5:29 AM
@Jasmine: Related question from the main site:
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The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter on "Light Waves", for the number of fringes that will shift in a Young's double slit experiment (YDSE) when one of slits is covered by a transparent sheet of thickness $t$ made of a material of refractive index $\mu... 2 hours later… 7:14 AM @JohnRennie Sir May I ask my doubt? @Knight hi, yes, go ahead 7:33 AM @JohnRennie I don’t know what happened to my network, internet stopped working in my device. I’m having trouble in understanding a symmetry argument. OK ... ? Given: We have cylindrically symmetrical volume current density$\mathbf J$. Conclusion: Our magnetic field will also be cylindrically symmetrical. Seems reasonable to me :-) Arguments: By Maxwells’ Law $$\nabla \times \mathbf B = \mu_0 \mathbf J$$ so if B is not cylindrically symmetrical then J cannot also be cylindrically symmetrical. What’s is this argument doing? Suppose we use cylindrical coordinates, so we have distance from the axis$r$, distance along the axis$z$, and the angle around the axis$\phi$. @Knight Are you comfortable with what is meant by cylindrical coordinates? 7:39 AM Yeah, but can we do it without mathematics. Because we can certainly check that by Biot-Savart Law I'm not going to use any maths in my argument. Okay, I understand cylindrical cooridantes Cylindrical symmetry means$\mathbf J$is a function only of$r$and$z$. It is not dependent on the angle$\phi$. Yes? I think that means it’s only a function of$r$, because a cylinder have same radius all along the z-axis. Am I right? Cylindrical symmetry only means independent of the equatorial angle. The system can still vary along the axis. 7:43 AM But the equation of a cylinder is $$x^2 + y^2 = c^2$$ Which is just $$r =c$$ e.g. a rotating sphere is cylindrically symmetric, but its radius still varies along the axis of rotation. So no dependence on variable z Cylindrically symmetric doesn't mean a cylinder. Okay, now I have got you. That sphere example has said it all OK. Since$\nabla \times \mathbf B = \mu_0 \mathbf J$that means$\nabla \times \mathbf B$is also dependent only on$r$and$z$i.e.$\nabla \times \mathbf B$is also cylindrically symmetric. 7:45 AM Okay And if curl of B is independent of angle that means the field lines have to be circles centred on the axis. Can you please explain a little more? I mean if the derivative of$f$is cyldircally symmetrical how does that say that$f$gonna be circular ? Roughly speaking the curl of a field is the rotation of that field. And if the rotation is independent of angle the rotation must be in a circle. rotation is independent of angle does that mean the magnitude of rotation is same in all directions and we know that the field is rotating therefore it must be a circle. Ha? Non-zero curl implies that the B field is rotating And it’s rotational magnitude is same in all directions (of course I mean in a plane) there fore our B field is a circle, Have I got your right? Yes, though this isn't a rotation in time i.e. curl isn't proportional to angular momentum. I'm trying to give you an intuitive feel for why the field must by cylindrically symmetric. To go any further I'll need to start writing some equations. 7:55 AM Curl gives us how much the field is twisted at a point, ha? If you're travelling long a field line at some velocity then the curl tells you how fast you are rotating given your speed. I need to work now. I'll be around an hour. Okay thank you sir 2 hours later… 9:55 AM @JohnRennie Sir are you here? Should I ask it here on in Bar? @Knight hi Hello First of all were you joking there or do you really suspect me that I cross check you? 😭 I'm just answering a question about lenses, I'll be a few minutes. Post the question here and I'll have a look as soon as I'm free. A uniform surface current$\mathbf K$is flowing in the$xy$plane and the direction of current is in positive$x$axis. So, by Maxwell’s Law we can write $$\left ( \nabla \times \mathbf B \right)_x = \mu_0K$$ $$\left ( \nabla \times \mathbf B \right)_y =0$$ $$\left ( \nabla \times \mathbf B \right)_z = 0$$ So, it means that moving along$y$or$z$axis wouldn’t change the rotation of B field (if x position is fixed). But how can we conclude and what can we conclude about the symmetry of B field ? 10:14 AM I'm not sure how to calculate the curl of a sheet of current ... Curl of current? Curl of magnetic field equal the surface current @Knight oops, yes. 😭 You know everything, you’re just punishing me by saying “I’m not sure how to do that” 😭 If you have an infinitely thin sheet then the curl is zero everywhere except in the sheet i.e. in the xy plane. Yes, I agree 10:19 AM But I'm not sure how to calculate the curl given an infinitely thin sheet. On the sheet our Magnetic field will be zero? The trouble is that for an infinite sheet current B is independent of distance from the sheet, but it has different directions either side of the sheet. So it changes discontinuously at the sheet. I'm not sure how you handle that. @Knight: Is finding magnetic field due the infinite plane sheet relevant to your question? @GuruVishnu No @JohnRennie What can we infer about the symmetry of B field? The way we inferred previously We can say that B will depend only on$z$. The current is a function only of$z$i.e. it is equal to something like$I(z) = I_0(\delta(z), 0, 0)$10:24 AM Yes I agree And that means the field can depend only on$z$. 😁 @JohnRennie Can you please explain once more, I mean how if curl is a function of$z$only then the field will also be a function of$z$I don't think using the curl helps here. Since K is a function of z and we know $$\nabla \times B = \mu_0 K$$ therefore curl is only the function of z, isn’t it? Yes, but now the curl is a weird delta function. 10:30 AM Is it like this $$\left ( \nabla \times \mathbf B\right)_x = K \delta (z)$$ I don't know to be honest. My vector calculus is too rusty for me to have any confidence in the matter. But it should be obvious that if the current doesn't depend on$x$or$y$the field can't either. Sir But Analysis by Biot Savart Law gives us that B field will point in negative y direction above the plane. Is this information deducable from symmetry alone? No That comes from the right hand rule I meant which direction it’s gonna point, it could point in z direction or it could point in y direction or both. But why it pointed in y direction only? The symmetry only tells us that the field cannot be a function of$x$or$y$i.e. it must have the form: $$\mathbf B = (f_x(z), f_y(z), f_z(z))$$ But it doesn't tell us what the functions are. 10:38 AM Can we deduce something from the info that y and z component of curl is zero therefore the field doesn’t have a component of rotation in y and z direction? Well let's try. The$y\$ component of the curl is:
$$(\nabla \times B)_y = \frac{dB_x}{dz} - \frac{dB_z}{dx}$$

Sir I’m getting some network problems
Parsing of MathJax is taking too much time

This going to be hard to discuss without MathJax ...

3 hours later…
1:38 PM
@JohnRennie Sir finally I got the network back