CSS Physics Paper-II 2024 Solved

Question 2: Poynting Theorem, Vector Potential and Loop Field

Q2(a): Poynting Theorem and Poynting Vector

Poynting theorem is the work-energy theorem of electrodynamics. It expresses conservation of electromagnetic energy. It shows that field energy can decrease because energy flows out of a region or because the field does work on charges.

The electromagnetic energy density in vacuum is:

u = (1/2)ε0E² + B²/(2μ0)

The Poynting vector is:

S = E×H

In vacuum, since H=B/μ0:

S = (1/μ0)E×B

The direction of S gives the direction of electromagnetic energy flow. Its SI unit is:

W/m²

Derivation of Poynting Theorem

Start with the Ampere-Maxwell law:

∇×B = μ0J + μ0ε0 ∂E/∂t

Take dot product with E:

E·(∇×B) = μ0E·J + μ0ε0 E·∂E/∂t

Faraday’s law is:

∇×E = -∂B/∂t

Take dot product with B:

B·(∇×E) = -B·∂B/∂t

Use the vector identity:

∇·(E×B) = B·(∇×E) - E·(∇×B)

Substitute the two Maxwell-equation results into the identity:

∇·(E×B) = -B·∂B/∂t - μ0E·J - μ0ε0E·∂E/∂t

Divide by μ0:

∇·[(1/μ0)E×B] = -(1/μ0)B·∂B/∂t - E·J - ε0E·∂E/∂t

Recognize:

S = (1/μ0)E×B

and:

∂/∂t[(1/2)ε0E² + B²/(2μ0)] = ε0E·∂E/∂t + (1/μ0)B·∂B/∂t

Therefore:

∇·S = -E·J - ∂u/∂t

Rearrange:

∂u/∂t + ∇·S = -J·E

Q2(b): Magnetic Vector Potential and Scalar Potential

Magnetic vector potential A is a vector field defined by:

B = ∇×A

This definition automatically satisfies Gauss law for magnetism:

∇·B = 0

because the divergence of a curl is always zero:

∇·(∇×A) = 0

Electric scalar potential V is related to the electrostatic field by:

E = -∇V

In time-varying electromagnetic fields, electric field can be expressed as:

E = -∇V - ∂A/∂t

Q2(c): Magnetic Field on the Axis of a Circular Loop

Consider a circular loop of radius R carrying steady current I. We need the magnetic field at a point on the axis of the loop at distance z from the center.

By symmetry, transverse components cancel and only the axial component remains. The Biot-Savart law gives:

dB = (μ0/4π)(I dl sinθ)/r²

For a point on the axis, every current element is at distance:

r = √(R²+z²)

The axial component of the field is:

dBz = dB cosα

where:

cosα = R/√(R²+z²)

After integrating around the loop, the magnetic field on the axis is:

B(z) = μ0IR²/[2(R²+z²)^(3/2)]

At the center of the loop, z=0:

B(0) = μ0I/(2R)

Question 3: Gauss Law, Induced EMF and Curl of Electric Field

Q3(a): Limitations of Gauss Law and Use in Dielectrics

Gauss law states that electric flux through a closed surface equals the charge enclosed divided by ε0:

∮E·dA = Qenc/ε0

In differential form:

∇·E = ρ/ε0

Limitations of Gauss Law in Practice

Gauss law is always true as a law of electromagnetism, but it is not always convenient for finding electric field. Its practical limitations are:

1. It is most useful only for high symmetry such as spherical, cylindrical or planar symmetry.
2. It gives total flux easily, but not local electric field in arbitrary charge distributions.
3. If E is not constant over the Gaussian surface, the integral becomes difficult.
4. It cannot directly determine the direction of electric field in irregular geometries without additional symmetry arguments.
5. In materials, bound charges complicate the use of E unless polarization is included.

Application in Dielectrics

In dielectric materials, charges may be free or bound. Bound charges appear due to polarization. Instead of writing Gauss law only in terms of E, we use electric displacement field D:

D = ε0E + P

where P is polarization.

Gauss law in a dielectric is:

∇·D = ρfree

Integral form:

∮D·dA = Qfree,enclosed

For a linear dielectric:

D = εE

where:

ε = Kε0

Q3(b): Induced EMF in Coil Inside Solenoid

Given:

n = 220 turns/m

I = 1.5 A

N = 130 turns

d = 2.1 cm = 0.021 m

r = d/2 = 0.0105 m

Δt = 25 ms = 0.025 s

The magnetic field inside a long solenoid is:

B = μ0nI

Substitute values:

B = (4π×10⁻⁷)(220)(1.5)

B ≈ 4.15×10⁻⁴ T

The area of the small coil is:

A = πr²

A = π(0.0105)²

A ≈ 3.46×10⁻⁴ m²

Magnetic flux through one turn is:

ΦB = BA

The induced emf magnitude is:

|ε| = N A |ΔB/Δt|

The field changes from 4.15×10⁻⁴ T to zero, so:

|ΔB| = 4.15×10⁻⁴ T

Now:

|ε| = 130(3.46×10⁻⁴)(4.15×10⁻⁴)/(0.025)

|ε| ≈ 7.47×10⁻⁴ V

Q3(c): Curl of Electric Field

The curl of electric field is given by Faraday’s law of electromagnetic induction:

∇×E = -∂B/∂t

This means that a time-varying magnetic field produces a circulating electric field.

Special Case: Electrostatics

For electrostatic fields, magnetic field is not changing with time:

∂B/∂t = 0

Therefore:

∇×E = 0

Question 4: Photoelectric Effect, Compton Shift and Pair Production

Q4(a): Photoelectric Effect and Planck Constant

The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

Einstein’s photoelectric equation is:

hf = φ + Kmax

where hf is photon energy, φ is work function and Kmax is maximum kinetic energy of emitted electrons.

Stopping potential Vs is the minimum reverse potential needed to stop the fastest photoelectrons. Therefore:

Kmax = eVs

So:

eVs = hf - φ

Rearrange:

Vs = (h/e)f - φ/e

This is a straight-line equation of the form:

y = mx + c

where Vs is plotted against frequency f. The slope is:

slope = h/e

Therefore:

h = e × slope

Q4(b): Compton Shift Numerical

Compton shift is given by:

Δλ = (h/mec)(1-cosθ)

The electron Compton wavelength is:

h/mec = 2.426 pm

Given:

θ = 85°

Substitute:

Δλ = 2.426(1-cos85°) pm

Since:

cos85° ≈ 0.0872

Then:

Δλ = 2.426(1-0.0872) pm

Δλ = 2.426(0.9128) pm

Δλ ≈ 2.21 pm

The scattered wavelength is:

λ' = λ + Δλ

λ' = 22 pm + 2.21 pm

λ' ≈ 24.21 pm

Q4(c): Role of Nearby Nucleus in Pair Production

Pair production is the conversion of a high-energy photon into an electron-positron pair:

γ → e⁻ + e⁺

The minimum photon energy required is the rest energy of electron and positron:

Ethreshold = 2mec²

Ethreshold = 2(0.511 MeV)

Ethreshold = 1.022 MeV

However, pair production cannot occur in empty space with only one photon, because both energy and momentum cannot be conserved simultaneously. A nearby nucleus is needed to take recoil momentum.

The nucleus plays three important roles:

1. It provides a strong electromagnetic field in which the photon can transform into a pair.
2. It absorbs recoil momentum so momentum conservation is satisfied.
3. It makes the process physically possible without violating conservation laws.

Question 5: Zeeman Effect, Quantum Operators and Particle in a Box

Q5(a): Zeeman Effect and Sodium Spectral Lines

The Zeeman effect is the splitting of atomic spectral lines when atoms are placed in an external magnetic field. It occurs because atomic magnetic moments interact with the applied magnetic field.

The interaction energy is:

ΔE = μB gJ mJ B

where μB is Bohr magneton, gJ is Landé g factor, mJ is magnetic quantum number and B is external magnetic field.

Sodium Spectral Lines

Sodium is famous for its yellow D-lines. In the absence of an external magnetic field, allowed transitions produce characteristic spectral lines. When a magnetic field is applied, energy levels split into sub-levels according to their magnetic quantum numbers. As a result, one spectral line splits into multiple components.

Quantum Explanation

1. Atomic electrons possess orbital and spin angular momenta.
2. These angular momenta produce magnetic moments.
3. In an external magnetic field, different mJ states have different energies.
4. Transitions between these split levels produce multiple spectral components.

The selection rule commonly used for Zeeman components is:

Δm = 0, ±1

Q5(b): Momentum and Energy Operators

In quantum mechanics, physical observables are represented by operators. The wave function contains the state information, and operators act on the wave function to extract measurable quantities.

Momentum Operator

In one-dimensional position representation, the momentum operator is:

p̂ = -iℏ d/dx

In three dimensions:

p̂ = -iℏ∇

Energy Operator

The energy operator is:

Ê = iℏ ∂/∂t

For time-independent systems, the Hamiltonian operator represents total energy:

Ĥ = -(ℏ²/2m)∇² + V

Significance of Operators

1. Operators connect wave functions with measurable physical quantities.
2. Eigenvalues of operators are possible measurement results.
3. They form the basis of Schrödinger equation.
4. Commutators of operators determine whether observables can be measured simultaneously.

If an operator Â satisfies:

Âψ = aψ

then ψ is an eigenfunction and a is the eigenvalue.

Q5(c): Discrete Energy Values of Particle in a Box

Consider a particle of mass m confined in a one-dimensional infinite potential well between x=0 and x=L.

The potential is:

V(x)=0, 0<x<L

V(x)=∞, x≤0 and x≥L

Inside the box, the time-independent Schrödinger equation is:

-(ℏ²/2m)d²ψ/dx² = Eψ

Rearrange:

d²ψ/dx² + (2mE/ℏ²)ψ = 0

Let:

k² = 2mE/ℏ²

Then:

d²ψ/dx² + k²ψ = 0

General solution is:

ψ(x) = A sin(kx) + B cos(kx)

Boundary condition at x=0:

ψ(0)=0

This gives:

B=0

So:

ψ(x)=A sin(kx)

Boundary condition at x=L:

ψ(L)=0

Thus:

A sin(kL)=0

For non-zero A:

sin(kL)=0

Therefore:

kL=nπ

k=nπ/L, where n=1,2,3,...

Since:

E=ℏ²k²/(2m)

we get:

En = n²π²ℏ²/(2mL²)

Using ℏ=h/(2π):

En = n²h²/(8mL²)

Question 6: Energy Bands, Sodium Electrons and Bravais Lattice

Q6(a): Formation of Energy Bands in Crystalline Solids

When isolated atoms come together to form a crystal, their atomic energy levels interact. Because of the Pauli exclusion principle, many closely spaced levels are formed instead of a single level. These closely spaced allowed levels form energy bands.

The two most important bands are:

• Valence band: The highest occupied band at low temperature.
• Conduction band: A higher band where electrons can move freely and conduct current.

The energy difference between them is the forbidden energy gap:

Eg = conduction band minimum - valence band maximum

Metals and Semiconductors

Q6(b): Number of Conduction Electrons in Sodium

Given:

Volume = 2×10⁻⁶ m³

Density = 0.968 g/cm³

Molar mass = 22.98 g/mol

Convert density into SI units:

0.968 g/cm³ = 968 kg/m³

Molar mass in kg/mol:

22.98 g/mol = 0.02298 kg/mol

Mass of sodium sample:

m = ρV

m = (968)(2×10⁻⁶)

m = 1.936×10⁻³ kg

Number of moles:

n = m/M

n = (1.936×10⁻³)/(0.02298)

n ≈ 0.0842 mol

Number of atoms:

N = nNA

N = (0.0842)(6.022×10²³)

N ≈ 5.07×10²² atoms

Sodium is monovalent, so each atom contributes one conduction electron.

Number of conduction electrons ≈ 5.07×10²²

Q6(c): Bravais Lattice and Its Types

A Bravais lattice is an infinite periodic arrangement of points in space such that every point has the same surroundings. It represents the translational symmetry of a crystal.

A lattice point can be generated by:

R = n1a1 + n2a2 + n3a3

where n1, n2 and n3 are integers, and a1, a2, a3 are primitive vectors.

14 Bravais Lattices in Three Dimensions

Question 7: Radiation Detectors, Cyclotron, Synchrotron and Deuteron Energy

Q7(a): Detection of Nuclear Radiation

Nuclear radiation is detected through its interaction with matter. Radiation can ionize gas, excite scintillating materials, produce tracks in vapour, blacken photographic plates or create electron-hole pairs in semiconductors.

Common Radiation Detectors

• Geiger-Muller counter
• Scintillation counter
• Cloud chamber
• Bubble chamber
• Ionization chamber
• Semiconductor detector
• Photographic plate

1. Geiger-Muller Counter

A Geiger-Muller counter contains a gas-filled tube with a central anode wire and cylindrical cathode. When radiation enters through a thin window, it ionizes gas atoms. The freed electrons accelerate toward the anode and produce further ionization, creating an avalanche. This produces a voltage pulse that is counted electronically.

Its advantages are:

1. Simple and sensitive.
2. Useful for detecting alpha, beta and gamma radiation with suitable window design.
3. Gives count rate quickly.

Its limitation is that it does not measure particle energy accurately.

2. Scintillation Counter

A scintillation counter uses a material that emits tiny flashes of light when radiation enters it. These flashes are detected by a photomultiplier tube and converted into electrical pulses.

Its advantages are:

1. Fast response.
2. Useful for gamma rays and charged particles.
3. Can provide energy information.
4. More suitable than GM counter for spectroscopy.

Q7(b): Cyclotron and Synchrotron

Cyclotron

A cyclotron accelerates charged particles using a perpendicular magnetic field and an alternating electric field between two D-shaped electrodes called dees.

The magnetic force provides centripetal force:

qvB = mv²/r

Thus:

r = mv/(qB)

The cyclotron angular frequency is:

ω = qB/m

So the cyclotron frequency is:

f = qB/(2πm)

The particle gains energy every time it crosses the gap between the dees.

Synchrotron

A synchrotron also accelerates charged particles but keeps them moving in a fixed circular path. As particle energy increases, magnetic field and radiofrequency acceleration are adjusted synchronously.

Q7(c): Magnetic Field and Kinetic Energy of Deuterons

Given:

f = 12 MHz = 12×10⁶ Hz

R = 53 cm = 0.53 m

m = 3.34×10⁻²⁷ kg

q = 1.60×10⁻¹⁹ C

Cyclotron frequency is:

f = qB/(2πm)

Rearrange:

B = 2πmf/q

Substitute values:

B = [2π(3.34×10⁻²⁷)(12×10⁶)]/(1.60×10⁻¹⁹)

B ≈ 1.57 T

Maximum speed at dee radius R is:

v = 2πfR

Substitute:

v = 2π(12×10⁶)(0.53)

v ≈ 3.996×10⁷ m/s

Kinetic energy:

K = (1/2)mv²

K = (1/2)(3.34×10⁻²⁷)(3.996×10⁷)²

K ≈ 2.67×10⁻¹² J

Convert into electron-volts:

K = (2.67×10⁻¹²)/(1.60×10⁻¹⁹) eV

K ≈ 1.67×10⁷ eV

K ≈ 16.7 MeV

Question 8: MOSFET and Fermi-Dirac Occupancy Probability

Q8(a): MOSFET Working Principle and Applications

MOSFET stands for Metal-Oxide-Semiconductor Field-Effect Transistor. It is a voltage-controlled semiconductor device used for switching and amplification.

Main Terminals

• Gate: Controls the channel.
• Source: Supplies charge carriers.
• Drain: Collects charge carriers.
• Body/Substrate: Semiconductor base.

Working Principle

The gate is separated from the semiconductor by a thin oxide layer. Because of this insulation, gate current is extremely small. When a suitable gate voltage is applied, it changes the charge distribution in the semiconductor and controls the conducting channel between source and drain.

For an n-channel enhancement MOSFET, a positive gate voltage attracts electrons near the oxide-semiconductor interface and forms an n-type channel. Once the channel is formed, current can flow from drain to source.

Applications

1. Digital logic gates.
2. Microprocessors and memory chips.
3. Power switching circuits.
4. Voltage regulators.
5. Amplifiers.
6. Motor control circuits.
7. Integrated circuits and CMOS technology.

Q8(b): Fermi-Dirac Occupancy Probability

The Fermi-Dirac distribution gives the probability that a quantum state of energy E is occupied by an electron:

f(E) = 1/[e^((E-EF)/kT)+1]

Given:

E - EF = 0.10 eV

T = 800 K

k = 8.617×10⁻⁵ eV/K

Calculate kT:

kT = (8.617×10⁻⁵)(800)

kT = 0.0689 eV

Now:

(E-EF)/kT = 0.10/0.0689

(E-EF)/kT ≈ 1.451

Therefore:

f(E) = 1/[e^(1.451)+1]

e^(1.451) ≈ 4.27

f(E) = 1/(4.27+1)

f(E) ≈ 0.190

Q8(c): Occupancy Probability and Density of States Graph

The Fermi-Dirac occupancy probability decreases as energy increases above the Fermi level. At E=EF, the probability is:

f(EF) = 1/2

At lower temperatures, the curve becomes sharper. At higher temperatures, the transition around EF becomes smoother.

Density of states tells how many available quantum states exist per unit energy interval. For free electrons in three dimensions, density of states increases with energy approximately as:

g(E) ∝ √E

The actual number of occupied states is obtained from the product:

Occupied states = g(E)f(E)