6:00 AM
@Sarabsrimt if it is easier you can do that. if you are worried about eyes use night light feature ( I am not sure it is called the same in apple..). I-pad is good only if you don't have to show your work to anyone or if you are attending online classes though, I doubt any teacher would allow you to use it in class..

@Buraian I don't understand what you are asking.
I've seen questions where we have an infinitely thin wire moving in a field. The motion creates a Lorentz force on the electrons in the wire, and this causes them to move to the end of wire. The charge separation creates a potential difference between the ends of the wire.
But in your question you have a rod where the radius is comparable to the length, and it's moving along its length not normal to its length.
That would be complicated as you'd get eddy currents.

7:11 AM
@JohnRennie Hmm right, then ignore fringe effects and eddie currents

@Buraian in that case doesn't it just behave like the wire?

Hmm yeah looks like it

If you consider an electron in the rod then because it's moving relative to the field it experiences a force v x B.
That means it will move right on your diagram.
So the right edge of your rod will get a net negative charge and the left edge a net positive charge. Yes?

yes that makes sense so far

That creates an electric field E inside the wire, and the field exerts a force eE on the conduction electrons in the wire. The electrons move until the net force on them is zero i.e. ev x B = eE

7:18 AM
so, here's where I am getting confused
you have a current till charge gets built up?

@Buraian Yes
When the rod is stationary the electrons are evenly distributed. If you now start accelerating the rod then the electrons start to move to the right, and moving electrons = a current.
However the electrons only flow for a very short time because very quickly their movement causes a charge separation, and that charge separation causes an electric field, and the electric field causes a force that stops the electrons moving.

7:51 AM
does faradays law remain same in accelerated motion?
what is the intuition that they create an electric field such that motion is cancelled

@Buraian faraday's law is simply $$\nabla \times \mathbf{E}= \dfrac{-\partial \mathbf{B}}{\partial t}$$
i dont see why acclereated/unaccelerated motion has any role to play

8:14 AM
that is the statement of the fields , I'm talking about the statement of the physical loops
qvXB integral thing, I ask how we can expect the behaviour of object to change under acceleration

2 hours later…
9:47 AM
@Buraian in theory electrons take a non-zero time to respond to the Lorentz force. They have a mass so when you apply a force they can only accelerate at a = F/m.
But in practice their mass is so small that they respond to changes in external forces effectively instantly.
So when you accelerate a loop the electrons don't do anything weird.

10:18 AM
@JohnRennie sir can you explain why this must occur
the charges gets built up , then still there is PD through which current should pass through , right?

1 hour later…
11:47 AM
@JohnRennie hi

@AshishAhuja hi :-)

Let's say we have two point charges $q_1$ and $q_2$ located somewhere in space

OK ... ?

I bring in another point charge from infinity to some place at a finite distance from these two charges
You would agree that the work I do is the sum of the work I would have done individually for the point charges, yup?

Yes

11:50 AM
Ok so now we can say that the net potential energy of the charge I bring is the sum of the potential energies I would have done individually
$$U_{net} = U_1 + U_2$$

Yes

Oh nevermind I spotted my error, I'm so sorry. I spent 20 mins trying to figure it out but just couldn't spot it till now :(

OK :-)

12:10 PM
Is there any proof given my Mr.maxwell in saying that energy imparted by light depends only on Amplitude and not frequency
@JohnRennie Sir
Hello

@Sarabsrimt Classically the energy carried by light is described by the Poynting vector.
In physics, the Poynting vector represents the directional energy flux (the energy transfer per unit area per unit time) of an electromagnetic field. The SI unit of the Poynting vector is the watt per square metre (W/m2). It is named after its discoverer John Henry Poynting who first derived it in 1884. Oliver Heaviside also discovered it independently in the more general form that recognises the freedom of adding the curl of an arbitrary vector field to the definition. The Poynting vector is used throughout electromagnetics in conjunction with Poynting's theorem, the continuity equation expressing...
I can't remember how you derive it, but it is derived from Maxwell's equations.
I suspect the derivation is probably quite complicated.

But in end his derivation or answer is wrong right @JohnRennie
Max Planck was right
Is there any of his derivation which I can understand
Also , Q 2 is that energy is F * displacement .
So , whose F are we talking about. EM field

12:37 PM
@JohnRennie Whenever you get time sir . Please check this

@Sarabsrimt I have run out of time and I need to go now. We can talk about it tomorrow.

Sure sir. All the best for the day@JohnRennie

2 hours later…
2:10 PM
anyone checked official key for mains?

@RishiNandhaVanchi Do you have 10 minutes?

does anyone wanna help me with a number based algorithm?

@RishiNandhaVanchi I did , but only for 26th feb shift 1.

@MáttheusSpoo shoot

2:25 PM
@Wolgwang Hi what's poppin'

Guys , so does photo electric effect work for all states of matter

ok, here it goes.

i got 2 number inputs.

the first one, the user can put any number between 0 and 99.

When the sum of the first with the second input reaches 10, the second one that was 0 adds 2
when the sum of the two reaches 20, adds 3

so, when the first one gets 18, since the second one will already have 2, it will sum 20, adding +3 into the second input making it five. That's fine and I already done it. The problem is that, if now i subtract one from the first input making it 17, it doesn't go back to 2 on the se

A question of my institute . There are two person(1 &2)falling down at a separation of 5 m from each other . The 1st person is falling with a velocity u or 0 m/s and 2nd person with velocity 2m/s . Both the persons fall into a deep of hole of 60 m .Find the amount of force exerted on each other . Taking into consideration that person 1 falls on person 2 . Take g as $10m/s^2$.