CSIR NET Physical Sciences Electromagnetic Theory Syllabus

Write the telegrapher's equations for a transmission line.

$$\frac{\partial V}{\partial z} = -\left(R + L\frac{\partial}{\partial t}\right)I, \qquad \frac{\partial I}{\partial z} = -\left(G + C\frac{\partial}{\partial t}\right)V.$$ They describe voltage and current as functions of position $z$ and time on the line.

Give the propagation constant $\gamma$ of a transmission line and identify its parts.

$$\gamma = \sqrt{(R + i\omega L)(G + i\omega C)} = \alpha + i\beta$$ where $\alpha$ is the attenuation constant (Np/m) and $\beta$ is the phase constant (rad/m). For a lossless line $\alpha = 0$, $\beta = \omega\sqrt{LC}$.

Define the reflection coefficient $\Gamma$ at a transmission-line load.

$$\Gamma = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}}$$ where $Z_{L}$ is the load impedance and $Z_{0}$ the line's characteristic impedance. $\Gamma = 0$ for a matched load, $\Gamma = -1$ for a short, $\Gamma = +1$ for an open.

What is the standing wave ratio (SWR/VSWR) and its relation to $\Gamma$?

$$\text{VSWR} = \frac{V_{\max}}{V_{\min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}.$$ It ranges from 1 (perfect match, $|\Gamma|=0$) to $\infty$ (total reflection, $|\Gamma|=1$).

State the input impedance of a lossless transmission line of length $l$.

$$Z_{\text{in}} = Z_{0}\frac{Z_{L} + iZ_{0}\tan(\beta l)}{Z_{0} + iZ_{L}\tan(\beta l)}.$$ Special cases: a quarter-wave line gives $Z_{\text{in}} = Z_{0}^{2}/Z_{L}$; a half-wave line gives $Z_{\text{in}} = Z_{L}$.

What is the impedance-transforming property of a quarter-wave line?

For $l = \lambda/4$, $\tan(\beta l)\to\infty$, giving $$Z_{\text{in}} = \frac{Z_{0}^{2}}{Z_{L}}.$$ A short appears as an open and vice versa; it is used as an impedance-matching transformer with $Z_{0} = \sqrt{Z_{\text{in}}Z_{L}}$.

Why is the dominant mode on an ideal two-conductor transmission line a TEM mode?

Because two conductors can support a transverse-electromagnetic mode in which both $\vec{E}$ and $\vec{B}$ are entirely transverse ($E_{z}=B_{z}=0$). TEM modes have no cutoff frequency and propagate at the medium speed $v = 1/\sqrt{LC} = c/\sqrt{\varepsilon_{r}}$.

Why can a hollow (single-conductor) waveguide NOT support a TEM mode?

A TEM mode requires a transverse static-like field with two conductors to terminate field lines; a hollow waveguide has only one conductor. Inside, Laplace's equation with a single equipotential boundary forces $\vec{E}_{\perp}=0$, so only TE and TM modes (with nonzero $E_{z}$ or $H_{z}$) exist.

Distinguish TE, TM, and TEM modes in a waveguide.

TEM: $E_{z}=H_{z}=0$ (both fields transverse). TE (transverse electric): $E_{z}=0$, $H_{z}\neq 0$. TM (transverse magnetic): $H_{z}=0$, $E_{z}\neq 0$. Hollow waveguides support only TE and TM modes.

Give the cutoff frequency formula for the $\text{TE}_{mn}$/$\text{TM}_{mn}$ mode of a rectangular waveguide of dimensions $a\times b$.

$$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}.$$ Only frequencies above $f_{c}$ propagate; below cutoff the mode is evanescent.

What is the dominant mode of a rectangular waveguide with $a > b$, and its cutoff frequency?

The dominant (lowest cutoff) mode is $\text{TE}_{10}$, with $$f_{c,10} = \frac{c}{2a}, \qquad \lambda_{c} = 2a.$$ TM modes require $m,n\geq 1$, so the lowest TM mode is $\text{TM}_{11}$, which has a higher cutoff.

Write the guide wavelength and phase velocity inside a waveguide for a mode with cutoff frequency $f_{c}$.

$$\lambda_{g} = \frac{\lambda}{\sqrt{1 - (f_{c}/f)^{2}}}, \qquad v_{p} = \frac{c}{\sqrt{1 - (f_{c}/f)^{2}}} > c.$$ The group velocity is $v_{g} = c\sqrt{1 - (f_{c}/f)^{2}} < c$, and $v_{p}v_{g} = c^{2}$.

What is the dispersion relation for a mode in a hollow waveguide?

$$\left(\frac{\omega}{c}\right)^{2} = k_{z}^{2} + k_{c}^{2}, \quad k_{c} = \frac{\omega_{c}}{c}.$$ Propagation ($k_{z}$ real) requires $\omega > \omega_{c}$; this is formally identical to the plasma/relativistic massive-particle dispersion relation.

Why is the $\text{TE}_{10}$ mode preferred for practical rectangular waveguides?

It has the lowest cutoff frequency (largest usable single-mode bandwidth), a simple field pattern, lowest attenuation in the dominant band, and allows a wide frequency range over which only one mode propagates, avoiding modal dispersion.

What are the fields radiated by a uniformly moving charge (constant velocity)?

A charge in uniform motion produces a 'pancake' field: the radial electric field is compressed toward the transverse plane. $$\vec{E} = \frac{q}{4\pi\varepsilon_{0}}\frac{(1-\beta^{2})}{(1-\beta^{2}\sin^{2}\theta)^{3/2}}\frac{\hat{R}}{R^{2}}.$$ It carries no radiation (no energy loss).

State the Liénard–Wiechert potentials for a point charge.

$$\phi(\vec{r},t) = \frac{1}{4\pi\varepsilon_{0}}\frac{qc}{(Rc - \vec{R}\cdot\vec{v})}, \qquad \vec{A}(\vec{r},t) = \frac{\mu_{0}}{4\pi}\frac{qc\,\vec{v}}{(Rc - \vec{R}\cdot\vec{v})}$$ evaluated at the retarded time; $\vec{R}$ points from the charge to the field point.

State the Larmor formula for the power radiated by a nonrelativistic accelerating charge.

$$P = \frac{\mu_{0}q^{2}a^{2}}{6\pi c} = \frac{q^{2}a^{2}}{6\pi\varepsilon_{0}c^{3}}.$$ Radiated power is proportional to the square of the acceleration $a$ and the square of the charge.

Write the relativistic (Liénard) generalization of the Larmor formula.

$$P = \frac{\mu_{0}q^{2}\gamma^{6}}{6\pi c}\left[a^{2} - \left(\frac{\vec{v}\times\vec{a}}{c}\right)^{2}\right]$$ where $\gamma = (1-\beta^{2})^{-1/2}$. It reduces to the Larmor formula when $v \ll c$.

What is the angular distribution of radiation from a nonrelativistic accelerated charge, and where is it maximum?

$$\frac{dP}{d\Omega} = \frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\sin^{2}\theta,$$ where $\theta$ is measured from the acceleration direction. It is zero along the acceleration axis and maximum perpendicular to it ($\theta = 90^{\circ}$).

Define cyclotron versus synchrotron radiation.

Both arise from charges accelerated in magnetic fields. Cyclotron radiation is emitted by nonrelativistic charges (narrow band at the gyrofrequency, broad dipole pattern). Synchrotron radiation is emitted by relativistic charges, beamed into a narrow forward cone of half-angle $\sim 1/\gamma$ with a broad spectrum.

What is the electric potential of an ideal electric dipole?

$$V(\vec{r}) = \frac{1}{4\pi\varepsilon_{0}}\frac{\vec{p}\cdot\hat{r}}{r^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{p\cos\theta}{r^{2}}$$ where $\vec{p} = q\vec{d}$ is the dipole moment. It falls off as $1/r^{2}$, faster than a monopole's $1/r$.

Write the electric field of an ideal electric dipole in spherical coordinates.

$$\vec{E} = \frac{p}{4\pi\varepsilon_{0}r^{3}}\left(2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta}\right).$$ The field strength falls off as $1/r^{3}$.

Give the torque and potential energy of a dipole $\vec{p}$ in a uniform field $\vec{E}$.

Torque: $\vec{\tau} = \vec{p}\times\vec{E}$. Potential energy: $U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$. The energy is minimized when $\vec{p}$ aligns with $\vec{E}$ ($\theta = 0$). Net force is zero in a uniform field.

What is the total power radiated by an oscillating electric dipole, and its frequency dependence?

$$P = \frac{\mu_{0}p_{0}^{2}\omega^{4}}{12\pi c} = \frac{p_{0}^{2}\omega^{4}}{12\pi\varepsilon_{0}c^{3}}.$$ It scales as $\omega^{4}$ (the $\omega^{4}$ law underlying Rayleigh scattering and the blue sky).

Give the radiation (far-field) electric field and the angular power distribution of an oscillating electric dipole.

Far field: $\vec{E} \propto \dfrac{\sin\theta}{r}\,\omega^{2}p_{0}$. Angular power: $$\frac{dP}{d\Omega} = \frac{\mu_{0}p_{0}^{2}\omega^{4}}{32\pi^{2}c}\sin^{2}\theta,$$ maximum at $\theta = 90^{\circ}$, zero along the dipole axis.

Compare the radiated power and field falloff of electric dipole, magnetic dipole, and electric quadrupole radiation.

Electric dipole radiation dominates (power $\propto \omega^{4}$). Magnetic dipole and electric quadrupole radiation are weaker by a factor $\sim (a/\lambda)^{2}$ or $(v/c)^{2}$; quadrupole power scales as $\omega^{6}$. All radiation far-fields fall off as $1/r$ so that $\langle S\rangle \propto 1/r^{2}$.

What is the retarded time, and why is it required in time-dependent potentials?

$$t_{r} = t - \frac{|\vec{r} - \vec{r}'|}{c}.$$ It is the earlier time at which the source must be evaluated so that its influence, traveling at speed $c$, reaches the field point $\vec{r}$ at time $t$. It enforces causality.

Write the retarded potentials (solutions of the inhomogeneous wave equation in Lorenz gauge).

$$\phi(\vec{r},t) = \frac{1}{4\pi\varepsilon_{0}}\int\frac{\rho(\vec{r}',t_{r})}{|\vec{r}-\vec{r}'|}\,d^{3}r', \quad \vec{A}(\vec{r},t) = \frac{\mu_{0}}{4\pi}\int\frac{\vec{J}(\vec{r}',t_{r})}{|\vec{r}-\vec{r}'|}\,d^{3}r'$$ with $t_{r} = t - |\vec{r}-\vec{r}'|/c$.

What inhomogeneous wave equations do the potentials satisfy in the Lorenz gauge?

$$\Box^{2}\phi = \nabla^{2}\phi - \frac{1}{c^{2}}\frac{\partial^{2}\phi}{\partial t^{2}} = -\frac{\rho}{\varepsilon_{0}},$$ $$\Box^{2}\vec{A} = \nabla^{2}\vec{A} - \frac{1}{c^{2}}\frac{\partial^{2}\vec{A}}{\partial t^{2}} = -\mu_{0}\vec{J}.$$ The retarded potentials are their causal solutions.

Why are the advanced potentials usually discarded in favor of the retarded potentials?

The wave equation admits both retarded (evaluated at $t_{r} = t - R/c$) and advanced (at $t_{a} = t + R/c$) solutions. The advanced solution has the field depending on the future state of the source, violating causality, so it is discarded on physical grounds.