Poisson's Ratio

  • When two equal and opposite forces are applied to a body in a certain direction , the body extends along that direction and at the same time it cintracts along the perpandicular direction.

  • The fractional change in length of the body in the direction of the applied forces is longitudinal strain and fractional change in the perpandicular direction of the force applied is called lateral strain.

  • The ratio of lateral strain to the longitudinal strain is called poisson's ratio which is constant for material of that body.


  • So when a body is subjected to strain , say an elongation, it also suffers contraction in parpandicular direction.

  • Within elestic limits, lateral strain β is is proportional to longitudinal strain α.hence
    σ =β/α

  • since,
    longitudinal strain = α = Δl/l and
    lateral strain = β = ΔD/D
    hence poisson's ratio is σ =lΔD/DΔl

    Hamiltonian Formulism of mechanics (part 1)

    Hamiltonian is H=T+V
    or,



    Hamilton's Canonical Equations of motion:-











    • Co-ordinates cyclic in Lagrangian will also be cyclic in Hamiltonian.
    • Canonical transformations are characterized by the property that they leave the form of Hamilton's equations of motion invarient.
    • Lagrange's equation of motion are covarient w.r.t. point transformations (Qj=Qj(qj,t) and if we define Pj as,


             
         
               the Hamilton's canonical equation will also be covarient.
    • Consider the transformations
             Qj=Qj(p,q,t)
             Pj=Pj(p,q,t)
             where Qj and Pj are new set of co-ordinates.
    • For Qj and Pj to be canonical they should be able to be expressed in Hamiltonian form of equations of motion i.e.,








       where,  K=K(Q,P,t) and is substitute of Hamiltonian H of old set in new set of co-ordinates.
    • Qj and Pj to be canonical must also satisfy modified Hamilton's principle i.e.,






    • Using same principle for old set qj and pj
                                 (1)

       where F is any function of phase space co-ordinates with continous second derivative.
    • Term ∂F/∂t in 1 contributes to the variation of the action integral only at end points and will therefore vanish if F is a function of (q,p,t) or (Q,P,t) or any mixture of phase space co-ordinates since they have zero variation at end points.
    • F is useful for specifying the exact form of anonical transformations only when half of the variables (except time) are from the old set and half from the new set.
    • F acts as bridge between two sets of canonical variables and is known as generating function of transformations.
    In next post I'll discuss more about generating functions , lagrange and poisson brackett.

    Drift velocity and mobility (concepts)

    Objective Type questions : Test Yourself

    1. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
    (a) x2
    (b) ex
    (c) x
    (d) loge x

    2. Which of the following statements is false for a particle moving in a circle with a constant angular speed.
    (a) The velocity vector is tangent to the circle
    (b) the acceleration vector is tangent to the circle
    (c) the acceleration vector points to the centre of the circle
    (d) the velocity and acceleration vectors are perpandicular to each other

    3. If A × B = B × A, then angle between both the vectors is
    (a) π
    (b) π/3
    (c) π/2
    (d) π/4

    4. A ball is released from top of the tower of hight h m. It takes t sec. to reach the ground. What is the position of ball in T/3 sec.
    (a) h/9 m from the ground
    (b) 7h/9 m from the ground
    (c) 8h/9 m from the ground
    (d) 17h/18 m from the ground

    5. A machine gun fires a bullet of mass 40g with a velocity 1200 m/s. the man holding it can exert a maximum force 144 N on the gun. How many bullets can he fire per second at the most.
    (a) one
    (b) four
    (c) two
    (d) three

    Answer
    1. a
    2. b
    3. a
    4. c
    5. d

    Constrains and constrained motion

    • A constrained motion is a motion which can not proceed arbitrary in any manner.
    • Particle motion can be restricted to occur (1) along some specified path (2) on surface (plane or curved) arbitrarily oriented in space.
    • Imposing constraints on a mechanical system is done to simplify the mathematical description of the system.
    • Constraints expressed in the form of equation f(x1,y1,z1,......,xn,yn,zn :t)=0 are called holonomic constraints.
    • Constraints not expressed in this fashion are called non-holonomic constraints.
    • Scleronomic conatraints are independent of time.
    • Constraints containing time explicitely are called rehonomic.
    • Therefore a constraint is either
              "Scleronomic where constraints relations does not depend on time or rheonomic where constraints relations depends explicitly on time "
              and either
               "holonomic where constraints relations can be made independent of velocity or non-holonomic where these relations are irreducible functions of velocity"

    Constraints types of some physicsl systems are given below in the table


    How to simplify circuits with resistors

    1. In any given circuit first of all recognize the resistances connected in series then by summing the individual resistances draw a new, simplified circuit diagram.

    For series combination of resistances equivalent resistance is given by the equation
    Req= R1 + R2+R3
    The current in each resistor is the same when connected in parallel combination.

    2. Then recognize the resistances connected in parallel and find the equivalent resistances of parallel combinations by summing the reciprocals of the resistances and then taking the reciprocal of the result. Draw the new, simplified circuit diagram.
    (1/R)=(1/R1)+(1/R2)+(1/R2)

    Remember that for resistors connected in parallel combination ‘The potential difference across each resistor is the same’.

    3. Repeat the first two steps as required, until no further combinations can be made using resistances. If there is only a single battery in the circuit, this will usually result in a single equivalent resistor in series with the battery.

    4. Use Ohm’s Law, V= IR, to determine the current in the equivalent resistor. Then work backwards through the diagrams, applying the useful facts listed in step 1 or step 2 to find the currents in the other resistors. (In more complex circuits, Kirchhoff’s rules can be applied).

    CSIR NET physics: Ferromagnetism (in short) Part 1

    For reading ferromagnetism for higher level visit the link given below
    CSIR NET physics: Ferromagnetism (in short) Part 1: "A ferromagnetic material has a spontaneous magnetic moment- magnetic moment even in zero applied magnetic field this means that electron s..."
    CSIR NET physics : Ferromagnetism (in short) part 2: "Nature of Ferromagnetic carriers : Entire magnetization must be essentially associated with electron spin, and not at all with the orbital motion of electrons....."
    Hope you like it
    physics expert

    Ferromagnetism (in short) Part 2

    Nature of Ferromagnetic carriers
    • Entire magnetization must be essentially associated with electron spin, and not at all with the orbital motion of electrons.
    • Argon core (1s22s22p63s23p6) of Fe , Co and Ni can be left out of account as a source of ferromagnetism.
    • 4s electrons are responsible for electrical conductivity and crystal binding.
    • Thus 3d electrons with unpaired spins are responsible for magnetization of these metals.
    • An effective number of magnetic moment carriers per atom should be non integral, despite that each atom has an integral number of electrons Fe - 4 electrons  :   Co - 3 electrons : Ni - 2 electrons and each electron contributes a magnetic moment of 1μB due to spin alone.
    • Above argument applies to free atom but here they are bound into solids where atomic levels are bounded into bands.
    • Non integral values 2.22μB, 1.72μB and 0.54μB for Fe, Co, and Ni  resp. of magnetic moment carriers which each atom supplies can be explained as follows : Wide 4s band of these metals overlaps with narrow 3d band. As a consequence , there is , on an average a certain fraction of total number of 3d plus 4s electrons in each band; the relative occupation of two bands being determined by fermi level EF
              for example in case of nickle 10 sd electrons are distributed in them in such a way that on the average 9.40 electrons are in 3d band and 0.6 in 4s band. In accordance with hund rule 5 electrons set their spins parallel to the field and remaining 4.40 set their spins anti parallel to the field. Thus net parallel spin of Ni atom would be 5-4.40 = 0.60 and hence magnetic moment is 0.60μB agrees fairly with the experimental value

    Origin of exchange interaction

    • Explanation of large value of molecular field is based on non magnetic interaction that is exchange interaction.
    • Exchange interaction arises as a consequence of Pauli's Exclusion principle. Because of this principle we can not change the relative orientation of two spins without changing the spatial distribution of charge , clearly indicating that interaction exists between two atoms.
    • This interaction depends on relative orientation of electron spins not on the magnetic moments.
    • The energy of this interaction between atoms i , j bearing spins Si , Sj is of the form Eex=-2JeSi.Sj where Je is the exchange integral , its value is related to the overlap of the charge distribution of atoms i and j i.e., on their inter atomic separation.
    • Energy of parallel configuration is lower than that of anti parallel by amount 2Je this implies that former configuration is more stable favouring magnetization to occur.
    • Note that exchange interaction is positive only for iron group and negative for others.

    Ferromagnetic domains
    • The fundamental problem of ferromagnetism is to explain why the elementary moments of a ferromagnetic material can be aligned so much more easily then those of paramagnetic materials.
    • Weiss suggested that these were forces of interaction between elementary magnetic moments tending to make each one parallel to their neighbours.
    • Such forces would cause all moments to be aligned in the same direction at absolute zero of temperature and this ordering of moments would continue when temperature is raised, though with increasing deviation from perfect alignment, until a critical temperature is reached , above which the moments are arranged in random, as in a paramagnetic material.
    • Weiss theory can thus account for the fact that ferromagnetic materials may be spontaneously magnetized even in the absence of external magnetic field ; it does not explain why the majority of ferromagnetic are not actually found in this spontaneously magnetized state , but are much more likely to have zero magnetization.
    • This difficulty can be met by introducing the hypothesis of domain theory.
    • Here supposed that the forces of interaction only maintained the parallel alignment of elementary moments over fairly small regions .
    • Actual specimen are composed of small regions called domains , within each of which the local magnetization is saturated.
    • The direction of magnetization of different domains need not be parallel.
    • The increase of gross magnetic moment of a ferromagnetic specimen in an applied magnetic field takes place by two independent processes : (1) In weak applied fields the volume of domains favorably oriented w.r.t. the field increases at the expense of unfavorably oriented domains         (2) In strong applied fields the domain magnetization rotates towards the direction of the field.

    Ferromagnetism (in short) Part 1

    • A ferromagnetic material has a spontaneous magnetic moment- magnetic moment even in zero applied magnetic field this means that electron spins and magnetic moments are arranged in regular manner.
    • Consider a paramagnet with a concentration of N ions of spin S. Given an internal interaction tending to line up the magnetic moments parallel to each other , we shall have a ferromagnet.
    • This internal interaction is called exchange field.
    • Orienting effect of exchange field is opposed by thermal agitation.
    • At elevated temperatures the spin order is destroyed.
    • Exchange field can be treated as equivalent to BE (magnetic field) also assume that the exchange field BE is proportional to the magnetization M.
    • Magnetization M is defined as the magnetic moment per unit volume.
    • In mean field approximation each magnetic atom experiences a field proportional to the magnetization

      BE=λM (1)
      Where λ a is constant independent of temperature.
    • Each spin sees average magnetization of all the other spins and more precisely of the neighboring spins.
    • Curie Temperature (Tc) is the temperature above which spontaneous magnetization vanishes.

    • Tc separates disordered paramagnetic phase at temperature T > Tc from ordered ferromagnetic phase at temperature T < Tc.
    • If Ba is the external magnetic field then the effective field acting on atom or ion is

      B= Ba+ BE = Ba+ λM
    • If χp is paramagnetic susceptibility then
      M= χp( Ba+ BE)
      χp=C/T from curie law for paramagnetic materials
      this implies that MT=C(Ba+ λM)
    • Susceptibility has singularity at T=Cλ.
    • At this temperature and below there exists a spontaneous magnetization , because if χ is infinite, we can have a finite M for zero Ba.

    • Curie-Weiss law is
      χ=C/(T-Tc) or Tc=Cλ
    • This spontaneous magnetization decreases very slowly as the temperature is first raised above absolute zero and drops more steeply at higher temperatures until finally falls to zero at curie temperature.

    Radiology and Physics – Career Options and Other Important Aspects

    Radiology and Physics – Career Options and Other Important Aspects Most people are familiar with the terms radiology and physics – the former is the branch of medicine that deals with the research in and application of imaging technologies for diagnostic and treatment purposes like X-rays, CT, MRI, PET, ultrasound and nuclear medicine, while the latter is a branch of science that deals with the properties of matter and energy and the relationships between them. Putting the two together, we get radiologic physics, a specialized branch of physics that has three fields: Therapeutic radiological physics – deals with the physical aspects of the therapeutic applications of X-rays, gamma rays, electron beams, charged particle beams, neutrons, and radiations from sealed radionuclide sources, the use of the equipment that produces them, and the safety aspects of using radiation in diagnostics and therapy. Diagnostic radiological physics – concerns the diagnostic applications of X-rays, gamma rays from sealed sources, ultrasonic radiation and magnetic resonance, the use of the equipment that produces them, and the safety aspects in using them for diagnostic and therapeutic purposes. Medical nuclear physics - is related to the therapeutic and diagnostic applications of radionuclides (from unsealed sources), the equipment associated with their production and use, and the safety aspects of radiation. You can choose to obtain certification from the American Board of Radiology (ABR) in one or more of the above areas of study. As a radiologic physicist, you are qualified to act in an advisory capacity to physicians regarding the physical aspects of radiation therapy, diagnostic radiation and/or nuclear medicine. You will be working directly with oncologists and physicians in planning the treatment of patients who require radiation therapy and in delivering the therapy using the right equipment. Besides this, you are also in charge developing and directing quality control programs for equipment and procedures; this means that you ensure that the equipment that delivers the radiation works properly and has been correctly calibrated and that you ensure that complicated therapy routines are tailored to the needs of each patient. You also supervise the work of dosimetrist (they work as part of oncology teams and analyze data to come up with the right course of therapy to deliver the right dosage of radiation to the right location, minimizing harm to neighboring organs and tissue). Radiologic physicists must have at least a master’s degree if they wish to find good jobs and challenging positions with healthcare centers, hospitals and research facilities.
    This guest post is contributed by Rachel Davis, she writes on the topic Radiology programs. She welcomes your comments at her email id: racheldavis65@gmail.com.

    Test Yourself (Miscellaneous MCQ's)

    Qiestion 1
    If relatibe permeability of iron is 5500 then its magnetic susceptibility is
    (a) 5500 x 107
    (b) 5501
    (c) 5499
    (d) 5500 x 10-7

    Question 2
    An atom is paramagnetic if it has
    (a) an electric dipole moment
    (b) no magnetic moment
    (c) a magnetic moment
    (d) no electric dipole moment

    Question 3
    Magnetic moment of a diamegnetic atim is
    (a) 0
    (b) infinity
    (c) negative
    (d) positive

    Question 4
    Mark the correct statement(s)
    (a) When a particle moves under uniform circular motion its acceleration is constant
    (b) Under the influence of uniform circular motion , instantaneous acceleration is perpandicular to the tangential velocity
    (c) Under the influence of non-uniform circular motion , instantaneous acceleration is perpandicular to the tangential velocity
    (d) All the above

    Qiestion 5
    In a meter bridge apparatus , the bridge wire should have
    (a) high resistivity and low temperature coefficent
    (b) high resistivity and high temperature coefficent
    (c) low resistivity and high temperature coefficent
    (d) low resistivity and low temperature coefficent

    Question 6
    Mark out the correct statement
    (a) Image formed by convex mirror can be real
    (b) Image formed by convex mirror can be virtual
    (c) Image formed by convex mirror can be magnified
    (d) Image formed by convex mirror can be inverted

    Answer
    1. c
    2. c
    3. a
    4. b .... When a particle moves under the influence of uniform circular motion magnitude of acceleration is always constant but direction of acceleration keeps changing, it always remains towards the centre . So acceleration is always perpandicular to the the tangential velocity in this case.
    5. a
    6. a,b,c,d

    Prepare yourself for Competetive Exams

    This time is mid of the november and can prove deciding time for those preparing for competetive examinations like IITJEE/PMT/AIEEE etc and various other examinations. Ideally you should nearly complete all your syllabus by this month and if you have not completed it then try to cover the syllabus quickly by the end of this month. After you have completed all your syllabus then start revising it keeping yourself cool and calm.
    Try an explore all the topics of your syllabus for the competetive exams. After having looked up and revised complete syllabus you should start practicing simulated exame type situations to increase your confidence there are various websites online where you can take such tests.
    So all the best for the preparations of your competetive exams. Stop panicing if you have little time left and large portions to cover just fix your goals and start preparing.

    Comparison between Coulomb’s laws and Biot Savart laws

    1. Both the electric and magnetic field depends inversely on square of distance between the source and field point .Both of them are long range forces


    2. Charge element dq producing electric field is a scalar whereas the current element Idl is a vector quantity having direction same as that of flow of current

    3. According to coulomb’s law ,the magnitude of electric field at any point P depends only on the distance of the charge element from any point P .According to Biot savart law ,the direction of magnetic field is perpendicular to the current element as well as to the line joining the current element to the point P

    4. Both electric field and magnetic field are proportional to the source strength namely charge and current element respectively. This linearity makes it simple to find the field due to more complicated distribution of charge and current by superposing those due to elementary changes and current elements

    Electrostatics :- Very Short answer type questions

    Question 1: Define following terms

    (a) Dielectric constant of a medium
    (b) Electric dipole moment
    (c) Electric flux
    Answer 1:
    (a)It is the ratio of the capacitance (Cr) of a capacitor with dielectric between the plates to the capacitance (Cr) of the same capacitor with vacuum between the plates i.e. K=Cr/C0
    (b) It is the product of the magnitude of one of the point charges constitting the dipole ant the distance of separation between two point charges.
    (c) Electric flux through an area is the product of the magnitude of the area and the component of electric field vector normal to this area element.
    Electric flux = E.ds
    SI unit of flux is NC1m2

    Question 2: Electric field inside a dielectric decreases when it is placed in an external field. Give reason to support this statement.
    Answer 2: An electric field EP is induced inside the dielectric in a direction opposite to the direction of external electric field E0. Thus net field becomes
    E=E0-EP

    Question 3: Electric field lines can not be discontinuous. Give reason.
    Answer 3: Electric field lines can not be discontinous because if they are discontinous then it will indicate the absence of electric field at the break point.

    Question 4: Why do electric field lines can never cross each other?
    Answer 4: It is so because if they cross each other then at the point of intersection there will be two tangents which is not possible.

    Question 5: What is the net amount of charge on a charged capacitor?

    Answer 5: The net charge of a charged capacitor is zero because the charges on its two plates are equal in number and opposite in sign. Even when the capacitor is discharged net charge of the capacitor remains zero because then each plate has zero charge.

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