CSIR NET (Physics)

Why India does not achieve academic excellence?

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While such an important question merits a much detailed and comprehensive approach, here is something that you might consider relevant especially if you are just on your morning coffee on a Sunday and you just want to ponder over it. In other words what I am going to write is perhaps a terse but lists one of the most important factors, and in my defense its just a blog.

Why India does not achieve academic excellence, I will just give example of one primary reason: arbitrary rules wrt qualifications, merit, skills, caste, age, time line.

For a glossary of certain terms which are nonetheless self evident (at-least for an Indian audience) you don’t need to scroll to the dead-end. Here they are; PG: Post Graduate, UG: Under Graduate, NET: National Eligibility Test, SET/SLET: State Eligibility Test, B.Ed.: Bachelor of Education, M.Phil.: Master in Philosophy.  

qualification: it has always been arbitrary. We are not sure whats enough or adequate to teach. Sometimes a person with PG isn’t considered good enough to even teach in high school or junior college, without additional degrees of any moiety of relevance. Sometimes lower qualifications than PG are fine for teaching such classes, with (mostly) irrelevant degrees or qualifications. eg someone with B.Sc.and B.Ed. is qualified to teach high-school but someone with PG, or PG+PhD may not be qualified to teach high-school or junior college, although its fine for him/her to teach senior college with a PhD. Also with PG qualification you can’t teach UG. But with PG, one entrance (NET/SLET) can decide your qualification to teach both UG and PG. I am talking about basic minimum qualifications. So you are not considered qualified without certain irrelevant factors. Its not like you are considered and given a time line to achieve any required additional qualifications.

To find the magnetic vector potential of a rotating, uniformly charged spherical shell. Vector potential, magnetic field intensity (mag induction) and current density are the most basic parameters of magnetostatics.

Magnetic vector potential of a rotating uniformly charged shell.

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Today we will solve the problem of finding magnetic vector potential of a rotating, uniformly charged spherical shell. We won’t discuss the general idea behind the vector potential (how it follows from Helmholtz theorem, and gauge freedom etc) and how its defined. That will be part of a conceptual lecture and will be available when the same would be created. The offline version is available, but the web version will call for a special priority to be assigned.

The problem is quite well defined. We just need to follow the straightforward method of implementing the basic definitions and carrying out the required steps. But we need to be mindful of the framework in which we need to accomplish these steps.

The framework I am talking about here is the coordinate system we need to set in order to solve the problem. Notice that the problem has been stated in the spherical coordinate system (which has been discussed couple of times in this website). But we need not worry about all the aspects of this coordinate system, we will only pick on those which are immediately applicable to our problem.

While this choice of the system where the polar axis (z-axis, wrt which the polar angle θ is measured in a r, θ, φ spherical coordinate system) coincides with the angular velocity vector ω is very natural, it isn’t the most convenient for carrying out the ensuing integral for the vector potential A.

A cylindrical wire carrying current I. Two situations arise. Case A: uniform surface current on the outside surface. Case B: a volume current density that depends upon radial distance from axis of wire. Whats the magnetic field of such a configuration?

Problem 5.13 Application of Ampere’s Law.

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Yesterday we saw an interesting application of the Ampere’s Law (– in magnetostatics and sometimes called Ampere’s circuital law also) for the infinite uniform surface current. Today we will see yet another display of the elegance and efficacy of this law in the following problem. This problem is inherited from Griffith’s text on Electrodynamics (3rd edition)

I have tried to be a bit more explanatory than the basic solution available (in instruction manual, if you have a copy). Thats the whole idea of this labor I have taken up. I also strongly suggest anyone who want to sharpen his saber to try the problem on his/her own effort before looking into the solution. That way one can prepare oneself for the pitfalls of one’s own understanding before taking up help and damaging the opportunity of developing of a better sense of solving such problems.

A steady current I flows down a cylindrical wire of radius a. What would be the magnetic field outside the wire and inside of it? We need to find the same in two different scenarios given.

Here are the two different scenarios.

A. Its a surface current density on the outside surface and its uniform across the surface.

B. Its a volume current density and its distributed in the volume of the  wire, but this time its not uniform. In-fact the volume current density J is directly proportional to s; the distance from the axis of the wire where we are referring the value of J.

Ampere’s law application: infinite surface current, diagram. An infinite sheet of current in xy plane with Amperian loop. We like to determine its magnetic field by Ampere's Law.

Example 5.7; Application of Ampere’s law.

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The following problem is an interesting application of Ampere’s law apart from usual applications found in honors syllabus (eg infinite straight conductor, Solenoid and Torroid). This is to be found the excellent book by Griffith on Electrodynamics. 

Find the magnetic field of an infinite uniform surface current K (vect) = K i-cap, flowing over the XY-plane.

Lets first visualize the problem. This will help us solve the problem. We chose a Cartesian coordinate system as shown. Our infinite surface current is a sheet that is concurrent with the XY-plane. We also show the Ampere loop which is a rectangle of length l parallel to the y-axis. This loop is half above the XY-plane and half below. 

CSIR NET 2018 December Physical Sciences: converting a RS flip-flop into a JK flip-flop. Photo Credit: mdashf.org

CSIR NET 2018 December Physical Science Solutions Part-B

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The CSIR NET 2018 held on December 16: Indian
Assistant Professor and PhD scholarship exam
solution, prepared by me. The answers and
detailed explanations are available for 18
out of 25 questions of Part-B. The detailed
explanations and answers to Part-A is also
available, see link below.

Please point out any inadvertent errors.

(this is entirely free stuff: help spread the word)

The article aims to make the best attempt at finding the answers for the recently concluded 2018 CSIR NET. Detailed explanatory answers for physical sciences section ( part-B ) is available ( for 18 out of 25 questions at the moment ). Also full explanation based solutions to part-A is available, check link above.

Q – 21. Consider the decay A → B + C of a relativistic
spin 1/2 particle A. Which of the following statements
is true in the rest frame of the particle A?

1. The spin of both B and C may be 1/2.
2. The sum of the masses of B and C is greater than
the mass of A.
3. The energy of B is uniquely determined by the
masses of the particles.
4. The spin of both B and C may be integral.

Explanation: obviously the second option is incorrect as it violates conservation of energy in relativistic kinematics, rest-masses of the product particles can not be more than that of the parent particle. Option 3 is explained with a diagram. The value of the energy of one of the daughter particle ( B ) is determined uniquely as evinced by the given formula for the same.

In a two body relativistic decay in the parent rest frame: EB = ( MA2 – MC2 + MB2 ) / 2 MA

Also the other options talk about the spin of the particles but we need not bother since option 3 is the correct option.