@SamyakMarathe the electric and magnetic fields are not separate things. They are both just parts of the electromagnetic field. The EM field is a single field but we see it as partly magnetic and partly electric depending on how we look at it.

So when we have a changing magnetic field we are actually seeing a changing EM field and the change affects both the magnetic part and the electric part.

You ask for an intuitive way to understand this, but there isn't one - or at least I don't know of one. The EM field is a tensor field and these are very different from the sorts of fields we learn about at school so we have no experience of them. If you go on to study physics you'll find the EM field is actually quite simple and easy to understand, but it involves principle that you simply haven't had a chance to learn yet.

@JohnRennie Hello sir :-) , free?

@LalitTolani Hi :-)

Yes I'm free

Sir inside a large hollow current carrying cylinder , magnetic field is zero

Outside it is $$\frac{\mu i }{2\pi r}$$

Yes, you can get this from Ampere's law

If we consider this cylinder to be 4-5 atoms thick

Then magnetic field just outside the outermost layer will be $$\frac{\mu i}{2\pi R}$$

where R is radius of cyclinder

Am i right?

Yes

And then it will gradually decrease to 0?

as we move inwards

It depends on how the current is distributed as a function of radial distance.

@JohnRennie I think it is not significant to tal about radial distance when our outer surface is 4-5 atoms thick

Ampere's law tells that 2πrB/μ₀ is equal to the current inside the radius r. Yes?

@JohnRennie yes

And if the current is evenly distributed throughout the metal of the cylinder then as r decreases from the outer to the inner surface the current enclosed decreases smoothly as well. Yes?

yes

So in that case the field will decrease smoothly to zero as we approach the inner surface.

yes

oh so you want to say that field will decrease gradually or not depends on how current flows in that small thickness of 4-5 atoms

When you get down to the sizes of atoms things get more complicated because now we get various quantum effects involved.

But the basic principle still applies.

yes

am i right?

Yes

Ok sir, it's clear

Is there a line missing from the diagram? Should V₀ be connected to the dot on the right?

yes sorry

here's the correct one

Option (a)?

But on that diagram V+ and V- are connected together so they will always be at the same voltage ...

@JohnRennie yes sir exactly that is my doubt , how can both be different

I’ve seen this question somewhere before v+ is only connected to the right, not to the left, diagram is wrong

There has to be a mistake in the diagram.

@AshishAhuja yes (a) is the correct answer

@AshishAhuja it is from recent kvpy sx

Ahh yes. Why don’t you look up the official paper? It’s there on the website. You’ll find the right diagram there

The question says V₀ can only be ±10V so it must be (a)

@JohnRennie yup exactly.

@AshishAhuja yes right , the book i have seems to have misprint it

@JohnRennie are you free for a question?

@JohnRennie yes got it that way , but what's the explanation of the graph ? why it oscillates between +10 and -10

I was unable to solve it so I looked up the solution in which they have assumed that the force exerted by the right half of the sphere on the left half passes through the center of the left half.. I’m highly skeptical of this, in fact I’m convinced that the net force would act at a point on the lower half of the left hemisphere (although my reasoning is a bit long). Although please answer Yash first.

@AshishAhuja find the accleration of centre of mass of left sphere , and then apply F(ext) = m*a(cm) should get you answer

@YashAgrawal Suppose V₀ is at +10V. That means V+ is at +5V because the two resistors on the right act as a voltage divider. Yes?

yes

Then current starts to flow onto the capacitor so the voltage on the capacitor increases with time.

yes

Eventually the voltage on the capacitor exceeds 5V, and when this happens V- > V+ so V₀ switches to -10V.

"when this happens V- > V+" how ?

Suppose we just turned the circuit on so the capacitor is uncharged. V- is connected to the capacitor so V- = 0. Yes?

yes

@AshishAhuja it doesn't matter where the force acts on the left part , although we are sure of it's direction (horizontal in this case ) by acceleration of centre of mass

Then if V₀ is at +10V the capacitor starts charging and its voltage increases.

yes

If V₀ stays at +10V the capacitor will eventually charge up to +10V, and since the capacitor is connected to V- this means V- will increase to +10V. Yes?

yes

And V+ is fixed at +5V by the voltage divider on the right, so as the capacitor charges that means V- must become greater than V+.

yes got it

now as Vo will become -10 , reverse would happen

Yes, now the capacitor starts discharging, and it discharges until it falls below -5V.

Once that happens V- < V+ again and the output switches back to +10V.

The end result is the output oscillates between +10V and -10V.

yes sir, got it thanks !

Also I remember the first time I saw this question I was confused by the triangle. Is the triangle a universal symbol for ground?

I'm not sure to be honest.

I'm used to the three lines meaning ground.

@YashAgrawal why does it not matter? I’m assuming that to calculate acceleration you mean centripetal acceleration so where the force acts must matter

But it's obvious it does mean ground in this case.

@AshishAhuja acceleration of centre of mass is dependent on net external force , how it matter's where external force acts ?

Ah ok. If you’re free could you have a look at the question I had posted above?

in all these 3 cases acceleration of cm is F/M in forward direction

Hmm

@YashAgrawal Yes, the linear acceleration is always a = F/M regardless of where on the body the force acts.

And the angular acceleration is always α = τ/I

Yes ok although intuitively using centripetal acceleration to calculate the force between the two halves seems a bit weird, it does make sense. Thanks.

You'll see lots of questions about beams being hit at their end and you have to work out how they translate and rotate, and you use the two principles above.

@JohnRennie there sir??

@LalitTolani Hi :-) I'm just answering another question ...

@LalitTolani Hi, I'm free now :-)

@JohnRennie Hi sir, I had a small doubt but I cleared myself, thankyou:-)

OK :-)

@JohnRennie Sir can we find magnetic field due to a finite length current carrying solid cyclinder using ampere's law

@LalitTolani There is no such thing as a finite length current carrying wire because current cannot just appear at one end of the wire and disappear at the other.

Either the wire is infinite, in which case we don't need to worry what happens at the ends, or it's part of a circuit in which case we would have to use the Biot-Savart law for the whole circuit.

@JohnRennie hi

Hi :-)

Are u free?

Yes

I wanted u to check a code just a sec I'll add a pic

I was trying loops

The error here is use of +=

It's because you've written:

for items in 'bill':

instead of:

for items in bill:

And can u tell what exactly ' ' is used for

( double and single quotes are the same in Python )

D:\rhs\Python>python
Python 3.9.6 (tags/v3.9.6:db3ff76, Jun 28 2021, 15:26:21) [MSC v.1929 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> for items in 'bill':
...     print(items)
...
b
i
l
l

Compared to:

>>> bill = [76, 79]
>>> for items in bill:
...     print(items)
...
76
79

Ok thank you

:-)

This will work:

bill = [76, 79]
total = 0
for items in bill:
    total += items
print(f"total: {total}")

Yup

@AshishAhuja in circuit simulators ground is mostly represented with a triangle like that.

I wish to discuss these two flowcharts

Anybody who can assist me with these?

Especially @JohnRennie Are you around?

Please ping me back when you are around.