Given:

• Diameter of tube, (d = 0.1\ \text{cm})
• Radius, (r = \frac{d}{2} = 0.05\ \text{cm})
• Length of tube, (l = 40\ \text{cm})
• Pressure head, (h = 20\ \text{cm}) of water
• Coefficient of viscosity, (\eta = 0.0089\ \text{poise})
• Time, (t = 10\ \text{min} = 600\ \text{s})
• Density of water, (\rho = 1\ \text{g/cm}^3)
• Acceleration due to gravity, (g = 980\ \text{cm/s}^2)

Formula: By Poiseuille’s formula, the volume of liquid flowing per second is

[ Q=\frac{\pi r^4 \Delta P}{8\eta l} ]

where

[ \Delta P = h\rho g ]

Mass of water flowing in time (t) is

[ m=\rho Q t ]

Calculation: [ \Delta P = 20 \times 1 \times 980 = 19600\ \text{dyn/cm}^2 ]

[ Q=\frac{\pi (0.05)^4 \times 19600}{8 \times 0.0089 \times 40} ]

[ Q=\frac{\pi \times 6.25\times 10^{-6}\times 19600}{2.848} ]

[ Q=\frac{0.384845}{2.848}=0.1351\ \text{cm}^3/\text{s} ]

Volume in 10 minutes:

[ V=Q t=0.1351 \times 600=81.06\ \text{cm}^3 ]

Therefore,

[ m=\rho V=1 \times 81.06=81.06\ \text{g} ]

[ \boxed{m \approx 81.1\ \text{g}} ]

So, the mass of water flowing through the tube in 10 minutes is approximately 81.1 g.

Magnetic Vector Potential

Meaning

The magnetic vector potential is a vector field $\vec{A}$ defined so that the magnetic field $\vec{B}$ is obtained from it by taking the curl:

$\vec{B} = \nabla \times \vec{A}$

It plays the same role for the magnetic field that the electric scalar potential $V$ plays for the electric field (where $\vec{E} = -\nabla V$). Since $\vec{B}$ is solenoidal ($\nabla \cdot \vec{B} = 0$), it can always be written as the curl of some vector field. The vector $\vec{A}$ is not unique — adding the gradient of any scalar $\nabla \psi$ gives the same $\vec{B}$, since $\nabla \times (\vec{A} + \nabla\psi) = \nabla \times \vec{A}$. This freedom is removed by the Coulomb gauge condition $\nabla \cdot \vec{A} = 0$.

Obtaining the expression

Start from Ampère's law in differential form:

$\nabla \times \vec{B} = \mu_0 \vec{J}$

Substitute $\vec{B} = \nabla \times \vec{A}$:

$\nabla \times (\nabla \times \vec{A}) = \mu_0 \vec{J}$

Using the vector identity $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$:

$\nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A} = \mu_0 \vec{J}$

Applying the Coulomb gauge $\nabla \cdot \vec{A} = 0$, the first term vanishes:

$\nabla^2 \vec{A} = -\mu_0 \vec{J}$

This is Poisson's equation for each Cartesian component, e.g. $\nabla^2 A_x = -\mu_0 J_x$. Its solution, by analogy with electrostatics, is

$\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}')}{|\vec{r} - \vec{r}'|}\, d\tau'$

where $\vec{r}'$ is the position of the source element, $\vec{r}$ the field point, and $d\tau'$ the volume element of the current distribution. For a thin current-carrying loop or wire, this reduces to

$\vec{A} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l}}{|\vec{r} - \vec{r}'|}$

Taking the curl of this expression and using the identity for $\nabla \times (\vec{J}/R)$ recovers the Biot–Savart law, confirming consistency:

$\vec{B} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l} \times \hat{r}}{r^2}$

Thus the magnetic vector potential is defined by $\vec{B} = \nabla \times \vec{A}$, satisfies $\nabla^2 \vec{A} = -\mu_0 \vec{J}$ in the Coulomb gauge, and its integral expression above gives $\vec{A}$ for any steady current distribution.

Magnetic vector potential is a vector field, usually denoted $\vec{A}$, from which the magnetic field $\vec{B}$ can be derived by taking the curl:

$\vec{B} = \nabla \times \vec{A}$

It plays the same role for the magnetic field that the electric scalar potential $V$ plays for the electric field (where $\vec{E} = -\nabla V$). Since $\vec{B}$ is solenoidal ($\nabla \cdot \vec{B} = 0$), it can always be written as the curl of some vector field, and that field is the magnetic vector potential. The vector potential is not unique — adding the gradient of any scalar function $\chi$ to $\vec{A}$ leaves $\vec{B}$ unchanged:

$\vec{A}' = \vec{A} + \nabla \chi \quad \Rightarrow \quad \nabla \times \vec{A}' = \nabla \times \vec{A} = \vec{B}$

This freedom is called gauge invariance, and we usually fix the gauge by the Coulomb condition $\nabla \cdot \vec{A} = 0$.

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Expression for the magnetic vector potential

1. From the Biot–Savart law. For a steady current distribution, the magnetic field is

$\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}')\times(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}\, d\tau'$

Using the identity $\nabla \times \left(\dfrac{\vec{J}}{R}\right) = \nabla\left(\dfrac{1}{R}\right)\times \vec{J}$ with $R = |\vec{r}-\vec{r}'|$, and noting that $\nabla$ acts only on $\vec{r}$ (not on $\vec{r}'$), this becomes

$\vec{B} = \frac{\mu_0}{4\pi} \nabla \times \int \frac{\vec{J}(\vec{r}')}{R}\, d\tau'$

Comparing with $\vec{B} = \nabla \times \vec{A}$, we get

$\boxed{\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|}\, d\tau'}$

For a line current $I$ in a thin wire ($\vec{J}\,d\tau' \to I\,d\vec{l}'$):

$\boxed{\vec{A}(\vec{r}) = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l}'}{|\vec{r}-\vec{r}'|}}$

2. Derivation from Ampère's law (alternative check). In magnetostatics, $\nabla \times \vec{B} = \mu_0 \vec{J}$. Substituting $\vec{B} = \nabla \times \vec{A}$ and using the vector identity

$\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$

with the Coulomb gauge $\nabla \cdot \vec{A} = 0$, we obtain the Poisson equation for each Cartesian component:

$\nabla^2 \vec{A} = -\mu_0 \vec{J}$

Its solution (by analogy with electrostatics) is exactly the integral expression obtained above. This confirms the result.

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Physical significance

• $\vec{A}$ is not directly measurable, but it is mathematically convenient: it simplifies the calculation of $\vec{B}$ for complicated current distributions.
• The magnetic flux through a surface $S$ bounded by a loop is

$\Phi_B = \int_S \vec{B}\cdot d\vec{a} = \int_S (\nabla \times \vec{A})\cdot d\vec{a} = \oint_C \vec{A}\cdot d\vec{l}$

so the vector potential directly gives the flux via a line integral.

• In quantum mechanics, $\vec{A}$ appears explicitly in the Schrödinger equation (via $\vec{p} \to \vec{p} - q\vec{A}$) and produces observable effects such as the Aharonov–Bohm effect, showing that $\vec{A}$ has physical (not merely mathematical) significance.

Damped and Forced Harmonic Oscillator, and N-Coupled Oscillators

1. Damped Harmonic Oscillator

A damped harmonic oscillator is a system that oscillates under the action of a restoring force and a resistive (damping) force proportional to its velocity. In practice, every real oscillator loses energy to friction, air resistance, or viscosity, so its amplitude decreases gradually with time.

Equation of Motion

Let a body of mass $m$ be attached to a spring of force constant $k$, and let it move in a medium offering a resistive force $F_d = -b v$, where $b$ is the damping coefficient.

By Newton's second law:

$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$

$\boxed{\frac{d^2x}{dt^2} + \frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0}$

Writing $\dfrac{b}{m} = 2\beta$ (damping constant) and $\dfrac{k}{m} = \omega_0^2$ (natural angular frequency):

$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0$

Solution and Cases

The auxiliary equation is $r^2 + 2\beta r + \omega_0^2 = 0$, giving $r = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$.

(i) Overdamped ($\beta > \omega_0$): Both roots are real and negative. The displacement decays exponentially without oscillation:

$x = A_1 e^{-(\beta - \gamma)t} + A_2 e^{-(\beta + \gamma)t}, \quad \gamma = \sqrt{\beta^2 - \omega_0^2}$

(ii) Critically damped ($\beta = \omega_0$): The roots are equal, $r = -\beta$. The system returns to equilibrium in the shortest possible time without crossing it:

$x = (A + Bt)e^{-\beta t}$

(iii) Underdamped ($\beta < \omega_0$): The roots are complex, $r = -\beta \pm i\omega$, where $\omega = \sqrt{\omega_0^2 - \beta^2}$. The solution is

$x = A e^{-\beta t}\cos(\omega t + \phi)$

This represents oscillations of continuously decreasing amplitude $A e^{-\beta t}$. The quantity $\omega$ is the angular frequency of the damped oscillator, slightly less than $\omega_0$.

Energy of a Damped Oscillator

The total mechanical energy is

$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2e^{-2\beta t} = E_0e^{-2\beta t}$

Energy decays exponentially at twice the rate of the amplitude. The quality factor $Q = \dfrac{\omega_0}{2\beta}$ measures how slowly the oscillator loses energy.

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2. Forced Harmonic Oscillator

A forced oscillator is one in which an external periodic driving force $F(t) = F_0\cos\omega_d t$ is applied in addition to the restoring and damping forces. After the initial transient dies out, the oscillator settles into steady-state oscillation at the driving frequency $\omega_d$, not at its natural frequency $\omega_0$.

Equation of Motion

$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt} + F_0\cos\omega_d t$

$\boxed{\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = \frac{F_0}{m}\cos\omega_d t}$

Steady-State Solution

The steady-state solution is

$x = A\cos(\omega_d t - \theta)$

where the amplitude is

$A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + 4\beta^2\omega_d^2}}$

and the phase lag $\theta$ satisfies

$\tan\theta = \frac{2\beta\omega_d}{\omega_0^2 - \omega_d^2}$

Resonance

The amplitude is maximum when the denominator is minimum. For small damping this occurs near $\omega_d \approx \omega_0$. This condition is called resonance, and the corresponding frequency is the resonant frequency:

$\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$

At resonance the amplitude becomes very large (limited only by damping). This explains why a swing is pushed at its natural frequency, why bridges can collapse under marching soldiers, and why tuning circuits respond sharply to one station.

Power Absorbed

The average power absorbed at steady state is maximum at $\omega_d = \omega_0$:

$P_{av} = \frac{F_0^2}{2b} \quad \text{(at resonance)}$

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3. N-Coupled Oscillators

When two or more oscillators are connected so that the motion of one influences the others (through springs or coupling), the system is called a coupled oscillator. A system of $N$ such oscillators is an N-coupled oscillator.

Equations of Motion

For $N$ identical masses $m$ joined by springs of force constant $k$, with fixed ends, the displacement $x_i$ of the $i$-th mass obeys

$m\ddot{x}_i = -k(x_i - x_{i-1}) - k(x_i - x_{i+1})$

$\boxed{\ddot{x}_i + \frac{k}{m}(2x_i - x_{i-1} - x_{i+1}) = 0, \quad i = 1,2,\dots,N}$

with $x_0 = x_{N+1} = 0$ at the fixed ends.

Normal Modes

Such a system has $N$ normal modes (independent modes of vibration), each with its own normal frequency. For the fixed-end chain,

$\omega_n = 2\sqrt{\frac{k}{m}}\sin\!\left(\frac{n\pi}{2(N+1)}\right), \quad n = 1,2,\dots,N$

The corresponding mode shape is

$x_i^{(n)} = A_n \sin\!\left(\frac{ni\pi}{N+1}\right)$

• $n = 1$: fundamental mode — all masses move in the same direction.
• $n = N$: highest mode — adjacent masses move in opposite directions.

General Solution and Beats

Any arbitrary motion is a superposition of all normal modes:

$x_i(t) = \sum_{n=1}^{N} A_n \sin\!\left(\frac{ni\pi}{N+1}\right)\cos(\omega_n t + \phi_n)$

For two weakly coupled oscillators ($N=2$), the superposition of two nearly equal normal frequencies produces beats — energy transfers periodically from one oscillator to the other. As $N$ increases, the number of normal modes increases and the frequency spectrum becomes denser. In the limit $N \to \infty$, the discrete modes merge into a continuous band, which is the basis of wave motion and lattice vibrations in solids.

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4. Significance

• Damping explains the gradual dying of real oscillations and the design of shock absorbers, galvanometers, and LCR circuits.
• Forced oscillation and resonance explain tuning, vibration isolation, and the destructive effects of resonance.
• N-coupled oscillators form the bridge between simple harmonic motion and wave propagation, and are fundamental in molecular spectroscopy and solid-state physics.

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OR

Equations of Motion for a Rotating Rigid Body and Kinetic Energy of Rotating and Rolling Bodies

1. Rigid Body and Rotational Motion

A rigid body is one in which the distance between any two particles remains constant. When such a body rotates about a fixed axis, every particle moves in a circle whose centre lies on the axis, and all particles have the same angular velocity $\omega$ and angular acceleration $\alpha$.

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2. Equations of Motion for a Rotating Rigid Body

(a) Torque–Angular Acceleration Relation

The torque about the axis of rotation is

$\tau = I\alpha$

where $I$ is the moment of inertia of the body about that axis. This is the rotational analogue of $F = ma$.

(b) Angular Momentum Form

Angular momentum about the axis is $L = I\omega$. Differentiating,

$\tau = \frac{dL}{dt} = \frac{d}{dt}(I\omega)$

If $I$ is constant, this reduces to $\tau = I\alpha$. If $I$ changes (as in a collapsing star or a skater pulling in her arms), then

$\tau = I\frac{d\omega}{dt} + \omega\frac{dI}{dt}$

(c) Work–Energy Theorem for Rotation

$W = \int_{\theta_1}^{\theta_2} \tau\, d\theta = \frac{1}{2}I\omega_2^2 - \frac{1}{2}I\omega_1^2 = \Delta K_{rot}$

(d) Kinematic Equations (for constant $\alpha$)

These are the rotational analogues of the linear equations:

$\omega = \omega_0 + \alpha t$ $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ $\omega^2 = \omega_0^2 + 2\alpha\theta$

(e) General Motion (Translation + Rotation)

For a body that both translates and rotates, the motion of the centre of mass is governed by

$\vec{F}_{ext} = M\vec{a}_{cm}$

and rotation about the centre of mass by

$\vec{\tau}_{cm} = I_{cm}\vec{\alpha}$

These two vector equations together give the complete equation of motion of a rigid body.

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3. Kinetic Energy of a Rotating Body

Consider a rigid body rotating about a fixed axis with angular velocity $\omega$. A particle of mass $m_i$ at distance $r_i$ from the axis has linear speed $v_i = r_i\omega$. Its kinetic energy is

$K_i = \frac{1}{2}m_i v_i^2 = \frac{1}{2}m_i r_i^2\omega^2$

Summing over all particles,

$K_{rot} = \sum \frac{1}{2}m_i r_i^2\omega^2 = \frac{1}{2}\left(\sum m_i r_i^2\right)\omega^2$

$\boxed{K_{rot} = \frac{1}{2}I\omega^2}$

where $I = \sum m_i r_i^2$ is the moment of inertia. This is the rotational analogue of $\frac{1}{2}mv^2$.

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4. Kinetic Energy of a Rolling Body

A rolling body (such as a cylinder or sphere rolling without slipping) has both translational and rotational motion. Its total kinetic energy is the sum of the translational KE of the centre of mass and the rotational KE about the centre of mass:

$\boxed{K = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2}$

For rolling without slipping, $v_{cm} = R\omega$, so

$K = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\frac{v_{cm}^2}{R^2} = \frac{1}{2}Mv_{cm}^2\left(1 + \frac{I_{cm}}{MR^2}\right)$

Writing $I_{cm} = Mk^2$, where $k$ is the radius of gyration,

$K = \frac{1}{2}Mv_{cm}^2\left(1 + \frac{k^2}{R^2}\right)$

Fraction of Energy in Rotation

$\frac{K_{rot}}{K_{total}} = \frac{k^2/R^2}{1 + k^2/R^2}$

Acceleration on an Inclined Plane

For a body rolling down an incline of angle $\theta$,

$a = \frac{g\sin\theta}{1 + k^2/R^2}$

This shows that a solid sphere ($k^2/R^2 = 2/5$) rolls down faster than a hollow cylinder ($k^2/R^2 = 1$), because more of its energy goes into translation. This is a classic demonstration that rolling speed depends on the distribution of mass, not on the mass or radius alone.

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5. Summary

• A rotating rigid body obeys $\tau = I\alpha$ and $\tau = dL/dt$, with kinematic equations analogous to linear motion.
• The kinetic energy of pure rotation is $\frac{1}{2}I\omega^2$.
• A rolling body has total kinetic energy $\frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$, and the sharing between translation and rotation depends only on the shape factor $k^2/R^2$.

Planck's Hypothesis (Quantum Theory of Radiation)

Statement of the hypothesis

Planck put forward the following postulates to explain the observed energy distribution in the spectrum of a black body:

1. Atomic oscillators. The walls of a black-body cavity contain a very large number of charged harmonic oscillators (atomic vibrators) which continuously emit and absorb electromagnetic radiation. Each oscillator has its own natural frequency $\nu$.

2. Quantisation of energy. An oscillator of frequency $\nu$ cannot have any arbitrary energy. It can possess only discrete energies given by $E_n = n\,h\nu,\qquad n = 0,1,2,3,\dots$ where $h = 6.626\times10^{-34}\ \text{J s}$ is a universal constant, called Planck's constant.

3. Quantised emission and absorption. Energy is emitted or absorbed not continuously but in discrete packets (quanta) of magnitude $h\nu$. An oscillator emits or absorbs a quantum $h\nu$ only when it jumps from one allowed energy level to another.

4. Frequency condition. The frequency of the radiation emitted or absorbed when an oscillator changes its energy from $E_m$ to $E_n$ is $\nu = \frac{E_m - E_n}{h} = (m-n)\frac{h\nu}{h}.$

Thus radiation behaves as if it were made up of discrete particles of energy $h\nu$ — the photons.

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Derivation of Planck's Radiation Law

Average energy of an oscillator. By Maxwell–Boltzmann statistics, the number of oscillators having energy $E_n = nh\nu$ at absolute temperature $T$ is proportional to $e^{-E_n/kT}$. Hence the average energy is

$\bar{E}=\frac{\displaystyle\sum_{n=0}^{\infty} nh\nu\,e^{-nh\nu/kT}}{\displaystyle\sum_{n=0}^{\infty} e^{-nh\nu/kT}}$

Put $x = \dfrac{h\nu}{kT}$:

$\bar{E}=h\nu\,\frac{\displaystyle\sum_{n=0}^{\infty} n\,e^{-nx}}{\displaystyle\sum_{n=0}^{\infty} e^{-nx}}$

Now

$$\sum_{n=0}^{\infty} e^{-nx}=\frac{1}{1-e^{-x}},\qquad \sum_{n=0}^{\infty} n\,e^{-nx}=-\frac{d}{dx}\left(\frac{1}{1-e^{-x}}\right)=\frac{e^{-x}}{(1-e^{-x})^{2}}$$

Therefore

$$\bar{E}=h\nu\,\frac{e^{-x}/(1-e^{-x})^{2}}{1/(1-e^{-x})} =\frac{h\nu\,e^{-x}}{1-e^{-x}} =\frac{h\nu}{e^{x}-1}$$

$\boxed{\bar{E}=\frac{h\nu}{e^{h\nu/kT}-1}}$

Number of modes. The number of standing-wave modes per unit volume of the cavity in the frequency interval $\nu$ to $\nu+d\nu$ is $g(\nu)\,d\nu=\frac{8\pi\nu^{2}}{c^{3}}\,d\nu$

Energy density. The energy density of radiation in that interval is $u(\nu)\,d\nu=\bar{E}\,g(\nu)\,d\nu$

$u(\nu)\,d\nu=\frac{8\pi h\nu^{3}}{c^{3}}\,\frac{1}{e^{h\nu/kT}-1}\,d\nu$

This is Planck's radiation law. In terms of wavelength, using $\nu = c/\lambda$ and $|d\nu| = (c/\lambda^{2})d\lambda$, the spectral emissive power is

$E_\lambda\,d\lambda=\frac{2\pi h c^{2}}{\lambda^{5}}\,\frac{1}{e^{hc/\lambda kT}-1}\,d\lambda$

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Wien's Law as a Special Case

For high frequencies (short wavelengths), $h\nu \gg kT$, so $x \gg 1$ and $e^{x} \gg 1$. Hence $e^{x}-1 \approx e^{x}$, and Planck's law becomes

$u(\nu)\,d\nu=\frac{8\pi h\nu^{3}}{c^{3}}\,e^{-h\nu/kT}\,d\nu$

which is exactly Wien's distribution law. In wavelength form,

$E_\lambda=\frac{2\pi h c^{2}}{\lambda^{5}}\,e^{-hc/\lambda kT}$

Wien's displacement law. To find the wavelength of maximum emission, set $\dfrac{dE_\lambda}{d\lambda}=0$. Let $y=\dfrac{hc}{\lambda kT}$, so that

$E_\lambda = A\,\frac{y^{5}}{e^{y}-1}$

$\frac{dE_\lambda}{dy}=A\,\frac{5y^{4}(e^{y}-1)-y^{5}e^{y}}{(e^{y}-1)^{2}}=0$

$5(e^{y}-1)=y\,e^{y}\quad\Rightarrow\quad e^{-y}=1-\frac{y}{5}$

The non-zero root is $y = 4.965$. Therefore

$$\lambda_m T=\frac{hc}{4.965\,k} =\frac{(6.626\times10^{-34})(3\times10^{8})}{4.965\,(1.38\times10^{-23})} =2.90\times10^{-3}\ \text{m K}$$

$\boxed{\lambda_m T = b = 2.90\times10^{-3}\ \text{m K}}$

This is Wien's displacement law, obtained as a limiting case of Planck's formula.

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Rayleigh–Jeans Law as a Special Case

For low frequencies (long wavelengths), $h\nu \ll kT$, so $x \ll 1$. Expanding,

$e^{x}-1 = x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \approx x$

Substituting in Planck's law,

$$u(\nu)\,d\nu=\frac{8\pi h\nu^{3}}{c^{3}}\cdot\frac{1}{x}\,d\nu =\frac{8\pi h\nu^{3}}{c^{3}}\cdot\frac{kT}{h\nu}\,d\nu$$

$u(\nu)\,d\nu=\frac{8\pi \nu^{2}kT}{c^{3}}\,d\nu$

which is precisely the Rayleigh–Jeans law. Thus Planck's formula correctly reproduces the classical result in the long-wavelength region, where the classical theory agrees with experiment.

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Conclusion

Planck's quantum hypothesis thus unifies the two classical laws as special cases and, in addition, on integrating over all frequencies gives the Stefan–Boltzmann law $E=\sigma T^{4}$, where $\sigma = \dfrac{2\pi^{5}k^{4}}{15h^{3}c^{2}}$. It also removes the ultraviolet catastrophe of the Rayleigh–Jeans law.

Plane Electromagnetic Waves in Isotropic Dielectrics and Coupled (Compound) Media

1. Wave Equation in an Isotropic Dielectric

For a non-conducting, isotropic dielectric with permittivity $\varepsilon$ and permeability $\mu$ (free charge $\rho=0$, free current $\mathbf{J}=0$), Maxwell's equations give

$\nabla^2 \mathbf{E} = \mu\varepsilon \frac{\partial^2 \mathbf{E}}{\partial t^2}, \qquad \nabla^2 \mathbf{H} = \mu\varepsilon \frac{\partial^2 \mathbf{H}}{\partial t^2}$

A plane harmonic solution is

$\mathbf{E} = \mathbf{E}_0 \, e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}$

with the dispersion relation

$k^2 = \omega^2 \mu \varepsilon \quad\Rightarrow\quad v = \frac{\omega}{k} = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{n}, \qquad n = \sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}} = \sqrt{\mu_r \varepsilon_r}$

For a non-magnetic dielectric $\mu_r \approx 1$, so $n \approx \sqrt{\varepsilon_r}$ (Maxwell's relation).

2. Behaviour in an Isotropic Dielectric

(i) Transverse nature. $\nabla\cdot\mathbf{E}=0$ and $\nabla\cdot\mathbf{H}=0$ give $\mathbf{k}\cdot\mathbf{E}=0$, $\mathbf{k}\cdot\mathbf{H}=0$: both fields are perpendicular to the direction of propagation.

(ii) Mutual perpendicularity. From $\nabla\times\mathbf{E} = -\mu\,\partial\mathbf{H}/\partial t$,

$\mathbf{H} = \frac{\mathbf{k}\times\mathbf{E}}{\mu\omega}, \qquad \frac{E}{H} = \sqrt{\frac{\mu}{\varepsilon}} = Z \ (\text{intrinsic impedance})$

(iii) Phase and energy travel together. $\mathbf{E}$, $\mathbf{H}$ and $\mathbf{k}$ form a right-handed triad; the Poynting vector

$\mathbf{S} = \mathbf{E}\times\mathbf{H}$

is along $\mathbf{k}$, and the energy velocity equals the phase velocity $v = c/n$.

(iv) Frequency dependence. In a real dielectric $\varepsilon_r$ is complex and frequency dependent:

$\varepsilon_r(\omega) = \varepsilon_r'(\omega) - i\varepsilon_r''(\omega), \qquad \tilde{k} = \frac{\omega}{c}\sqrt{\varepsilon_r(\omega)} = \frac{\omega}{c}(n - i\kappa)$

so the wave is attenuated as $e^{-\kappa\omega x/c}$ and $n$ varies with $\omega$ — normal/ anomalous dispersion.

(v) Reflection and refraction. At a dielectric boundary the wave obeys Snell's law $n_1\sin i = n_2\sin r$, Fresnel equations, and total internal reflection when going from denser to rarer medium.

3. Behaviour in Coupled (Compound) Media

A "coupled" or compound medium means two or more dielectrics in contact (layered stack, composite, or a medium with inclusions), so that waves in one region are coupled to waves in the next through boundary conditions. The essential new feature is multiple reflection and interference.

(i) Boundary conditions couple the fields. At each interface,

$E_{1t}=E_{2t}, \quad H_{1t}=H_{2t}, \quad D_{1n}=D_{2n}, \quad B_{1n}=B_{2n}$

so the transmitted wave in medium 2 becomes a source term for the reflected wave in medium 1 — the two media are dynamically coupled.

(ii) Superposition and interference. Inside a layer of thickness $d$, forward and backward waves coexist:

$\mathbf{E} = \mathbf{E}^+ e^{i(kz-\omega t)} + \mathbf{E}^- e^{i(-kz-\omega t)}$

giving standing-wave patterns, constructive/destructive interference, and Fabry–Perot type transmission maxima when $2nd = m\lambda$.

(iii) Effective medium behaviour. For a composite of two dielectrics with volume fractions $f_1, f_2$, the wave sees an effective permittivity (Maxwell–Garnett / Bruggeman):

$\frac{\varepsilon_{\text{eff}}-\varepsilon_h}{\varepsilon_{\text{eff}}+2\varepsilon_h} = f \frac{\varepsilon_i-\varepsilon_h}{\varepsilon_i+2\varepsilon_h}$

The wave speed, impedance and refractive index are those of this effective medium, not of either component.

(iv) Impedance matching and mismatch. Reflection at each interface is governed by the impedance mismatch

$R = \left|\frac{Z_2-Z_1}{Z_2+Z_1}\right|^2$

Perfect matching ($Z_1=Z_2$) gives zero reflection; a quarter-wave layer ($nd=\lambda/4$) acts as an antireflection coating.

(v) Anisotropy and mode coupling. If the coupled medium is anisotropic or periodically modulated (photonic crystal, grating), $\mathbf{k}$ and $\mathbf{E}$ are no longer simply related; polarization coupling and band gaps appear, and phase velocity becomes direction dependent.

4. Major Differences

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OR

Clausius–Mossotti Relation and Debye's Modification

1. Derivation of the Clausius–Mossotti Relation

Setup. Consider a dielectric of molecular number density $N$, each molecule having polarizability $\alpha$. In an applied field $\mathbf{E}$, the induced dipole moment per molecule is

$\mathbf{p} = \alpha \mathbf{E}_{\text{loc}}$

where $\mathbf{E}_{\text{loc}}$ is the local (internal) field actually acting on the molecule.

Step 1: Macroscopic relation. The polarization is

$P = N p = N\alpha E_{\text{loc}} \quad (1)$

and the electric displacement is

$D = \varepsilon_0 E + P = \varepsilon_0 \varepsilon_r E \quad\Rightarrow\quad P = \varepsilon_0(\varepsilon_r - 1)E \quad (2)$

Step 2: Lorentz local field. For a cubic (or isotropic) arrangement of dipoles, the field at a reference molecule is the sum of the applied field, the depolarization field of the surface charges, and the field of the dipoles inside a Lorentz sphere. The latter two combine to give the Lorentz result

$E_{\text{loc}} = E + \frac{P}{3\varepsilon_0} \quad (3)$

Step 3: Combine. Substituting (2) and (3) into (1):

$P = N\alpha\left(E + \frac{P}{3\varepsilon_0}\right)$

Using $E = P/[\varepsilon_0(\varepsilon_r-1)]$:

$P = N\alpha\left(\frac{P}{\varepsilon_0(\varepsilon_r-1)} + \frac{P}{3\varepsilon_0}\right)$

Cancelling $P$ and multiplying by $\varepsilon_0$:

$1 = N\alpha\left(\frac{1}{\varepsilon_r-1} + \frac{1}{3}\right) = N\alpha\,\frac{3 + \varepsilon_r - 1}{3(\varepsilon_r-1)} = N\alpha\,\frac{\varepsilon_r + 2}{3(\varepsilon_r-1)}$

Hence

$\boxed{\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha}{3\varepsilon_0}\ }$

This is the Clausius–Mossotti relation. In terms of molar mass $M$ and density $\rho$ (with $N = N_A\rho/M$):

$\frac{\varepsilon_r - 1}{\varepsilon_r + 2}\cdot\frac{M}{\rho} = \frac{N_A \alpha}{3\varepsilon_0} = P_m \ (\text{molar polarization})$

Significance. The right-hand side depends only on the molecule ($\alpha$), the left only on measurable bulk quantities ($\varepsilon_r$, $\rho$). It therefore allows $\alpha$ to be determined from experiment, and predicts that $\varepsilon_r$ increases with density — as observed for gases and non-polar liquids.

2. Debye's Modification

Limitation of Clausius–Mossotti. It assumes the molecules are non-polar (only induced polarization) and that the field is static. It fails for polar molecules (H₂O, HCl, etc.), whose permanent dipoles can rotate, and it ignores the finite time needed for dipoles to orient.

Debye's idea. Debye treated the molecule as a rigid permanent dipole of moment $\mu_0$ that can rotate in the field, in competition with thermal agitation. Using Boltzmann statistics, the average orientation polarization per molecule is (for $\mu_0 E_{\text{loc}} \ll k_BT$)

$p_{\text{orient}} = \frac{\mu_0^2 E_{\text{loc}}}{3 k_B T}$

so the total effective polarizability becomes

$\alpha_{\text{eff}} = \alpha_{\text{induced}} + \frac{\mu_0^2}{3 k_B T}$

Substituting into the Clausius–Mossotti result gives the Debye equation:

$\boxed{\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N}{3\varepsilon_0}\left(\alpha_{\text{induced}} + \frac{\mu_0^2}{3 k_B T}\right)\ }$

Molar form:

$\frac{\varepsilon_r - 1}{\varepsilon_r + 2}\cdot\frac{M}{\rho} = \frac{N_A}{3\varepsilon_0}\left(\alpha_{\text{induced}} + \frac{\mu_0^2}{3 k_B T}\right)$

Key consequences.

1. Temperature dependence. The orientation term $\propto 1/T$, so polar dielectrics show $\varepsilon_r$ decreasing with rising temperature — a behaviour Clausius–Mossotti cannot explain. A plot of molar polarization vs $1/T$ is a straight line whose slope gives $\mu_0$ and whose intercept gives $\alpha_{\text{induced}}$.

2. Frequency dependence (relaxation). For an alternating field, dipoles cannot follow the field at high frequency. Debye introduced a relaxation time $\tau$ and wrote the complex permittivity

$\varepsilon_r(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 + i\omega\tau}$

where $\varepsilon_s$ is the static and $\varepsilon_\infty$ the high-frequency (optical) permittivity. This gives the characteristic Debye relaxation with a peak in $\varepsilon_r''$ at $\omega = 1/\tau$.

3. Unified picture. At optical frequencies only $\alpha_{\text{induced}}$ survives (dipoles cannot respond), so $\varepsilon_\infty = n^2$; at low frequencies the full orientation contribution appears. Thus Debye's modification extends Clausius–Mossotti from static, non-polar dielectrics to polar dielectrics over a wide frequency and temperature range.

Twisting couple on a cylindrical rod or wire

When one end of a cylindrical rod or wire is fixed and a couple is applied at the free end, the free end rotates through an angle $\theta$ while the fixed end remains stationary. Each cross-section twists relative to the neighbouring one, and shearing stress is produced in the material. The couple required to maintain this twist is called the twisting couple or torsional couple.

For small twist within the elastic limit, the twisting couple is directly proportional to the angle of twist:

$$\tau \propto \theta$$

or

$$\tau = C\theta$$

where $C$ is the couple per unit twist or torsional rigidity of the rod.

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Expression for the twisting couple

Let a cylindrical rod of length $l$, radius $r$, and modulus of rigidity $\eta$ be fixed at the upper end. Let the lower end be twisted through an angle $\theta$.

Consider a thin coaxial cylindrical shell of radius $x$ and thickness $dx$.

A point on this shell at the free end is displaced through an arc length $x\theta$. Hence the angle of shear is

$$\phi = \frac{x\theta}{l}$$

Therefore, the shearing stress at radius $x$ is

$$\text{stress} = \eta \phi = \eta \frac{x\theta}{l}$$

The area of the annular shell is

$$dA = 2\pi x\,dx$$

So the shearing force on the shell is

$$dF = \text{stress} \times dA = \left(\eta \frac{x\theta}{l}\right)(2\pi x\,dx)$$

The twisting couple due to this shell is

$$d\tau = dF \times x = \left(\eta \frac{x\theta}{l}\right)(2\pi x\,dx)(x)$$ $$d\tau = \frac{2\pi \eta \theta}{l}x^3\,dx$$

Integrating from $x=0$ to $x=r$:

$$\tau = \frac{2\pi \eta \theta}{l}\int_0^r x^3\,dx$$ $$\tau = \frac{2\pi \eta \theta}{l}\cdot \frac{r^4}{4}$$ $$\boxed{\tau = \frac{\pi \eta r^4 \theta}{2l}}$$

This is the expression for the twisting couple on a cylindrical rod or wire.

The couple per unit twist is

$$C = \frac{\tau}{\theta} = \frac{\pi \eta r^4}{2l}$$

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Work done in twisting a rod or wire

Let the rod be twisted gradually from $0$ to $\theta$. At any intermediate stage, let the angle of twist be $\theta'$. The twisting couple at that stage is

$$\tau' = C\theta'$$

For a further small twist $d\theta'$, the small work done is

$$dW = \tau'\,d\theta'$$ $$dW = C\theta'\,d\theta'$$

Total work done in twisting the rod through angle $\theta$ is

$$W = \int_0^\theta C\theta'\,d\theta'$$ $$W = C\int_0^\theta \theta'\,d\theta'$$ $$W = C\left[\frac{\theta'^2}{2}\right]_0^\theta$$ $$W = \frac{1}{2}C\theta^2$$

Since $C\theta = \tau$, we can also write

$$W = \frac{1}{2}\tau\theta$$

Substituting $C = \dfrac{\pi \eta r^4}{2l}$:

$$W = \frac{1}{2}\left(\frac{\pi \eta r^4}{2l}\right)\theta^2$$ $$\boxed{W = \frac{\pi \eta r^4 \theta^2}{4l}}$$

Thus, the work done in twisting a cylindrical rod or wire through an angle $\theta$ is

$$\boxed{W = \frac{1}{2}\tau\theta = \frac{\pi \eta r^4 \theta^2}{4l}}$$

This work is stored in the rod as elastic potential energy. If the rod is twisted from angle $\theta_1$ to $\theta_2$, the work done is

$$W = \frac{1}{2}C(\theta_2^2 - \theta_1^2)$$

Entropy. Entropy $S$ is a state function that measures the degree of disorder (randomness) of a system, or the part of its thermal energy unavailable for work. For a reversible change it is defined by

$dS=\frac{dQ_{rev}}{T}\qquad\Longrightarrow\qquad \Delta S=S_2-S_1=\int_1^2\frac{dQ_{rev}}{T}$

Its unit is J K$^{-1}$. Because $S$ is a state function, $\oint dS=0$ for any cycle, and $\Delta S$ depends only on the end states, not on the path.

Relation to the second law. The second law states that heat cannot flow by itself from a colder to a hotter body, and that no cyclic engine can convert heat wholly into work. In terms of entropy: in any natural (irreversible) process the total entropy of an isolated system increases, and in an ideal reversible process it stays constant.

$\Delta S_{isolated}\ge0$

Entropy thus gives the direction of natural processes, and fixes why some processes (heat flowing hot to cold, gas expanding into vacuum, mixing) occur and their reverse does not.

Proof that entropy cannot decrease. Take a cycle in which the system goes from state 1 to state 2 by an irreversible path $I$ and returns from 2 to 1 by a reversible path $R$. By the Clausius inequality, $\oint\dfrac{dQ}{T}\le0$ for any cycle, so

$\int_{1}^{2}\frac{dQ_{irr}}{T}+\int_{2}^{1}\frac{dQ_{rev}}{T}\le0$

The reversible part equals $S_1-S_2=-\Delta S$. Hence

$\Delta S\ge\int_1^2\frac{dQ_{irr}}{T}$

with equality only for a reversible process. For an isolated system no heat crosses the boundary, $dQ=0$, so

$\Delta S\ge0$

Thus the entropy of an isolated system never decreases. Since the system together with its surroundings (the universe) forms an isolated system,

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}\ge0$

Illustration. Let heat $Q$ pass from a body at $T_1$ to a body at $T_2<T_1$. The change is $\Delta S=-\dfrac{Q}{T_1}+\dfrac{Q}{T_2}=Q\,\dfrac{T_1-T_2}{T_1T_2}>0$. The reverse flow would give $\Delta S<0$, which is impossible.

Remark. The statement holds for the substance plus its surroundings (an isolated system). A substance alone can lose entropy (for example water freezing, or a gas compressed isothermally) only when its surroundings gain at least as much, so the total never decreases.

Attempt any TWO questions.

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a) Two examples of conservation of angular momentum

Angular momentum is conserved when the net external torque on a system is zero, i.e. (L = I\omega = \text{constant}).

1. A spinning skater pulling in her arms: When a skater brings her arms close to her body, the moment of inertia (I) decreases. Since (L = I\omega) is constant, the angular velocity (\omega) increases — she spins faster. When she stretches her arms out, (I) increases and (\omega) decreases.

2. A planet moving in an elliptical orbit around the Sun (Kepler's second law): The gravitational force on the planet is directed towards the Sun, so the torque about the Sun is zero. Hence (L) is constant. As a result, the planet moves faster when nearer the Sun (perihelion) and slower when farther away (aphelion), so that equal areas are swept in equal times.

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b) Inverse square law of force

Statement: The magnitude of the force between two point objects (or between a point source and a point object) is inversely proportional to the square of the distance between them.

$F \propto \frac{1}{r^{2}} \quad \Rightarrow \quad F = \frac{k}{r^{2}}$

where (k) is a constant depending on the nature of the interaction and the strength of the sources.

Explanation:

• If the distance is doubled ((r \to 2r)), the force falls to one-fourth (\left(\dfrac{1}{2^2}\right)) of its original value.
• If the distance is tripled, the force becomes one-ninth.
• This arises because the influence of a point source spreads uniformly over the surface of a sphere of radius (r), whose area is (4\pi r^{2}). The same "effect" is distributed over an area growing as (r^{2}), so the intensity (force per unit area) falls as (1/r^{2}).

Examples:

• Gravitational force: (F = \dfrac{G m_1 m_2}{r^{2}})
• Electrostatic force (Coulomb's law): (F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^{2}})
• Intensity of light from a point source: (I \propto \dfrac{1}{r^{2}})

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c) Law of equipartition of energy

Statement: In thermal equilibrium at absolute temperature (T), the total energy of a system is equally distributed among all its degrees of freedom, each degree of freedom carrying an average energy of (\dfrac{1}{2}k_B T), where (k_B) is the Boltzmann constant.

Explanation:

• A degree of freedom is an independent way in which a molecule can store energy — translational, rotational or vibrational.
• For a molecule with (f) degrees of freedom, the average energy per molecule is $E = \frac{f}{2}k_B T$
• For one mole of gas ((N_A) molecules), the internal energy is $U = \frac{f}{2}N_A k_B T = \frac{f}{2}RT$ where (R = N_A k_B) is the universal gas constant.

Applications:

• Monatomic gas (He, Ne, Ar): only 3 translational degrees of freedom, (f = 3), so $C_V = \frac{3}{2}R, \qquad C_P = \frac{5}{2}R, \qquad \gamma = \frac{C_P}{C_V} = \frac{5}{3} = 1.67$
• Diatomic gas (H₂, O₂, N₂) at ordinary temperature: 3 translational + 2 rotational = 5, so $C_V = \frac{5}{2}R, \qquad \gamma = \frac{7}{5} = 1.40$
• Polyatomic gas: (f = 6) (3 translational + 3 rotational), so (C_V = 3R) and (\gamma = \dfrac{4}{3} = 1.33).

These predicted values agree well with experiment, which is the main significance of the law.

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d) Significance of complex numbers for ac analysis

In alternating-current (ac) circuits, voltages and currents vary sinusoidally with time and differ in phase because of inductance and capacitance. Complex numbers (phasors) allow both the magnitude and the phase of a sinusoidal quantity to be handled in a single algebraic number, converting differential equations into simple algebra.

Key points:

1. Representation of ac quantities: A sinusoid (v = V_0\sin(\omega t + \phi)) is written as the phasor $\tilde{V} = V_{\text{rms}}\, e^{j\phi} = V_{\text{rms}}(\cos\phi + j\sin\phi)$ where (j = \sqrt{-1}). The modulus gives the amplitude (rms value) and the argument gives the phase.

2. Impedance as a complex quantity: Resistance, inductive reactance and capacitive reactance are combined into a single complex impedance $Z = R + j\left(\omega L - \frac{1}{\omega C}\right) = R + jX$

• (|Z| = \sqrt{R^{2} + X^{2}}) gives the impedance magnitude.
• (\phi = \tan^{-1}!\left(\dfrac{X}{R}\right)) gives the phase difference between voltage and current.

3. Ohm's law in complex form: (\tilde{V} = \tilde{I} Z), so the current and its phase follow directly from simple division — no separate trigonometric treatment is needed.

4. Series and parallel combinations: Impedances add like complex numbers in series ((Z = Z_1 + Z_2 + \cdots)) and combine as reciprocals in parallel, exactly as resistances do in dc circuits. This makes the analysis of LCR circuits systematic.

5. Power calculation: Real power is (P = I_{\text{rms}}^{2} R = I_{\text{rms}}^{2},\text{Re}(Z)), while the imaginary part gives reactive power. The power factor (\cos\phi) follows at once from the argument of (Z).

6. Resonance: At resonance the imaginary part of (Z) vanishes ((\omega L = 1/\omega C)), giving minimum impedance and maximum current — easily identified from the complex expression.

Conclusion: Complex-number (phasor) algebra converts the awkward phase-and-amplitude problem of ac circuits into ordinary algebraic operations on (R), (L) and (C), making ac circuit analysis simple, compact and general.