Considering an Inductor in a DC circuit: when the switch is first closed there is a change in current through the inductor, which induces an emf in the opposite direction (Lenz's Law). My question is why does that emf cause a decrease in the rate of change of current through the inductor? (which would then cause a reduced emf and then an increased rate of change of current?)
[Physics] Rate of change of current in an Inductor
electromagnetisminductance
Related Solutions
The neon bulb needs 60V across it to strike (turn on) and obviously the 2V emf of the accumulator is not enough.
The induced emf in the circuit depends on the inductance and the rate of change of current.
The important parameter for the rate of change of current is the time constant of the circuit $\frac LR$ where $R$ is the resistance of the circuit.
When the switch is closed the resistance of the circuit is relatively low so the time constant of the circuit is large which in turn means that the rate of change of current is small leading to an induced emf which is less than 60V; not enough to strike the neon bulb.
On switching switching off the resistance of the circuit is very high. Think about the resistance of an air gap.
So the time constant of the circuit is very small, the rate of change of current very high which in turn means that the induced emf is high enough (greater than 60V) to strike the neon bulb.
So the graphs might look like this.
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Update to address the second question.
With the switch closed the current through the inductor reaches a steady value $I$ and the energy stored in the inductor is $\frac 12 LI^2$.
When the switch is opened the inductor is the only source of energy in the circuit and so the current must decrease.
If it increased that would mean that energy would need to be supplied to the circuit but there is no source of energy.
So the current needs to decrease.
Lena’s law states that the decrease in current would be opposed by an induced emf.
However that induced emf cannot stop the decrease in current it can only slow it down.
In this case that induced emf is large enough to make the neon bulb a conductor.
Now you have a series circuit consisting of the inductor and a resistor (the conducting neon bulb).
The current in this circuit decreases from its original value being opposed by a smaller induced emf because the resistance in the circuit is much lower.
Going back to the time immediately after the switch is opened.
You can think of the neon bulb as a capacitor.
The moving charge carriers (current) cannot stop instantaneously rather they start to charge the capacitor (neon bulb).
So the charge stored on the capacitor (neon bulb) increases and thus increases the potential difference across the capacitor (neon bulb).
Eventually the potential difference (electric field) across the dielectric of the capacitor, the neon gas, becomes so large (breakdown potential) that the neon becomes a conductor and lights up.
Note that during all this time the current in the circuit is decreasing as the increasing potential difference across the capacitor (neon bulb) is opposing further charges arriving at the plates of the capacitor (neon bulb).
So there are two stages when the switch is opened both of which are characterised by a decrease in the current.
The first is the build up in voltage across the neon bulb and the second is when the neon bulb becomes a conductor.
If the voltage across the neon bulb did not become large enough to make it conduct then you would have an series inductor, capacitor and resistor (resistance of the wires) undergoing damped oscillations with the energy originally stored in the inductor being dissipated as heat in the connecting wires.
If there is no resistance in the circuit, the rate of change of current in the circuit does not change, as explained here: inductor back EMF.
Thus, the decrease in the magnitude of the rate of change of current as time progresses must relate to there being resistance in the circuit.
In the circuit, you have $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$.
As time progresses, the current in the circuit increases, so $V_{\rm resistance}$ increases. Because $\mathcal E_{\rm battery}$ stays the same, $\mathcal E_{\rm back} $ must decrease, as does the magnitude of $\frac {dI}{dt}$.
To my knowledge, if there is a back emf, it should oppose the applied emf and the sum of these "emfs" should result in a net emf. Current should only increase if the net emf increases... No?
To understand what is happening one needs to consider where the equation $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$ come from and the use of the terms potential difference and emf.
From what you have written I assume that you have never queried as to what happens in a simple circuit which consists of an ideal battery and a resistor.
In such a situation the potential difference across the battery $V_{\rm battery}$ is equal in magnitude to the potential difference across the resistor $V_{\rm resistor}$ so that $V_{\rm battery}-V_{\rm resistor} =0$.
If the equation is multiplied by the current $i$ then the equation becomes $V_{\rm battery}i-V_{\rm resistor}i =0$ and it perhaps now becomes apparent that the equation is a restatement of the law of conservation of energy for electrical circuits, electrical power output from the battery is equal to electrical power dissipated in the resistor.
The only thing to add is that $V_{\rm battery}$ is also called the emf of an ideal battery, $\mathcal E_{\rm battery}$.
A slightly more complex example is an ideal battery used to power an ideal electric motor.
The steady state condition is that there is no current in the circuit because the emf of the battery, $\mathcal E_{\rm battery}$, is equal in magnitude to the back emf produced by the motor coil rotating in a magnetic field $\mathcal E_{\rm back,motor}$ thus $\mathcal E_{\rm battery} - \mathcal E_{\rm back,motor}=0$.
Now suppose that the motor is made to do some work by you pinching the axle of the motor.
The speed of revolution of the motor decreases and hence the back emf produced by the rotating coil, $\mathcal E_{\rm back,motor,load}$, decreases and what is the net input power, $\mathcal E_{\rm battery}i_{\rm load}-\mathcal E_{\rm back,motor,load}i_{\rm load}$, equal to?
In this case is is equal to the mechanical power output from the motor, the work done per second against the frictional force that you have applied to the axle.
Returning to your question one has to consider a dynamic situation.
Points to note are that connecting a battery to a length of wire will results in a current flowing in the wire.
If the circuit has inductance an emf will be induced because of the changing current (Faraday) and that induced emf will oppose the change producing it (Lenz) which in this case is the changing current.
So connect an ideal battery to and ideal (no resistance) inductor.
At the start even with no current flowing the current must change otherwise there would be no back emf from the inductor and then there is an unnatural situation with a battery being connected to a piece of wire with no resistance and no current is flowing.
Now we come to the situation which you are unsure about.
Two equal and opposing emf, $\mathcal E_{\rm battery}$ and $\mathcal E_{\rm back}$, and yet the current changes and noting that if there is no change in the current $\mathcal E_{\rm back}=0$.
$\mathcal E_{\rm battery}i-\mathcal E_{\rm back}i$ relates two powers.
$\mathcal E_{\rm battery}i$ is the instantaneous power delivered by the battery.
So what is $\mathcal E_{\rm back}i$?
It is the rate of change of the magnetic energy stored in the inductor.
So in this situation the instantaneous electrical power delivered by the battery is equal to the instantaneous rate of increase in magnetic magnetic energy stored by the inductor.
In a time $\Delta t$ the battery delivers electrical energy equal to $\mathcal E_{\rm battery}i\Delta t$.
During that time the magnetic energy stored by the inductor changes by $\frac 12 L(i+\Delta i)^2 - \frac 12 Li^2 = Li\Delta i + \mathcal O \Delta i^2$ and you will note that $Li\Delta i = L \frac{\Delta i}{\Delta t} i \Delta t= \mathcal E_{\rm back} i \Delta t \,(=\mathcal E_{\rm battery}i\Delta t)$.
Perhaps do not think of the emfs as battling it out for domination rather that one emf is the source of electrical energy and the other emf is the sink (user) of electrical energy and they are equal because energy is a conserved quantity.
Best Answer
The emf doesn't 'cause' a change in the rate of change of inductor current but it is, however, consistent with the rate of change in inductor current.
The fact is that the rate of change of current, the associated rate of change of magnetic field, and the associated non-conservative electric field must satisfy Maxwell's equations (the governing equations of classical electromagnetism) at each instant of time.
For the case that there is a constant, non-zero voltage across the inductor, Maxwell's equations are satisfied when
So it isn't true that the emf causes a change in rate of change of inductor current since, as pointed out above, there is a perfectly consistent solution in which the emf is non-zero and the rate of change of current is constant.