Electromagnetic Waves revision notes

• Conduction Current ($i_c$): The current carried by conductors due to the physical flow of charges.

• Symmetry of Electromagnetism: Faraday’s law states that a time-varying magnetic field induces an electric field. Maxwell's displacement current completes the symmetry: a time-varying electric field induces a magnetic field. Time-dependent electric and magnetic fields give rise to each other.

• Transverse Nature: The electric field ($\vec{E}$) and magnetic field ($\vec{B}$) in an EM wave are perpendicular to each other, and both are perpendicular to the direction of wave propagation.

• Direction of Propagation: JEE Tip The direction of propagation of an EM wave is strictly given by the direction of the cross product vector $\vec{E} \times \vec{B}$.

• Energy Transfer: EM waves carry energy from one place to another, making life possible on earth.

The spectrum ranges from $\gamma$-rays (wavelength $\sim 10^{-12}$ m) to long radio waves (wavelength $\sim 10^6$ m). There is no sharp division between types; classification is based roughly on production and detection methods. The interaction of these waves with matter heavily depends on their wavelength and the nature of the atoms/molecules in the medium.

1. Radio Waves

• Wavelength: $> 0.1$ m.
• Frequency: $500 \text{ kHz}$ to $\sim 1000 \text{ MHz}$.
• Production: Rapid acceleration and deceleration of electrons in conducting wires/aerials.
• Detection: Receiver's aerials.
• Uses: AM band (530-1710 kHz), Short wave (up to 54 MHz), TV waves (54-890 MHz), FM band (88-108 MHz), Cellular phones (UHF band).

2. Microwaves

• Wavelength: $0.1$ m to $1 \text{ mm}$.
• Production: Special vacuum tubes like klystrons, magnetrons, and Gunn diodes.
• Detection: Point contact diodes.
• Uses: Radar systems for aircraft navigation and speed guns (due to short wavelength), Microwave ovens (frequency matches resonant frequency of water molecules to transfer kinetic energy efficiently).

3. Infrared (IR) Waves (Heat Waves)

• Wavelength: $1 \text{ mm}$ to $700 \text{ nm}$.
• Production: Hot bodies and vibrations of atoms and molecules.
• Detection: Thermopiles, Bolometers, Infrared photographic film.
• Uses & Properties: Physical therapy, maintaining Earth's warmth (greenhouse effect), Earth satellites (military/crop observation), remote controls (using Light Emitting Diodes). They vibrate entire molecules (like $H_2O$, $CO_2$, $NH_3$), increasing internal energy and temperature, hence the name "heat waves".

4. Visible Light

• Wavelength: $700 \text{ nm}$ to $400 \text{ nm}$.
• Frequency: $4 \times 10^{14} \text{ Hz}$ to $7 \times 10^{14} \text{ Hz}$.
• Production: Electrons in atoms emitting light when transitioning from a higher to a lower energy level.
• Detection: The human eye, photocells, photographic film. Human eyes evolved to be most sensitive to this range because the center of our eye's sensitivity coincides with the center of the strongest wavelength distribution emitted by the Sun.

5. Ultraviolet (UV) Rays

• Wavelength: $400 \text{ nm}$ to $1 \text{ nm}$ (or $0.6 \text{ nm}$).
• Production: Special lamps, very hot bodies (like the Sun), inner shell electron transitions.
• Detection: Photocells, photographic film.
• Uses & Properties: Causes skin tanning (melanin production), absorbed by ordinary glass (can't get tanned through a window), absorbed by the ozone layer (40-50 km altitude), used in LASIK eye surgery, and used to kill germs in water purifiers.

6. X-Rays

• Wavelength: $1 \text{ nm}$ ($10^{-8}$ m) to $10^{-3} \text{ nm}$ ($10^{-13}$ m).
• Production: Bombarding a metal target with high-energy electrons (X-ray tubes) or inner shell electron transitions; emitted by heavy atoms.
• Detection: Photographic film, Geiger tubes, ionization chambers.
• Uses: Medical diagnostics, cancer treatment (destroys living tissue).

7. Gamma ($\gamma$) Rays

• Wavelength: $< 10^{-3} \text{ nm}$ ($<10^{-12}$ m).
• Production: Nuclear reactions and radioactive decay of atomic nuclei.
• Detection: Photographic film, Geiger tubes, ionization chambers.
• Uses: Medicine to destroy cancer cells.

• Displacement Current Existence: $i_d$ exists only in regions where the electric field is changing with time. For steady electric fields (like a steady DC current in a wire), $i_d = 0$.
• Applicability of $c = 1/\sqrt{\mu_0\epsilon_0}$: Strictly limited to free space (vacuum). In any material medium, the velocity depends on the specific electric permittivity ($\epsilon$) and magnetic permeability ($\mu$) of the medium ($v = 1/\sqrt{\mu\epsilon}$).
• Antenna Emission Efficiency: A transmitting antenna can most efficiently radiate waves if its physical size is about the same size as the wavelength ($\lambda$) of the wave it is trying to emit. Visible radiation emitted by atoms, however, is much longer in wavelength than atomic size.

• JEE Tip Kirchhoff's Laws and Capacitors: It is a misconception that Kirchhoff's junction rule fails at the plates of a charging capacitor. It remains valid as long as one recognizes that the "current" leaving the plate is the displacement current.
• JEE Tip Edge Case - Perfect Media: In theoretical problems, we assume perfectly conducting wires ($i_d = 0$) and perfectly insulating capacitor gaps ($i_c = 0$). However, in reality, there exist no perfectly conducting or perfectly insulating media. Thus, in most real regions of space, both $i_c$ and $i_d$ may be present simultaneously.
• JEE Tip Sign Convention Rules: The standard wave equation $E_x = E_0 \sin(kz - \omega t)$ strictly applies for propagation in the $+z$ direction. If a wave propagates in the $-z$ direction, the argument becomes $(kz + \omega t)$.
• JEE Tip Cross Product Direction Rule: The direction of EM wave propagation is exclusively given by the vector cross product $\hat{E} \times \hat{B} = \hat{c}$. Reversing the order to $\hat{B} \times \hat{E}$ will point opposite to the actual propagation direction.
• JEE Tip Frequency vs Medium: The frequency ($\nu$) of an EM wave is determined entirely by its source (the oscillating charge) and remains constant when moving between media. Only speed ($v$) and wavelength ($\lambda$) change when entering a material medium.

1. Displacement Current Calculations: Calculating $i_d$ given the rate of change of voltage or charge across a capacitor.
2. Direction of Propagation and Field Vectors: Identifying the direction of $\vec{B}$ given $\vec{E}$ and the wave propagation vector $\vec{k}$ using $\hat{k} = \hat{E} \times \hat{B}$.
3. Energy Density: Equating $u_E$ and $u_B$ or finding the total average energy density.
4. Wave Equation Parameter Extraction: Extracting frequency, wavelength, and speed from the general equation $E_x = E_0 \sin(kz - \omega t)$.
5. EM Spectrum Matching: Matching ranges of wavelength/frequency to their respective wave types (Radio, UV, X-Ray) or their production mechanisms (e.g., Klystron for Microwaves, inner shell transitions for X-Rays).

1. Finding Displacement Current in a Charging Capacitor

• Step 1: Identify the electric field between the plates of a parallel plate capacitor: $E = \frac{Q}{A\epsilon_0}$.
• Step 2: Calculate the Electric Flux ($\Phi_E$): $\Phi_E = E \cdot A = \frac{Q}{A\epsilon_0} \times A = \frac{Q}{\epsilon_0}$.
• Step 3: Take the time derivative of the flux: $\frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0} \frac{dQ}{dt}$.
• Step 4: Recognize that $\frac{dQ}{dt}$ is the conduction current $i$ charging the plates.
• Step 5: Substitute into the displacement current formula: $i_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \left( \frac{1}{\epsilon_0} i \right) = i$.
• Conclusion: The displacement current through the gap precisely equals the conduction current in the connecting wires ($i_d = i_c$).

2. Calculating Magnetic Field Vector from Electric Field Vector

• Problem: $E = 6.3 \hat{j}$ V/m propagating along $+x$ axis. Find $\vec{B}$.
• Step 1: Use magnitude relation $B_0 = E_0/c \implies B = 6.3 / (3 \times 10^8) = 2.1 \times 10^{-8}$ T.
• Step 2: Determine the vector direction. Propagation is $+x$ ($\hat{i}$). $\vec{E}$ is $+y$ ($\hat{j}$).
• Step 3: Apply $\hat{E} \times \hat{B} = \text{Propagation Direction}$. Therefore, $\hat{j} \times \hat{B} = \hat{i}$.
• Step 4: Using vector cross product rules, $\hat{j} \times \hat{k} = \hat{i}$. So $\hat{B}$ must be in the $+z$ direction ($\hat{k}$).
• Result: $\vec{B} = 2.1 \times 10^{-8} \hat{k}$ T.

3. Extracting Wave Properties from Equations

• Problem: $B_y = (2 \times 10^{-7} \text{ T}) \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$.
• Step 1: Compare with standard equation $B_y = B_0 \sin(kx + \omega t)$.
• Step 2: Extract $k = 0.5 \times 10^3 \text{ rad/m}$. Calculate wavelength $\lambda = 2\pi/k = 1.26 \text{ cm}$.
• Step 3: Extract $\omega = 1.5 \times 10^{11} \text{ rad/s}$. Calculate frequency $\nu = \omega/2\pi = 23.9 \text{ GHz}$.
• Step 4: Find $E_0$ using $E_0 = B_0 c = (2 \times 10^{-7})(3 \times 10^8) = 60 \text{ V/m}$.
• Step 5: Write Electric field equation. It must be orthogonal to $\hat{j}$ (direction of B) and $\hat{-i}$ (direction of propagation due to $kx + \omega t$). $\hat{E} \times \hat{j} = -\hat{i} \implies \hat{E} = \hat{k}$. Thus, $E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \text{ V/m}$.