w}\right)\mathbf{F} + \left(\mathbf{w}\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf{F} - \left(\mathbf{F}\boldsymbol{\cdot} \boldsymbol{\nabla}\right)\mathbf{w}\right] \boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{A-01a} \end{equation}
or from this equivalently \begin{equation} \bbox[#E6E6E6,8px]{\dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{F}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}\:=\:\int\limits_{S(t)} \left[\dfrac{\partial \mathbf{F}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \mathbf{F}\right)\mathbf{w} - \boldsymbol{\nabla} \boldsymbol{\times} \left( \mathbf{w} \boldsymbol{\times} \mathbf{F}\right)\right]\boldsymbol{\cdot} \mathrm d\mathbf{S}} \tag{A-01b} \end{equation} Equation (A-01a) or (A-01b) is known as the Helmholtz transport theorem.
['Generalized Vector and Dyadic Analysis', Chen-To Tai, IEEE PRESS, 2nd Edition 1997 equations (6.11),(6.12) page 119.]
Equation (A-01b) is important for the expression and understanding of Faraday's Law in electromagnetics \begin{equation} \mathrm{emf} \: = \: - \: \dfrac{\mathrm d\Phi}{\mathrm dt} \tag{A-02} \end{equation} that is, the electromotive force $\:(\mathrm{emf})\:$ along a closed path $\:C \:$ equals to the opposite of the time rate of change of the magnetic flux $\:\Phi \:$ passing through any surface $\:S\:$ whose perimeter is the closed path. Note that the surface need not be necessarily stationary but can be in motion and/or under deformation. Now, the magnetic flux $\:\Phi \:$ passing through a surface $\:S\:$ is defined as \begin{equation} \Phi\: \equiv \: \int\limits_{S}\mathbf{B}\left(\mathbf{x},t\right)\boldsymbol{\cdot}\mathrm d\mathbf{S} \tag{A-03} \end{equation} where $\:\mathbf{B}\left(\mathbf{x},t\right)\:$ the magnetic-flux density vector. Replacing $\:\mathbf{F}\:$ by $\:\mathbf{B}\:$ in equation (A-01b) yields \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S}\mathbf{B}\boldsymbol{\cdot} \mathrm d\mathbf{S}=\int\limits_{S} \left[\dfrac{\partial \mathbf{B}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \mathbf{B}\right)\mathbf{w} - \boldsymbol{\nabla} \boldsymbol{\times} \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right] \boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{A-04} \end{equation} and using Maxwell's equations in empty space \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & \:=\:-\:\dfrac{\partial \mathbf{B}}{\partial t} \tag{A-05.1}\\ \nabla \boldsymbol{\cdot}\mathbf{B} & \:=\: 0 \tag{A-05.2} \end{align}
equation (A-04) yields \begin{equation} \int\limits_{S} \boldsymbol{\nabla} \boldsymbol{\times}\left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right] \boldsymbol{\cdot} \mathrm d\mathbf{S} \:=\:-\:\dfrac{\mathrm d}{\mathrm dt}\int\limits_{S}\mathbf{B}\boldsymbol{\cdot} \mathrm d\mathbf{S}\: = \: - \: \dfrac{\mathrm d\Phi}{\mathrm dt} \tag{A-06} \end{equation}
By Stoke's Theorem \begin{equation} \int\limits_{S} \boldsymbol{\nabla} \boldsymbol{\times}\left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right] \boldsymbol{\cdot} \mathrm d\mathbf{S} \: = \: \oint\limits_{C} \left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right]\boldsymbol{\cdot} \mathrm d\mathbf{L} \tag{A-07} \end{equation} so \begin{equation} \oint\limits_{C} \left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right]\boldsymbol{\cdot} \mathrm d\mathbf{L} = \: - \: \dfrac{\mathrm d\Phi}{\mathrm dt} \tag{A-08} \end{equation}
and the electromotive force $\:(\mathrm{emf})\:$ along a closed path $\:C \:$ is \begin{equation} \bbox[#FFFF88,5px,border:1px solid black]{\mathrm{emf}\;=\: - \: \dfrac{\mathrm d\Phi}{\mathrm dt}=\;\oint\limits_{C} \left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right] \boldsymbol{\cdot} \mathrm d\mathbf{L}} \tag{A-09} \end{equation}
Note that the expression $\:\left[\mathbf{E} + \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right)\right] \:$ is the Lorentz force on a unit electric charge moving with velocity $\:\mathbf{w} \:$.
The electromotive force $\:(\mathrm{emf})\:$ along a closed path $\:\:C\:\:$ defined by (A-09) may be separated into two parts
(1) one part called transformer electromotive force $\:(\mathrm{emf})_{transformer}\:$ \begin{equation} (\mathrm{emf})_{transformer}\;\stackrel{\text{def}}{\equiv}\;\oint\limits_{C} \mathbf{E}\boldsymbol{\cdot} \mathrm d\mathbf{L} =\;-\;\int\limits_{S}\dfrac{\partial \mathbf{B}}{\partial t} \boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{A-10} \end{equation} due to the time rate of change of $\:\mathbf{B}\:$ and
(2) another part called motional electromotive force $\:(\mathrm{emf})_{motional}\:$ \begin{equation} (\mathrm{emf})_{motional}\;\stackrel{\text{def}}{\equiv}\;\oint\limits_{C}\left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right) \boldsymbol{\cdot} \mathrm d\mathbf{L} = \int\limits_{S} \boldsymbol{\nabla} \boldsymbol{\times} \left( \mathbf{w} \boldsymbol{\times} \mathbf{B}\right) \boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{A-11} \end{equation} due to the motion and/or deformation of the closed path.
This separation of the emf into the two parts, one due to the time rate of change of $\:\mathbf{B}\:$ and the other to the motion of the closed path, is somewhat arbitrary in that it depends on the relative velocity of the observer and the system (and in any case isn't Lorentz invariant).
'The Feynman Lectures on Physics' - Mainly Electromagnetics & Matter,Volume 2, New Millennium Edition, Basic Books . Chapter 17 : The Laws of Induction, 17-2 Exceptions to the "flux rule", page 17-2
"We begin by making an important point: The part of the emf that comes from the $\mathbf{E}$-field does not depend on the existence of a physical wire (as does the $\mathbf{w} \boldsymbol{\times} \mathbf{B}$ part). The $\mathbf{E}$-field can exist in free space, and its line integral around any imaginary line fixed in space is the rate of change of the flux of $\mathbf{B}$ through that line. (Note that this is quite unlike the $\mathbf{E}$-field produced by static charges, for in that case the line integral of $\mathbf{E}$ around a closed loop is always zero.)"