@sammygerbil Are you interested in seeing how it is done

@Abcd Yes. I still think option B is incorrect. But I could be wrong.

First of all recall the equivalent batteries in parallel formula:

$\dfrac{E_{eq}}{R_{eq}} = \dfrac{E_1}{R_1} + \dfrac{E_2}{R_2}$

Now here we have two batteries ach with internal resistance r/2

$R_{eq} = r/4$

and E for each is $B\omega a^2/2$

Hello @sammygerbil Are you free to clear my thermodynamics doubt?

Thus, $E_{eq} = \dfrac{B\omega a^2}{2}$

Now after replacing battery with equivalent battery

We have two resistances in series

@user64829 Sorry I am busy discussing a question with Abcd and Ice Inkberry. I will let you know when I am available.

internal resistance of equivalent battery and r

Okay

Thus, $I = \dfrac{E_{eq}}{r/4+ r}$

$\implies I = \dfrac {2B \omega a^2}{5r}$

@sammygerbil @IceInkberry Are you happy now

@Abcd Yes I think you are correct.

@sammygerbil i have more questions

@Abcd ok. You have the conn.

@sammygerbil what is conn

@Abcd It means you have the right to steer the ship. forum.wordreference.com/threads/you-have-the-conn.2991119 Nautical reference often used in Star Trek.

@sammygerbil based on intuition I got the right answer.

But I am not FULLY sure why option D is also correct.

Is induced electric field due to rectangular magnetic field variation rectangular?

Just like in a cylindrical region of varying magnetic field the induced electric field is circular

@Abcd I don't understand your last 2 sentences. I think you have to look at the currents induced in the 3 loops of the network.

@sammygerbil ??

how to do it?

@sammygerbil I just used basic Lenz's law

anticlockwise currents is what the loops want

to create downward magnetic field

But cant apply this to point R

@Abcd I think the induced currents are clockwise. $B$ points into the page and is reducing. Induced current tries to keep $B$ the same. To create downward $B$ it should be clockwise.

@sammygerbil anti clockwise was a "mistype" in hurry

yes clockwise, I agree

@Abcd Sorry there are 2 loops not 3. One loop is central rectangle. 2nd loop is between outer and inner rectangles.

@sammygerbil OK, but what about my question?

@Abcd Current through R is zero, I think.

@sammygerbil Even I think that but whats the physics reason behind it

@Abcd There same current flows all the way around the outer loop. This current overlaps with itself at R, travelling in opposite directions. Therefore it cancels.

@sammygerbil didnt get

@sammygerbil ??

@Abcd Starting at P trace the current clockwise around the outer edge of a continuous area. Current in this loop travels right at P then left through R then through Q. It continues anti-clockwise round the inner rectangle and back through R (this time travelling right through R) then clockwise again round the outer perimeter back to P.

So the same current flows left then right through R. Resultant current through R is zero.

@sammygerbil But answer is it is clockwise at point Q $:($

So I think A and D are correct.

@sammygerbil Answer is ACD

I haven't finished with Q yet. I was only answering for P and R.

@sammygerbil You have written "continues anticlockwise after Q"

B and C are mutually exclusive.

@sammygerbil what do you mean??

There are 2 loops. I have only considered the outer loop on which PQR are all located. There is also an inner loop on which only Q is located.

Oho I am not getting anything

why are you considering such complex loops

Current in the inner loop is clockwise (right) through Q, whereas current in outer loop is anticlockwise (left) through Q.

A current loop is a continuous perimeter of conducting material which encloses an area. There are 2 enclosed areas in this diagram.

Therefore there are 2 current loops to consider.

The 2 loops seem to have the same area therefore also the same rate of change of flux and the same induced EMF.

@sammygerbil why cant we trace anticlockwise when starting at P?

Current through P must be clockwise to try to stop the $B$ field from decreasing.

@sammygerbil but that results in current through Q being anticlockwise

@Abcd Q is situated on 2 different current loops. The currents in these 2 loops flow in opposite directions through Q.

@sammygerbil how do we know which loop has more pronounced effect

That's what I was just getting to.

The loops have the same EMF but different perimeters and therefore (presumably) different resistances.

Outer loop has greater resistance so smaller current.

So the clockwise current through Q is stronger than the anticlockwise current.

Resultant current through Q (ie sum of current from both loops) is clockwise.

Answer : ACD.

Follow?

However, I had to assume that the areas of the two loops are the same (we are not told). Maybe you are expected to measure them? Also the perimeters?

@sammygerbil its an objective question

@Abcd Meaning that you wouldn't be required to make measurements from a diagram?

No

Every question is supposed to be done in 3 minutes

Using all possible tricks/ shortcuts/ smartness

@sammygerbil How can we say clockwise at P > anticlockwise current at P ?

@Abcd These are pre-University questions? Some (like this) are quite tricky. We don't get such tricky questions in UK at this pre-University level. Maybe in 1st or 2nd year of University, but not before.

@sammygerbil Yes, they are high school questions in India.

@Abcd There is only one current at P and it is clockwise. P is located on only one current loop, whereas Q is located on 2, and R is located at the overlap of one loop with itself.

@sammygerbil Got it thanks. Very nice analysis and question.

@sammygerbil SO in such questions we are supposed to check all possible loops?

@sammygerbil Cant we just directly that it should be clockwise at both P and Q because both those loops want to obey Lenz's law? Our result to this question matches that but will it always work?

@Abcd Yes all possible loops through the points in question. In 2D, a point on a wire cannot be located at the intersection of more than 2 loops. In 3D it can be located on an infinite number of loops!

@Abcd That's a good question. It is possible (I think) that the areas and perimeters could work out such that the resultant current at Q is anticlockwise instead of clockwise. That is, the current in the outer loop is larger instead of smaller than that in the inner loop.

@sammygerbil OK, more questions

@IceInkberry Me also. The question ought to say, instead of leaving you to guess.

@IceInkberry same here

@IceInkberry where did you get them

@IceInkberry ok.

@sammygerbil please C^^

@IceInkberry Is that a book "Pathfinder"?

@IceInkberry "JEE advanced physics" is a book? Which author? Publication?

@IceInkberry is it nice

@IceInkberry does it have theory

@Abcd This looks difficult. Any ideas? @IceInkberry

@IceInkberry lolll I thought thats 2 different books

@sammygerbil Induced electric field. No more thoughts. Thought A LOT about it but couldn't land on any definite conclusion.

@Abcd In space between the 2 solenoids there is an increasing B field due to outer solenoid only, and a constant E field due to inner solenoid only. (I don't quite see how this results in circular motion.)

@sammygerbil I didn't think about constant E field.

@Abcd If B field is changing at a uniform rate then induced E field is constant.

@sammygerbil Oh I know that. you should have written constant induced E field. I thought there's some other field I have missed out.

@sammygerbil I think constant E field is due to outer solenoid

@Abcd There is an E field around any closed loop through which B field changes. Between the solenoids a loop encloses B fields from both inner and outer solenoids. (My earlier comment was wrong.)

@IceInkberry it's an integer type question

@Abcd Oh pity Ice didn't tell us how to solve it!

It seems rather complicated. I think I shall have to come back to this question later.

Shall we look at something else?

@IceInkberry HOW CAN SOMEONE MAKE SUCH AN ACCURATE GUESS !?

@sammygerbil Many questions are piling up today :(

@Abcd :(

@sammygerbil She wasn't able to solve it. She just guessed. Maybe the number 8 was appearing around very neatly in her rough answer.

@Abcd So 8 is the official answer?

@sammygerbil yes

@Abcd I will work on it later this evening (also the other postponed question). Another question? Or shall I look at user's question?

@sammygerbil Okay, you can see user's questions. Ill be back in a bit.

@Abcd ok

Hi @sammygerbil

@user64829 Hi. I am available for your question now.

I didn't understand how C = Cv +R/(1-n) for a polytropic process

C= molar heat heat capacity

n=polytropic index

@user64829 I think there is a derivation on the main site. Have you seen it?

I searched a lot, couldn't find proper explanation anywhere

Let me look into this

@sammygerbil Question with attempt

@Abcd ok

Got it, thanks

Hadn't checked out SE for it.

One more question

@user64829 ok.

Why is electric field due to infinitely charged plane sheet σ/2ε for non-conducting sheet and σ/ε for sonducting sheet

Tried to google this also, couldn't find satisfactory answer

@user64829 The same formula applies to both. The difference is how $\sigma$ is defined.

Could you please explain a bit more?

If there is a charge density $\sigma$ on a surface then the electric field due to this charge distribution is $\sigma/2\epsilon_0$.

If you have a conducting plate with finite thickness then you have charge distributions with surface density $\sigma$ on both sides. Both charge distributions have an electric field $\sigma/2\epsilon_0$. So the total field outside the plate is the sum (which is $\sigma/\epsilon_0$) and the total field inside the plate is zero.

For non-conducting sheet there is charge only on one face. If you place charge on one face it does not divide between both faces.

Hmm...so charge is distributed on both the faces of a conducting sheet, area is doubled. Shouldn't that reduce the value of $σ$ to half and we get the same result as that for non-conducting sheet?

@user64829 If you place a charge Q on a conducting sheet, and the same charge on a non-conducting sheet of the same dimensions, the electric field will be the same.

The only difference is that the charge stays on one side of the non-conducting sheet, so charge density on that face is Q/A. Whereas the charge divides between the 2 faces of the conducting sheet, so the charge density on each face is Q/2A.

(Actually you have the same average charge density in both cases. For the conductor there is charge Q/2 on each of 2 faces, for the non-conductor there is charge Q on one face and no charge on the other face.)

@Abcd Sorry you were interrupted. Do you have another question ready?

Oh, ok got it. Thank you.

@sammygerbil I was able to do that question finally. Now I am trying another difficult question. Not getting. Trying a bit more and if I dont get ill ask.

@Abcd ok. (I will take a break and get some food.)

@IceInkberry Is NCERT exemplar good for Mains-ONLY chapters like diffraction EM waves? What did your teacher suggest for Mains-ONLY chapters

@AvnishKabaj @Jasmine Please suggest the book to be used for Mains-ONLY chapters like diffraction.

@sammygerbil please let me know when you are free.

@Abcd Just finished eating.

Ready now.

@sammygerbil please C the question w/ attempt^

My answer is ABCD. (the values of $\lambda$ that result in integral m)

@sammygerbil Please reply

@Abcd thinking ...

I think the mistake is that you have measured angle $\theta$ from the lower of the 2 image sources, instead of from their midpoint.

@sammygerbil why is measuring it from lower part wrong !?

@Abcd You get a different value of $\theta$ if you measure the height of O above the lower image source than if you measure from the midpoint. If O is opposite the midpoint then the path difference from the 2 sources is zero, so we must have $\theta=0$ also.

But in this case O is not opposite the midpoint.

@sammygerbil ??????

@sammygerbil I have absolutely no clue what you are saying in this message $:($

hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html

@sammygerbil ???

See diagram in link above. The angle $\theta$ is measured from the midpoint of the two slits (sources). The angles measured from each of the 2 slits are slightly different.

@sammygerbil my question is why cant we measure it from lower slit??

@Abcd Because that angle is different from the one shown in the diagram. For use in the formula, $\theta$ is measured from the midpoint of the 2 slits/sources.

In the linked diagram there are 3 lines going to the same point on the screen. One from each slit and one from their midpoint. Each of these lines makes a different angle with the horizontal. Only one of these angles will give the correct answer in the formula : the middle one, from the midpoint.

So I think only A and B are correct.

@Abcd If the distance of O above the midpoint of the 2 sources was large (>>d) then the angle $\theta$ would be relatively large and approximately the same whether you measured from the midpoint or from either slit. However in this case the distance of O above the midpoint is comparable with d, so the angle $\theta$ is very small because D >> d. When $theta$ is very small it makes a significant difference whether you measure it from either source/slit or from the midpoint.

@sammygerbil is path length difference 3d sin theta or not?

@Abcd Yes.

@sammygerbil So we HAVE taken theta from below slit. So we should measure it also from lower slit. I don't understand the problem with that Seriously.

@Abcd If you are unsure, you can calculate the path length from each image source without using the formula and without taking a value for $\theta$. Then apply the approximation $d \ll D$. You will get the same path difference as when you use the formula with $\theta$ measured from the midpoint.

@sammygerbil Are you trying to say: $D>>d$ is valid ONLY from midpoint?

The same problem occurs with electric dipoles. The electric field is sometimes given in polar co-ordinates. You always measure the polar angle $\theta$ from the midpoint of the dipole.

@sammygerbil please reply to next message

@Abcd No. I am saying that when the distance $y$ on the screen is small and of the same order of magnitude as the separation of the slits then it matters which slit you measure from. Measuring from the midpoint gives the least error compared with the exact calculation which I outlined above.

@sammygerbil still not getting.

In this case measuring from the lower source gives $\theta \approx 2d/D$. Measuring from the midpoint gives $\approx d/2D$. And measuring from the upper source gives $\approx d/D$ below the horizontal.

@Abcd All I can suggest is that you draw the other angles on your diagram. Ask yourself, Are they the same or are they different? If different, which one is correct? Why the lower source? Why not the upper source?

@sammygerbil I think you mean $\sin \theta \approx \tan \theta$ is not valid for the lower and upper slit measurements. Do you mean this?

@Abcd No. $\sin\theta \approx \tan\theta$ is valid for all 3 slits because we are told that $D \gg d$. These are 3 different values of $\theta$. All 3 are very small. So small that it makes a significant difference whether we measure from the upper or lower slit or the midpoint.

@sammygerbil why has my book taken all as theta then?

@Abcd Usually the distances $y$ measured on the screen are much larger than the separation of the slits. EG $y$ is typically several mm or more, compared with $d$ which is around $1\mu m$ - ie same order of magnitude as wavelength of light.

But in this case the distance $y$ between O (which is P in the book diagram) and a point level with the midpoint of the slits (where the path difference is zero) is comparable with slit separation $d$.

In the diagram on the right the 2 parallel lines go (in theory) to the same point P. But parallel lines never meet. Their separation on the screen will be $d$ same as the separation of the slits.

That is ok if $y \gg d$. Then the error of assuming the rays are parallel is small. However in your case $y \approx d$. If you assume the rays are parallel (in which case it does not matter which source you measure from) then the error $d$ on the screen is large compared with $y$.

@Abcd Neither do I!

@sammygerbil can the rays be assumed to be parallel if theta is chosen from midpoint?

I have the same problem occasionally. I cannot see what is obvious to somebody else.

@Abcd Yes that is approximately true. If you assume the rays are parallel and measure $\theta$ from the midpoint then $d\sin\theta$ is a good approximation to the path difference between the actual rays which are not exactly parallel.

@sammygerbil Okay, quite satisified now.

@sammygerbil More doubts.

@Abcd ok

@Abcd What is your doubt here?

@sammygerbil no idea how to approach it

@Abcd Did you choose C? If so why? If not, which option(s) do you think is correct? Even if you're not sure, have a guess - as I hope you would do in the exam!

@sammygerbil lol there's negative marking for wrong answers.

Sometimes it is: +4, -2

somtimes +5, -3

Sometimes, +4, -1

So guessing is risky$^\infty$

@Abcd Oh well in that case ignore my advice in the exam. But at least have a guess now. Sometimes I do cryptic crosswords. Often I guess the answer first then check the clues to see if it is correct.

@sammygerbil Just replying to your question in a minute

@sammygerbil See I cant recall why I chose option C (did it 3 days back)

@sammygerbil FIrst of all energy has to be conserved

@Abcd good

So if >13.6 eV, electron will be released from atom

I think the answer should be b.

Because:

electron rises to higher energy level then returns

so releases same amount of energy in radiation.

therfore H+ neutron system should have 0 energy

= inelastic collision.

(sorry mistake)

@sammygerbil My final answer is B for aformentioned reason.

But answer given is A $:($

@Abcd I would go for B also.

You are more likely to be correct if you say X is possible than if you say X will definitely happen!

@sammygerbil Now what ...

@sammygerbil I even think C is correct.

@Abcd Well I agree with your reasoning that the collision can be inelastic (emission of photon) if the neutron energy is less than 13.6eV.

@sammygerbil but answer...

@Abcd Your reasoning also shows that B is correct.

@Abcd C is attractive but I'm not sure it is true. I don't see why ionization is necessary.

@sammygerbil when will there not be an ineleastic collision for KE of neutron <13eV

@Abcd Removing the double negative, you are asking when will there be an elastic collision for KE of neutron < 13 eV?

@sammygerbil why is intermediate not possible

that is $0<e<1$

@Abcd That reminds me that there are worldwide differences in terminology.

@sammygerbil please answer the question then...

Elastic means (everywhere) that $e=1$. But inelastic means $e\ne 1$ in some places (UK) and $e=0$ in others (eg USA).

@sammygerbil okay sorry. I should have said perfectly inelastic

@Abcd I am not sure what answer is intended. Like you I would go for B. I think we can eliminate D.

@sammygerbil you misinterpreted my message

@sammygerbil Context + message^

@Abcd Sorry I am confused about what you are asking.

@sammygerbil Okay my question is: When will there be a partially inelastic collision for neutron striking with KE less than 13.6 eV

Perfectly inelastic here means that the proton and neutron merge to become a deuteron with the emission of gamma radiation.

First excited state of hydrogen is -3.4eV so the neutron needs to have a KE of at least 10.2eV to excite an electron and cause emission of a photon.

@sammygerbil by partially inelastic I mean that both move in opposite directions after striking each other just like two balls when they strike each can move in diff directions or lose contact and move in same drcn.

@Abcd I think my last comment answers your last question.

no

@Abcd If neutron KE is less than 10.2eV the collision must be elastic - ie it is impossible to excite the atom. If neutron KE is more than 10.2eV if may be inelastic - ie it is possible to excite the atom. But it will not necessarily excite the atom.

I am not sure that we can say any more than this without a greater knowledge of atomic physics.

@sammygerbil why is it not possible for nucleus to get velocity and neutron's velocity to get reduced if neutron's KE is less than 13.6 or 10.2

I am still in favour of option B. I don't see why it is A.

@Abcd I think this is possible. So I cannot explain why it is not possible.

Neutron and proton have approximately the same mass. If this was a rigid body elastic collision (like billiard balls) then the neutron would stop and the proton would move with the velocity which the neutron had initially. The electrons have very little mass compared with the proton and neutron.

Maybe @DavidZ can answer this question?

@sammygerbil He doesn't take part in the activities here...

@sammygerbil One more doubt.

@Abcd ok.

@Abcd eh I pop in from time to time

@DavidZ By "activities" I meant "answering questions"

True, for the most part. I think I have answered a question or two here though.

sammy gerbil and john rennie have answered a thousand questions till now XD :D

@sammygerbil I dont understand why he hasn't considered the rest of the surface in $\int B.dl$ at 6:20

He has just considered the circular part.

But the entire surface encloses current. Why not consider it too?

@Abcd Ouch! That voice is grating!

@Abcd The surface must be bounded by a closed curve. It cannot be a closed surface.

@sammygerbil I'll get back to you and @Abcd if I think of it, though for the moment I have a meeting to go to

@sammygerbil I know, can't help it because his videos are good so I have to watch them ...

@sammygerbil ????

@sammygerbil Are you saying that he is wrong

@Abcd You don't recognise sarcasm, but you know how to use it?!

@sammygerbil no man. I was serious. His videos (not sure about this one) are quite helpful for physics theory

So it's like I'll have to bear his voice for the sake of physics' theory

@Abcd No I don't think the lecturer is wrong. (Except his voice, which should be banned!) Given an Amperian Loop, you can choose any surface you like to find the integral of current across that surface. It will be the same for all such surfaces because charge (and therefore current) is conserved.

@Abcd I agree. If he didn't have that voice he would be an excellent lecturer. And at least I can understand him. Some Indian videos I cannot tell if they are talking Hindi or English with a strong accent.

@Abcd You don't have to use a flat surface in the same plane as the loop. (The loop doesn't have to be planar anyway.)

Have I answered your doubt?

No

Given an Amperian Loop, you can choose any surface you like to find the integral of current across that surface. It will be the same for all such surfaces because charge (and therefore current) is conserved

Didn't get this

First of all I don't understand how he can use a SURFACE for amperean loop . Surfaces are for Gauss Law

@Abcd Returning to the last question, I have an idea : Suppose the neutron has momentum $p$ in the laboratory frame of reference. Then its KE is $K=p^2/2m$ where $m$ is its mass. In the centre-of-mass frame its momentum is approximately $p/2$ because it has approx the same mass as the hydrogen atom. Same for the H atom. So the initial KE is now $2\times \frac{(p/2)^2}{2m}=p^2/4m=K/2$.

So if $K<13.6eV$ in the lab frame then the energy available to excite the atom is only half that, ie $6.8eV$. This is too small to reach the 1st excited state, which requires $10.2eV$.

So a neutron energy less than $13.6eV$ guarantees that the collision will be elastic. Option A.

@sammygerbil momentum is zero in COM frame IIRC

@Abcd Yes, total momentum is zero. Neutron has momentum $p/2$ to right and H atom has momentum $p/2$ to left.

@sammygerbil i dont see how the above analysis guarantees elastic collision.

@Abcd In COM frame the collision can be totally inelastic. This happens when both particles come to rest. Initial KE is then converted in total to internal energy. In this frame initial KE is $K/2=6.8eV$. This is too small to excite H atom above ground state energy level. The next lowest level is $10.2eV$ above it. The gap is too high.

@sammygerbil why was previous analysis wrong

@sammygerbil i think excitation energy is frame dependent?

@sammygerbil I have to go for 3 hours to attempt questions. Will you be there after 3 hours?

@Abcd The problem before is that we did not ensure momentum was conserved in the collision. In lab frame not all of the incident KE can be converted to internal energy, because this would not conserve momentum. However, in COM frame 100% of incident energy can be converted to internal energy, because total momentum is already zero.

@Abcd Probably. :)

@sammygerbil wow that's brilliant. But now I have found something in my notebook which is irritating me. Basically we did a similar question in class when my teacher was teaching, here is the question and the answer:

@Abcd ok.

@sammygerbil Answer to the above question is 0. Why?

@sammygerbil Ill be back in 3 hours. I'll see your response when I come back.

But I am just going to finish with your qn about Ampere's Law. A common definition is "electric current passing through the loop". But this is ambiguous. We cannot know if current has "passed through a loop" until it has crossed a surface. Which surface? As explained in wikipedia article under Ambiguities and Sign Conventions #2, it doesn't matter which surface you choose. See en.wikipedia.org/wiki/…

@Abcd Your teacher's question is slightly different. We have just learnt that the collision must be elastic - because KE in the COM frame is less than $10.2eV$. We know that the neutron and H atom have approx the same mass. So they behave like billiard balls of equal mass which collide elastically. In the lab frame the neutron stops dead (zero KE) while the H atom moves off with the same velocity (and KE) which the neutron had initially.

@Abcd Ampere's Law : Walter Lewin also addresses this ambiguity at the start of the following lecture, especially from 3:35 onwards. youtube.com/watch?v=1dCChkEGi_c