Induced current due to induced current

Say we have a square coil and a change in magnetic flux is created through the coil. Thus inducing a current in the coil whose magnetic field opposes this change inside the coil. If we assume(for simplicity) that the external field stops changing as soon as a current is induced. Then shouldn't the magnetic field due to the current induce another current which again results in a change in magnetic flux thus again inducing a current in opposite direction and so on. Thus we should always keep getting induced currents and the process should never stop.

Can somebody please point out where I am getting it wrong or add to my knowledge a concept I am missing?

Thanks

• A monotonous increasing function need not to diverge. In other words, it isn't because a current consists of infinitely many terms that the current is infinite (or diverge). Mar 17, 2019 at 20:03
• I will edit my question. By infinite I meant that the process of current induction will go on forever Mar 17, 2019 at 20:05
• Changing flux creates a voltage, not a current. This is a very common mistake. Apr 25, 2020 at 2:12
• Inductance is an objects ability to resist a change in current due to its own induced emf acting in itself. If a current is changing it wil induce an emf on itself to act against this curent. Ie slowing its change down. If the induced emf slows its chnge down, thinking about it in terms of "this chnge then induced another emf", as you go down the rabbit.hole of infinite loops, the emf acts to slow the change down. Causing a point at which the change and induced emf is zero. Solving the equations you see it stops once the initial B field is turned off ( onceTheInitial change inCurrent stops) Aug 15 at 20:11

The first thought that comes to my mind on reading this is the fact that through this, we are making a current source which derives energy out of itself, and has a seemingly perpetual life. This is what makes me feel something is wrong.

So let’s assume that the process of making the magnetic field constant happens instantaneously. Initally, when the magnetic field is swiched on, it will oppose its increasing flux. Now when you will make it constant, basically the source of induced current dies,and the current starts its process of decrement. Just when it decreases, the flux through the ring by its own current decreases, so it generates some current in the same direction to increase its flux.

But at this point in time, the value of current is smaller than its initial value and since current induced by self induction in this case is proportional to the rate of change of current in the circuit, for approximately similar time intervals, a smaller current will have a smaller change(w.r.t Zero). This would mean that the induced current will not be able to make up for the intial decrememt.
This process will only magnify itself as time passes by and the curcuit will finally drop to almost zero.

Just like in discharging events it takes an infinite time to reach that absolute zero level, it possibly will be the same case here, but any observable current will die out almost instantaneously.
Also in any case a wire will have atleast some amount of resistance, which will lead to energy loss.

This situation would be like discharging a capacitor through a resistor. The energy from the field is dissipated in the resistance of the circuit. If the coil were a superconductor, (open circuit when the external field is applied but closed before it is removed), then the current could persist.

The phenomena you describe is called self inductance. When a current flows through a circuit, it creates a magnetic field whose variations induce an emf. This emf is given by: $$e=-L \frac{dI}{dt}$$.Where L is a constant that depends on the geometry of the circuit. It is called inductance and its unit is the Henry (H). The emf has thus two components, one due to the variations of the external magnetic field: $$e_{ext} =- \frac{d \Phi }{dt}$$ and one due to the self inductance: $$e_{int}=-L \frac{dI}{dt}$$ Thefore, the equation for the current reads:$$- \frac{d\Phi }{dt}-L \frac{dI}{dt} =R.I$$ The solution of which is:$$I(t)=I_{0} e^{- \frac{R}{L}t }- \frac{1}{L} \int_0^t \frac{d \varphi (t')}{dt'} e^{- \frac{R}{L} \big(t-t'\big)}dt'$$

As you have guessed, the self inductance opposes the disappearance of the external magnetic field but on the contrary of what you were thinking, it is not strong enough to completely take over and perpetuate the cycles of induction. In fact we can easily calculate the amount of energy lost by Joules effect when the external magnetic flux remains constant. It is just: $$E(t)=R.I^{2}=RI_{0}^{2} e^{- \frac{2.R}{L}t }$$ The magnetic energy stored into the circuit at any time is given by: $$E_{m}(t)= \frac{1}{2}L .I^{2}=\frac{1}{2}L .I_{0}^{2} e^{- \frac{2.R}{L}t }$$ As you can see, it is decreasing pretty fast.