Problem Solving Strategies

@JohnRennie I'm thinking that whenever there is a change in current it induces an EMF and that's an E-field..and E=-dV/dr I gave this reason...is it correct sir?

John Rennie

It is certainly true that $E = -dV/dx$

user8718165

@JohnRennie thank you sir....Can I give this reason?

John Rennie

I think I'd need to see the question to be sure

user8718165

5:49 AM

@JohnRennie No sir...I was just thinking about it...

John Rennie

I'm not sure it helps to focus on the back EMF i.e. I'm not sure it helps to understand what is going on.

Suppose you were somehow able to magically make the magnetic field disappear, then an (ideal) inductor would just be a zero resistance and there couldn't be any voltage drop across it.

user8718165

@JohnRennie okay sir...

John Rennie

The voltage drop happens because the work done on the electrons by the power source is then transferred to the magnetic field by the electrons doing work on the field.

Just like in a resistor the work done on the electrons by the power source is then transferred to the resistor as heat when the electrons do work on the resistor.

In both cases a potential gradient exists because as the electrons travel through the inductor/resistor they do work on it and hence lose energy.

user8718165

@JohnRennie okay sir...got it...

@JohnRennie I've another question

John Rennie

Yes?

user8718165

6:00 AM

@JohnRennie Sir is it correct to say the if a coil has n turns and V induced then the gradient is V/n ?

John Rennie

You could do, but what would you be trying to achieve? $dV/dx$ has a physical interpretation as the field, but $dV/dn$ isn't obviously useful.

user8718165

@JohnRennie got it fully sir...thank you

@JohnRennie sir.. is the concept of voltage defined for non conservative fields?

John Rennie

As a general rule a non-conservative field is non-conservative for paths that form loops. As long as you avoid these types of paths you can still define a voltage. However the quantity you define is only of limited use.

user8718165

@JohnRennie Sir which quantity will be of limited use? Voltage Sir?

John Rennie

If you move a charge $Q$ along some path then the voltage change is just the work done on the charge divided by $Q$. You can always measure the work done, so you can always calculate a voltage change.

user8718165

6:08 AM

@JohnRennie okay Sir...

John Rennie

The problem is that in a non-conservative field you can move a charge round a circle and back to its starting point and have to do work on it the whole way.

That means using our definition of the voltage change there must be a potential difference between the strt and end points, but the start and end points are the same point.

So you end up concluding that the same point must have two different voltages.

But then you can go round the loop again, and now you find the same point must have three different voltages. And so on round the loop until you get bored.

The conclusion is that you cannot assign a potential to any point in space because the potential can have an infinite number of different values there.

user8718165

@JohnRennie Thank you so much sir for explaining this so nicely...

John Rennie

But if you move a small distance $dx$ in a straight line, or at least not in a loop, it still makes sense to think of a potential difference as the work done divided by the charge.

user8718165

@JohnRennie I understood sir...Can I ask just last 3 questions?

@JohnRennie okay sir...got it

John Rennie

Yes, I don't need to work for a while so you're welcome to ask.

user8718165

6:15 AM

@JohnRennie got it sir...the problem arises because in a we're at the same point everytime but the voltage has changed...right?

John Rennie

Yes

OK, what did you want to ask about it?

user8718165

@JohnRennie sir when a stick breaks on impact means the thing which was hit applied enough force on the stick to break it and by 3rd law the stick also applied that force to the object...but if the stick doesn't break...that means F applied isn't as high, so by 3rd law the stick applied less force on object

@JohnRennie As others, I also have a feel that if it breaks...less force is applied...but reason is coming opposite. Please help sir

John Rennie

I don't think the question is well enough defined to be answered. It depends on the details of exactly how the stick breaks.

user8718165

@JohnRennie In general what will happen sir? Why is my reason wrong? Can you please tell? :-)

just your thoughts :)

John Rennie

The impact is not just a sudden force. The force starts increasing from zero when the stick first contacts the object you are hitting, then the force increases as the object and the stick deform under the impact.

user8718165

6:26 AM

@JohnRennie yeah sir

John Rennie

There will be work done, which is the integral of Fdx. If for example you are hitting something that plastically deforms the size of the dent will be related to the work done not the force.

user8718165

@JohnRennie okay sir...got it...Is that why my reason is incorrect sir?

John Rennie

So how much damage you do will depend on how the force changes with time during the impact, and that probably depends on whether the stick breaks or not. But that details are going to be complicated.

user8718165

@JohnRennie okay sir...could you please tell a qualitative fact why work done if stick breaks is less....

John Rennie

I don't know if the work done really is less or not.

It's tempting to believe that physicists should be able to figure everything out just by thinking about it, but in messy and complicated situations like a stick breaking this is risky at best. The only sure way to know is to do the experiment.

You just need to find a volunteer for you to hit with a stick :-)

user8718165

6:36 AM

@JohnRennie okay Sir... Got it

@JohnRennie LOL LOL

@JohnRennie Sir why are inductors more difficult to grasp than resistors and capacitors?

John Rennie

@user8718165 They really aren't when you get used to them.

Both inductors and capacitors are reactive i.e. the current and voltage are out of phase, so they are more complicated than resistors.

But once you get the hang of phasor diagrams they are straightforward to deal with.

user8718165

@JohnRennie hello sir...sorry for late reply...My phone died :(

@JohnRennie okay sir got it. Thanks a lot for telling :)

user8718165

7:11 AM

@JohnRennie hello sir

John Rennie

@user8718165 hi

user8718165

@JohnRennie sir what is your opinion about HC Verma ?

John Rennie

I've never read it

user8718165

@JohnRennie no sir, the person :)

John Rennie

I've never met him

Kevin N

7:21 AM

@JohnRennie Remember this yesterday?

John Rennie

@KevinN hi, yes I remember that

Kevin N

When the strip hits the first teeth does that create a full wavelength?

John Rennie

Do you mean if you start with everything stationary then start turn the wheel, does the first motion of the strip create one cycle of the sound wave?

Kevin N

Yes

John Rennie

As a general rule if you have a system that can oscillate then make a sudden change to it the motion will consist of an initial transient motion that quickly settles down to the long term cyclic motion.

So it seems likely that the first few motions of the strip would be different to the motion once the system has settled into a steady state.

But exactly what happens would require a detailed model of the mechanics of the strip and wheel.

Kevin N

7:26 AM

Ahh I see. Alright then!

@JohnRennie

I got this correct

But I was wondering something

It is said that the Anmeter $A$ reads $0$ if the potential across those 2 nodes (beside) are the same.

Isn't the potential across the Ammeter always $0$, no matter the resistance? What am I missing?

John Rennie

Current always flows in loops, so for the ammeter to read zero the current in the loop shown by the blue arrows has to be zero.

Kevin N

physics-ref.blogspot.com/2019/06/…

This was the suggested solution

@JohnRennie

John Rennie

You'd have to do a Kirchoff type analysis to work out the currents, but I don't think the result would be that the current is always zero through the ammeter.

Kevin N

Have you seen the link I sent?

John Rennie

Actually this would be best done using superposition.

@KevinN yes, I saw the link

Kevin N

7:36 AM

Superposition?!

Of circuits?

I've never heard of that

John Rennie

Have you not come across the idea of superposition for circuits?

Kevin N

No

John Rennie

It's a neat trick when you have multiple batteries.

Kevin N

Tell Me :)

John Rennie

You remove all the batteries except one and replace them with a wire. Then you calculate the currents in all the links due to that one remaining battery. You do this or each of the batteries in turn, then add up all the individual currents and it gives you the total current with all batteries present. Note that currents in opposite directions cancel.

Kevin N

7:40 AM

But is that really necessary for this problem?

The solution provided was very simple.

John Rennie

In this case the answer is obvious (or at least it seems so to me) but suppose the problem asked you to calculate the current in the ammeter as a function of R.

You calculate the current through the ammeter for these two circuits:

And just add them.

Then that gives you an equation for the current in terms of R that works for any R.

Kevin N

@JohnRennie For current to be zero through the Ammeter $A$, the potential difference between the nodes (left &right) should be zero, is this true?

John Rennie

No, that is not safe reasoning.

We assume this is an ideal ammeter and has a resistance of zero.

Then the potential difference between the two sides of the ammeter is always zero regardless of what current flows through it.

Kevin N

physics-ref.blogspot.com/2019/06/…

So this is incorrect?

John Rennie

Yes, the article starts with an incorrect statement. Fortunately that doesn't matter and it gets the correct result anyway.

Kevin N

7:50 AM

Yes that was what I was thinking. The potential difference across the ammeter should always be zero

John Rennie

You can easily come up with a simplified circuit that disproves the statement made in the article you link.

It just goes to show you can't believe everything you read on the Internet :-)

Kevin N

So how do you properly resolve this?

John Rennie

I would do it using superposition. That takes a little longer but avoids mistakes like the article makes.

There is a quick way to do it though the argument is a bit subtle ...

Kevin N

But why would Cambridge publish such a question that requires the participants to use Circuit Superposition , a technique that has never been taught or mentioned in the syllabus?

John Rennie

Suppose I put a switch in the circuit like this:

With the switch open it is, I trust, obvious that no current flows through A. Yes?

Kevin N

7:56 AM

Yes

Wait... Why though?

John Rennie

So the question is, what is the condition required for closing the switch to make no difference? Care to speculate?

@KevinN why does no current flow through A with the switch open?

Kevin N

Yes

Is it because the current cancels out?

John Rennie

Current always flows in loops. A common analogy is water flowing in pipes.

With the switch open A is not part of a loop, so how can any current flow through it?

Kevin N

When, let's say the electrons pass the 200 ohms resistor (now the electron is below it), doesn't it have a choice to go $A$?

John Rennie

Suppose there was a current flowing from right to left through A as you suggest.

That would mean electrons are continuously flowing right to left through A, so the right side would be getting increaingly positively charged and the left side would be getting increasingly negatively charged.

You end up with a charge separation appearing in your circuit and that charge separation would increase linearly with time, with no upper limit. Does that seem physical to you?

Kevin N

8:04 AM

Ah I see...

John Rennie

So are we agreed that with the switch open no current can flow through A?

Kevin N

Yes

John Rennie

So the question is, what is the condition required for closing the switch to make no difference? Care to speculate?

Kevin N

@JohnRennie Placing a resistor $R$ so that the potential difference across the nodes between the open switch (now closed) is zero?

Which is $C$ (400 Ohms)

John Rennie

Correct. If the potential difference between the ends of the open switch is zero then closing it will make no difference because with a potential difference of zero no current will flow through it.

So what we require is that with the switch open the value of $R$ makes the potential the same at either side of the (open) switch. And that gives you $R = 400$ ohms.

Kevin N

8:12 AM

@JohnRennie Assume that we eliminate the wire connecting the middle part (switch segment). And suppose we change the position of the left node on the Ammeter and place it diagonally on top of resistor $50 ohms$

Current would still be zero through the ammeter right?

Like that you mean?

Yes

Is A part of a loop?

No

So can any current flow through it?

Kevin N

8:17 AM

No :)

John Rennie

Correct :-)

Kevin N

But is it possible for the battery on the left loop to somehow attract the electrons from the right loop?

Or rather is there any alternative explanation on why current is absent on $A$? I didn't quite grasp the charge concept you told me earlier

(I'm awful at circuits) :)

John Rennie

A current is a flow of charge yes?

Kevin N

Yes

John Rennie

So if a current is flowing though A that means charge is flowing through A i.e. there is a net flow of electrons from one side to the other.

Kevin N

8:28 AM

Yes

John Rennie

So the number of electrons on one side must be decreasing and the number of electrons on the other side must be increasing.

Kevin N

What about in a regular circuit? Or because it's a loop the electron isn't lost anywhere

John Rennie

> Or because it's a loop the electron isn't lost anywhere

Correct.

Because we have a loop for every electron that flows from left to right another flows from right to left.

So the net flow of electrons is zero.

Kevin N

Ok continue :)

So if electrons were to move through $A$ it's as if the left loop was "stealing" electrons?

John Rennie

Yes, and since every electron carries a negative charge that means the left side must be accumulating a negative charge.

Kevin N

8:32 AM

Yes

John Rennie

Likewise the right side must be getting more positive.

Kevin N

And is that impossible?

John Rennie

It means you are starting with something uncharged and you get a charge increasing linearly with time. So if you wait long enough the charge will become infinite.

Kevin N

But isn't the number of electrons limited?

yuvraj singh

@JohnRennie hi

alesha here

John Rennie

8:36 AM

The charge is $Q = \int I dt$. Yes? If there is a steady current then this integral is just $Q = It$.

@yuvrajsingh hi Alesha :-)

yuvraj singh

i have a question.

John Rennie

@KevinN there could be a transient current, but not a steady current.

@yuvrajsingh yes?

yuvraj singh

sir actually today i learn about shordinger wave equation.

Kevin N

What is a transient current? A current not of electrons?

yuvraj singh

and i saw the equation.and how it derive the result.

of hydrogen

John Rennie

8:39 AM

@yuvrajsingh can I get you to wait just a moment while I answer Kevin's question?

@KevinN have you studied capacitors?

yuvraj singh

yes sir.

Kevin N

Not yet

John Rennie

In that case don't worry about transient currents for now. I'd have to explain how capacitors work and that would turn into a long explanation.

Kevin N

But why is the charge built up infinity, shouldn't it run of electrons ?

John Rennie

@KevinN let me answer Alesha's question then we can pick this up later if necessary.

Kevin N

8:41 AM

Alright :)

John Rennie

@yuvrajsingh what did you want to ask?

yuvraj singh

ok sir

first is i saw the equation

and it derives the result for si 1

John Rennie

OK ... ?

yuvraj singh

and the eqution is all about the energy where k.e +p.e=total eenrgy.

when i was derving this equation

there is term come eigrn value.

what is it.

John Rennie

Kind of. You can think of the Schrodinger equation as just saying the total energy is the sum of the kinetic and potential energies, but that isn't really true.

The time independent form of the Schrodinger equation looks that way, but it isn't really true of the full time dependent form of the Schrodinger equation.

The SE can't be derived in the usual sense. It is a postulate i.e. Schrodinger assumed it applied and checked whether it predicted the correct results.

yuvraj singh

8:48 AM

let me write the line which i do not understand.

John Rennie

OK ...

yuvraj singh

del squared +v and product of them with psi,is not taken as product ,instead of this it is an operator which operate on wave fuction

waht kind of operator it is an how it works.

John Rennie

Suppose you have a function $f(x)$. This can be any function even as simple as $f(x) = ax$.

If we differentiate the function we write it as $df/dx$. Yes?

yuvraj singh

ok.

John Rennie

We can write this as:

$$ \frac{d}{dx} ( f(x)) $$

yuvraj singh

8:52 AM

yes,,,,

John Rennie

i.e. we have $d/dx$ acting like a function that takes the argument $f(x)$ and returns the new function $df/dx$

yuvraj singh

ok

John Rennie

Written this way $d/dx$ is an operator i.e. it is something that takes a function as its argument and returns a new function.

yuvraj singh

ok

John Rennie

So when we write $\nabla^2 \psi$ the $\nabla^2$ is an operator that takes the function $\psi$ as its argument and returns a new function.

$$ \nabla^2 = \frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2} $$

So the operator $\nabla^2$ acting on $\psi$ produces the new function:

yuvraj singh

8:56 AM

but we should include v also.

that ,s what my book says .

John Rennie

$$ \nabla^2 \psi = \frac{d^2\psi}{dx^2} + \frac{d^2 \psi}{dy^2} + \frac{d^2 \psi}{dz^2} $$

The Schrodinger equation is normally written as:

$$ \frac{-\hbar^2}{2m} \nabla^2\psi + V\psi = i\hbar \frac{d\psi}{dt} $$

And with a minor rewrite we get:

$$ \left( \frac{-\hbar^2}{2m} \nabla^2 + V \right) \psi = i\hbar \frac{d\psi}{dt} $$

So the operator that takes $\psi$ as its argument is really $\frac{-\hbar^2}{2m} \nabla^2 + V$

yuvraj singh

ok sir can i ask second doubt

John Rennie

@yuvrajsingh yes?

yuvraj singh

what would be the probability of finding an electron in universe @JohnRennie

John Rennie

If you have an electron it must be somewhere. That is if you search the entire universe for it you are guaranteed to find it somewhere. Yes?

yuvraj singh

9:05 AM

ok

yes

John Rennie

So the probability of finding the electron somewhere in the universe is one.

yuvraj singh

yes ,but the problem is let say si star

John Rennie

The way we write this is that if $\psi$ is the wavefunction describing the electron (i.e. the solution to the Schrodinger equation) then the propability of finding the electron in some volume of space $V$ is given by:

yuvraj singh

is the complex conjugate

John Rennie

$$ P(V) = \int_V \psi^*\psi dV $$

yuvraj singh

9:07 AM

of si.

John Rennie

Where $\int_V$ means we do the integration over the region of space $V$.

yuvraj singh

ah,got it sir

John Rennie

And what we are saying is that if we take $V \to \infty$ i.e. integrate over the whole universe this probability must be one.

$$ \int_\infty \psi^*\psi dV = 1 $$

yuvraj singh

@JohnRennie one last question

John Rennie

@yuvrajsingh yes?

yuvraj singh

9:11 AM

i understand theuncertainity priciple but i do not know how can we prove it.

@JohnRennie

John Rennie

The uncertainty principle is really simple to understand once you have the required maths. In particular you have to understand Fourier transforms.

But without that mathematical background there is no good way to explain it. I'm afraid you will just have to accept it for now.

yuvraj singh

ok sir.

John Rennie

Sorry :-(

yuvraj singh

can you take a simple example

for derving a result from shordinger equation.

like psi 1

@JohnRennie

John Rennie

The example QM courses usually start with is the free particle i.e. the case where the potential $V$ is zero everywhere.

Let's restrict ourselves to one dimension to keep the algebra simple, then for the free particle the Schrodinger equation is:

yuvraj singh

9:16 AM

ok

John Rennie

$$ \frac{-\hbar^2}{2m} \frac{d^2\psi}{dt^2} = i\hbar\frac{d\psi}{dt} $$

@yuvrajsingh OK so far?

yuvraj singh

yes

Kevin N

@JohnRennie I still have a lot of question to ask. But we'll do it tomorrow. I got a Chemistry exam. Just remember the image I sent earlier. Bye :)

John Rennie

Now when we're solving equations like this we usually try guessing first. You can often solve an equation just by guessing and checking to see if the solution works.

In this case let's try the guess:

$$ \psi = A e^{i(\omega t - kx)} $$

where $A$, $\omega$ and $k$ are constants.

So what we need to do is take this guess and substitute it into the SE and see if it solves it. Shall I do the algebra or do you want to have a go?

yuvraj singh

can you take this .

John Rennie

9:21 AM

Oops, I wrote the SE down wrong. It should be:

$$ \frac{-\hbar^2}{2m} \frac{d^2\psi}{dx^2} = i\hbar\frac{d\psi}{dt} $$

yuvraj singh

yes.

John Rennie

@yuvrajsingh OK, let's set $A=1$ for now. In fact we won't need it. Then $\psi = e^{i(\omega t - kx)}$

yuvraj singh

yes

John Rennie

$$ \frac{d^2\psi}{dx^2} = -k^2 e^{i(\omega t - kx)} = -k^2\psi $$

yuvraj singh

yes

John Rennie

9:24 AM

$$ \frac{d\psi}{dt} = i\omega e^{i(\omega t - kx)} = i \omega \psi $$

@yuvrajsingh OK so far?

yuvraj singh

yes

John Rennie

So now we just have to substitute these back into the SE. This gives:

$$ \frac{-\hbar^2}{2m} (-k^2\psi) = i\hbar (i\omega\psi) $$

yuvraj singh

got it sir.

that ,s easy

John Rennie

So you can immediately see that this is a solution if:

$$ \frac{\hbar^2 k^2}{2m} = -\hbar \omega $$

yuvraj singh

actually QM which deal with micro particel right sir.

@JohnRennie

John Rennie

9:30 AM

We haven't specified what particle we are dealing with here. It could be an electron, a proton, even a whole atom.

yuvraj singh

is there a term force in qm .

John Rennie

The only difference would be the value of $m$ in the equation because that is the mass of the particle.

yuvraj singh

becasue in classical mechanics ,there is term force.

John Rennie

Force gets a bit complicated in quantum mechanics. Suppose we are dealing with a hydrogen atom not a free particle. Now the potential $V$ is not zeo.

We write it using the classical equation:

$$ V(r) = - \frac{ke^2}{r} $$

But it isn't really a classical equation even though it looks like it is. This is actually an operator that acts on $\psi$ to return the potential energy.

And force is $-dV/dr$, so force also isn't as simple as it looks in classical mechanics.

yuvraj singh

ok

John Rennie

9:36 AM

This all seems utterly confusing when you first start learning QM, but it's because you don't have the knowledge of the maths that explains it all. The maths required is linear algebra, and that is generally not taught unless you do a QM course at university.

So once again I have to tell you to just accept what the QM textbook tells you.

yuvraj singh

ok.

John Rennie

QM is absolutely fascinating, but I'd be cautious about spending too much time on it. You won't need it for the JEE.

yuvraj singh

yes ,sir.

thanks for your time.

John Rennie

You're welcome :-)