A BEAUTIFUL IRODOV PHYSICS CHALLENGE MODIFIED … CHECK YOUR UNDERSTANDING OF AMPERE'S LAW

[@physicsisdope4892]

Done #9 April 9 Revised whole *Magnetism* with this lecture.

[@riitk69]

wah sir dil jeetliye ....

[@rakshitrajiitkgp9662]

Great video sir....cleared a lot of concepts

[@exploreinfinityiitkanpur7289]

This question revise my all ampere law Mind blowing ❤️❤️

[@swathik8422]

sir @11:11 you took magnetic field at every dl to be contsant,but actually in the modified version only b spread is symmetrical,b due to wire is not,so magnetic field due to straight wire at 37 degree is not same for every point,so how can we take magnetic field at all points same?? (IN ACTUAL QUESTION WE CAN TAKE B NET OUTSIDE BECAUSE ITS SYMMETRICAL)

[@PHYSICSSIRJEE]

I understand the problem you are pointing out. After a lot of thought, I think, it should have been mentioned in the question that in the modified problem the current spreads in the plane in the same way as the irodov problem. That is , the incoming wire current and the spreading plane current are not interacting with each other. Otherwise, symmetry fails on the plan and my solution at the end cannot be correct. With that added condition of non-interaction ( in a hypothetical situation ) the solution can be considered correct . Thank you for pointing out 🙂🙏, keep visiting the channel videos and provide your feedback 👍🏼. It would be helpful

[@jeeboi347]

@@PHYSICSSIRJEE Even if they are not interacting, how can we say that the magnetic field due to the angled wire will be the same as the magnetic field due to perpendicular wire? Thanks

[@PHYSICSSIRJEE]

@@jeeboi347 magnetic fields due to two wires (slant and perpendicular) are not same. We didn't say that in the solution

[@perjurer7011]

Thanks Sir ❤️..... awesome explanation🙏

[@ilovemrbeast2380]

Sir in this case as well, if Bspread+Bline is 0(from amperes law), then for Bspread, Y did we take B due to straight line from that irodovs qn and not B due to straight line which is inclined at an angle in the given qn. I am a bit confused in this part of the question, thank you sir

[@linxuser897]

That is because the question was to find the field produced only by the current on the plane, not the net field (read carefully at 2:08). We could actually find the net field in the modified problem by adding the two fields. Why we're taking the wire in the first problem is because, it is not possible to solve the second problem using Ampere's law, so we're modifying it back to simpler case where the field produced by the field would be the same.

[@jeeboi347]

@@linxuser897 I don't quite get you. I did read the questions carefully and I do comprehend the fact that we are required to find only the B produced by the sheet. But the B of the sheet is (magnitude wise) equal to the field produced by the straight wire, so here for the straight wire we are supposed to find the B produced by the one at an angle right?

[@neerujat1978]

Thank you very much sir

[@Chole67767bhature]

Thank u sir 🙏🙏👌👌

[@simisimi9362]

Thank you sir 😇🙌🙌

[@AbjSir]

Just beautiful ❤️😍

[@rithikradhan367]

Sir I have a doubt regarding how the current disperses, (7:30) You claimed that it is radially symmetric in both cases We know that the net field at required point (P say,) is 0. => B(wire)=B(plane) (amplitude) By assumption that in both cases current in the plane is same, B(plane)1=B(plane)2. (At P) =>B(wire1)=B(wire2). (At P) which is false I may be wrong sir, Thanks for clarifying. (Big fan of yourself)

[@jonathan3372]

B(plane1) = B(plane2) does not imply B(wire1) = B(wire2), since the relation B(plane) = -B(wire) is derived from Ampere's law by exploiting the symmetry in the second case. In the original version, the net magnetic field at P is zero, whereas in the modified version, the magnetic field at P by superposition is non-zero.

[@apratimtewari4288]

Sir this question was recently asked in the prescore test for Allen SRG batches

[@ss-wx6pl]

the full syllabus pre score test

[@as-tb9jn]

Sir I have a doubt in 11:07. Can't we compare the value B due to the spread on the plane with a large sheet?( Not In magnitude atleast in the direction wise) if this is true then we can say that B due to spread will be along -k direction. Then the dot product of the B due to spread with dl would be zero. Is the analogy/comparison incorrect?(mostly wrong ig). If yes why is it wrong sir?

[@user-nw9xc2do1x]

Sir at 11:16 why is it that the magnetic field due to spread taken to be tangential to the loop? How can we say that the radial component of the magnetic field due to spread is 0?

[@PHYSICSSIRJEE]

Gauss of magnetic field forbids the presence of radial component

[@vatsal941]

@@PHYSICSSIRJEE sir can you please elaborate a bit.. I am also having the same doubt

[@mr.rachetphilanthrophist601]

Because divergence of Magnetic field is zero. Unlike electric field where Divergence Of electric field is non zero and is equal to Volume charge density divided by epsilon.

[@jitendrapandey1085]

Sir can you find the moment of inertia of ring about a tilted axis ( bent at angle theta) .... axis passing through center...

[@PHYSICSSIRJEE]

Yes I can 👍😊. You may solve by simple Integration by taking each element dm and apply basic formula for moment of inertia

[@anuragtiwari2854]

@@PHYSICSSIRJEE by vector method!

[@puspalpaul185]

@Aryan Pandey Sir has already taken it in his one video ..... but without integration 😁😁

[@anuragtiwari2854]

@@puspalpaul185 begani shaadi m abdullah deewana....

[@pranavmaniyar4592]

@@anuragtiwari2854 ok

[@endlesseducation2845]

#8 For my revision

[@AbjSir]

Sir I have doubt you remember that formula (u/4π)i/d(sintheta2-sintheta1) when we are writing magnetic field at a point due to a non-infinite or half infinite wire shifted up a distance d (formula becomes (u/4π)i(1-sintheta1) .Sir in this case when we apply ampere law at that point , There's actually no current enclosed in the loop but B.dl Integration is not zero. How's that possible?

[@k.nishanthkamath8665]

sir...at 11:20 you took the magnetic field out of the integral by arguing that there is symmetry....but how do we know that it is tangential to the amperian loop....i mean there might be some angle theta between the dl of the loop and magnetic field....

[@PHYSICSSIRJEE]

I understand the problem you are pointing out. After a lot of thought, I think, it should have been mentioned in the question that in the modified problem the current spreads in the plane in the same way as the irodov problem. That is , the incoming wire current and the spreading plane current are not interacting with each other. Otherwise, symmetry fails on the plan and my solution at the end cannot be correct. With that added condition of non-interaction ( in a hypothetical situation ) the solution can be considered correct . Thank you for pointing out 🙂🙏, keep visiting the channel videos and provide your feedback 👍🏼. It would be helpful

[@shivx3295]

what about point aboove x-z plain

[@chakradhar5884]

Nice sir

[@modernphysics7474]

Awesome explaination sir 🔥🔥... Sir there is one doubt that how the magnetic field of plane will cancel out the Magnetic field due to wire ,you have proved it using Ampere's Law but how to understand it physically?

[@PHYSICSSIRJEE]

Ampere's law as explained in the video for situation-1 from IRODOV

[@modernphysics7474]

@@PHYSICSSIRJEE ok sir 👍

[@riitk69]

sir , what will be the magnetic field due to the radially spreading current in the plane of containing it only due to this spreading curent

[@sohamadak6811]

Sir in Ampere's law, does the enlosed current have to be an infinite wire or a finite length wire works as well??

[@purushottamjha.]

For finite wires it works magnetic field due to leading wires should also be taken

[@ayush2559]

Sir ek spring hai jo ki fix nai hai ar usse h height upper block of mass m Is dropped ab wo block spring pe gire ga ar usse compress kre ga phir jab spring wapas apne natural length me aye ga uske bade spring ground se Kitna height tak bounce kre ga . Please explain sir

[@PHYSICSSIRJEE]

H=0 for massless spring. Please post similar questions in discord server and not here below this video

[@vedantvardhan2590]

When B is not symmetric in the modified version is it allowed to take it out of the integral

[@PHYSICSSIRJEE]

I understand the problem you are pointing out. After a lot of thought, I think, it should have been mentioned in the question that in the modified problem the current spreads in the plane in the same way as the irodov problem. That is , the incoming wire current and the spreading plane current are not interacting with each other. Otherwise, symmetry fails on the plan and my solution at the end cannot be correct. With that added condition of non-interaction ( in a hypothetical situation ) the solution can be considered correct . Thank you for pointing out 🙂🙏, keep visiting the channel videos and provide your feedback 👍🏼. It would be helpful

[@vibhashukla5729]

Sir how do u know that angle is not 90 degree for your result Bst =-Bsp?please reply 🙏🙏🙏

[@jeeboi347]

Same doubt

[@amirthya]

I guess the line integral from Ampere’s law only implies the axial net magnetic field is zero. However to comment on the radial component we would have to rely on symmetry arguments. Also I am wondering about induction effects in the modified version, since the radial symmetry is broken - like for instance due to the magnetic field of the wire the currents would no longer be radial straight lines - whether the original arguments are still valid.

[@PHYSICSSIRJEE]

In the original Question , radial component of the field is argued to be zero from Gauss law of magnetism. In the modified Question , magnetic field only due to spread current is asked and not due to the straight segment

[@amirthya]

PHYSICS SIR JEE I mean the spread current patterns are different due to induction effects.

[@PHYSICSSIRJEE]

@@amirthya yeah, this is a direct steady state current circuit. I should have mentioned to neglect the self inductance of the plane on which it is spreading and it's mutual inductance with surroundings

[@amirthya]

Sir it would make sense in the no inductance case but I am thinking on the following lines. Due to the current in wire the field on z axis is ~ i/z. And due to this field, considering a small sector around z axis, there is a force along negative x axis which causes a circular motion locally with radius of curvature m_e v_drift / (B e) ~ z / i. Thus for sufficiently small z we would have high radius of curvature and thus it seems would be impossible to achieve a no inductance as the plane being a free conductor.

[@PHYSICSSIRJEE]

@@amirthya valid point 💓🙂

[@Prashant-fh6px]

Nice question on amperes law