Sir, @JohnRennie this was my doubt yesterday but now that I think about it Lenz's law only slows it down and doesnt stop it altogether. And if the magnetic field is large enough then we can see that the acceleration would tend to zero. Is that correct sir?
I was solving this question and the acceleration comes out to be a constant. But from my rudimentary understanding of Lenz's Law the wire should stop accelerating after a while and just come to a halt to resist any change in flux. And thus, acceleration should be a variable which tends to zero after some amount of time. This is what I'm thinking intuitively but it's not matching the result. Where am I going wrong?
@JohnRennie That's what I was going to ask after seeing @KavinIshwaran's solution
What's happening is that when a current flows in the loop that current generates a field that opposes the external field.
We get simple harmonic motion.
The equilibrium point is when the capacitor is discharged and the wire has its maximum velocity. Then the EMF in the loop starts charging the capacitor and the capacitor potential opposes the EMF induced in the loop.
Sir @JohnRennie , I checked my notes again and it does say that when you charge a capacitor there would be a loss even if you take a superconductor. The loss would be in the form of electromagnetic pulse.
The only mistake that I can see in the derivation is that while deriving the work of the battery (Cε^2) we assume that the battery will have a constant emf but the emf of the battery will change as we charge the capacitor.
Could there be a reason why my notes make that assumption and say the loss is accounted by EMP?
Now lets find the energy of the equilibrium state.
The change in gravitational PE is -mgx and the change in the elastic energy of the spring is ¹⁄₂kx².
So the total change is the sum of these. And we expect the total change to be zero because energy is conserved so the total energy cannot have changed. Yes?
Now let's see how my analogy applies to a battery and a capacitor.
At time zero we connect the battery to the uncharged capacitor and the capacitor starts to charge. This time the equilibrium state is when the capacitor voltage V equals the battery EMF E: E = V = Q/C
When the capacitor starts charging the battery is accelerating the electrons in the wire, so when we get to the equilibrium state the electrons have a momentum that keeps them moving past the equilibrium state.
It's like the spring. If we have a truly ideal battery, wire and capacitor then the current would not stop flowing at the equilibrium state. It would keep flowing at charge the capacitor to Q = 2CE, just like the spring extends to 2mg/k.
Then the current would stop and start flowing in the reverse direction i.e. charge would now be flowing off the capacitor and charging the battery.
The result is that in an ideal system the current would never settle to zero. It would just keep oscillating forever.
The only way the current would ever settle at zero is if there is some resistance to dissipate the KE of the electrons flowing along the wire.
And if there is any resistance, no matter how small, you'll find half the energy is dissipated in the resistance just like in the spring half the energy is lost to friction.
The size of the resistance just affects how fast the energy is lost i.e. how quickly the current decreases to zero.
Well even if the resistance was zero (which it is in a superconductor) oscillating electrons do radiate energy as EM. That's exactly what a radio transmitter does.
So if you could construct a superconducting circuit like this it wouldn't oscillate forever - it would eventually radiate away the energy as radio waves.
It doesn't matter how the energy is lost. The lost energy is always equal to the work done by the battery minus the equilibrium state energy, and the energy lost is always a half of the work done.
@Swan It took me a long time to get this as a student, so I'm really keen to help new students understand it!
