Electromagnetic Waves

Given:
- Radius of each plate: $r = 12\text{ cm} = 0.12\text{ m}$
- Separation between plates: $d = 5.0\text{ cm} = 0.05\text{ m}$
- Charging current: $I = 0.15\text{ A}$ (constant)

---

(a) Capacitance and rate of change of potential difference:

Formula for capacitance of a parallel plate capacitor:
$$C = \frac{\varepsilon_0 A}{d}$$

Area of circular plates:
$$A = \pi r^2 = \pi \times (0.12)^2 = \pi \times 0.0144 = 4.524 \times 10^{-2}\text{ m}^2$$

$$C = \frac{8.854 \times 10^{-12} \times 4.524 \times 10^{-2}}{0.05}$$

$$C = \frac{4.006 \times 10^{-13}}{0.05} \approx 8.0 \times 10^{-12}\text{ F} = 80\text{ pF}$$

Rate of change of potential difference:

Since $Q = CV$, differentiating with respect to time:
$$I = C\frac{dV}{dt}$$

$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-12}}$$

$$\boxed{\frac{dV}{dt} = 1.875 \times 10^{10}\text{ V s}^{-1} \approx 1.9 \times 10^{10}\text{ V s}^{-1}}$$

---

(b) Displacement current across the plates:

The displacement current $i_d$ is given by:
$$i_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d(EA)}{dt} = \varepsilon_0 A \frac{d}{dt}\left(\frac{V}{d}\right) = \frac{\varepsilon_0 A}{d}\frac{dV}{dt} = C\frac{dV}{dt}$$

Therefore:
$$i_d = I = 0.15\text{ A}$$

$$\boxed{i_d = 0.15\text{ A}}$$

The displacement current equals the conduction current.

---

(c) Validity of Kirchhoff's junction rule:

Yes, Kirchhoff's first rule (junction rule) is valid at each plate of the capacitor, provided we generalise the concept of current to include displacement current. At each plate, the conduction current arriving (or leaving) equals the displacement current leaving (or arriving) through the gap between the plates. Thus, the total current (conduction + displacement) is continuous, and the junction rule holds in its generalised form.

Given:
- Radius of plates: $R = 6.0\text{ cm} = 0.06\text{ m}$
- Capacitance: $C = 100\text{ pF} = 100 \times 10^{-12}\text{ F}$
- rms voltage: $V_{rms} = 230\text{ V}$
- Angular frequency: $\omega = 300\text{ rad s}^{-1}$
- Point of interest: $r = 3.0\text{ cm} = 0.03\text{ m}$ from axis

---

(a) rms value of conduction current:

The capacitive reactance is:
$$X_C = \frac{1}{\omega C} = \frac{1}{300 \times 100 \times 10^{-12}} = \frac{1}{3 \times 10^{-8}} = 3.33 \times 10^{7}\text{ }\Omega$$

rms conduction current:
$$I_{rms} = \frac{V_{rms}}{X_C} = \frac{230}{3.33 \times 10^7}$$

$$\boxed{I_{rms} = 6.9 \times 10^{-6}\text{ A} \approx 6.9\text{ }μA}$$

---

(b) Is conduction current equal to displacement current?

Yes. At any instant, the displacement current between the plates equals the conduction current in the external circuit. This is a consequence of Maxwell's modification of Ampere's law — the displacement current $i_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ ensures continuity of current. So:
$$i_d = i_c = 6.9\text{ }μA\text{ (rms)}$$

---

(c) Amplitude of $\mathbf{B}$ at $r = 3.0\text{ cm}$ from the axis:

Since r = 3.0\text{ cm} < R = 6.0\text{ cm}, the point lies inside the capacitor plates.

Using Ampere's law with displacement current, for a circular path of radius $r$ inside the capacitor:
$$B \times 2\pi r = \mu_0 \times i_d(\text{enclosed})$$

The displacement current enclosed within radius $r$ (assuming uniform distribution):
$$i_d(\text{enclosed}) = i_d \times \frac{\pi r^2}{\pi R^2} = i_d \times \frac{r^2}{R^2}$$

Amplitude of current:
$$I_0 = \sqrt{2}\, I_{rms} = \sqrt{2} \times 6.9 \times 10^{-6} = 9.76 \times 10^{-6}\text{ A}$$

Amplitude of $B$:
$$B_0 = \frac{\mu_0 I_0 r}{2\pi R^2}$$

$$B_0 = \frac{4\pi \times 10^{-7} \times 9.76 \times 10^{-6} \times 0.03}{2\pi \times (0.06)^2}$$

$$B_0 = \frac{4\pi \times 10^{-7} \times 9.76 \times 10^{-6} \times 0.03}{2\pi \times 3.6 \times 10^{-3}}$$

$$B_0 = \frac{2 \times 10^{-7} \times 9.76 \times 10^{-6} \times 0.03}{3.6 \times 10^{-3}}$$

$$B_0 = \frac{5.856 \times 10^{-14}}{3.6 \times 10^{-3}} = 1.63 \times 10^{-11}\text{ T}$$

$$\boxed{B_0 \approx 1.63 \times 10^{-11}\text{ T}}$$

Concept: All electromagnetic waves, regardless of their wavelength or frequency, travel through vacuum (free space) with the same speed.

The physical quantity that is the same for all three is the speed in vacuum (free space).

$$c = 3 \times 10^8\text{ m s}^{-1}$$

This is because X-rays, visible light (red light), and radio waves are all electromagnetic waves, and all electromagnetic waves travel with the same speed $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8\text{ m s}^{-1}$ in vacuum, irrespective of their wavelength or frequency.

Direction of field vectors:

In an electromagnetic wave, the electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ are both perpendicular to the direction of propagation and perpendicular to each other.

Since the wave travels along the $z$-direction, the electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ lie in the $xy$-plane (i.e., they are transverse to the $z$-direction). $\mathbf{E}$ and $\mathbf{B}$ are mutually perpendicular to each other and to the $z$-axis.

For example, $\mathbf{E}$ could be along the $x$-direction and $\mathbf{B}$ along the $y$-direction.

---

Wavelength:

Given: Frequency $\nu = 30\text{ MHz} = 30 \times 10^6\text{ Hz} = 3 \times 10^7\text{ Hz}$

Formula: $c = \nu\lambda$

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3 \times 10^7}$$

$$\boxed{\lambda = 10\text{ m}}$$

Given:
- Frequency range: $\nu_1 = 7.5\text{ MHz} = 7.5 \times 10^6\text{ Hz}$ to $\nu_2 = 12\text{ MHz} = 12 \times 10^6\text{ Hz}$
- Speed of electromagnetic waves: $c = 3 \times 10^8\text{ m s}^{-1}$

Formula: $\lambda = \dfrac{c}{\nu}$

Wavelength corresponding to $\nu_1 = 7.5\text{ MHz}$:
$$\lambda_1 = \frac{3 \times 10^8}{7.5 \times 10^6} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40\text{ m}$$

Wavelength corresponding to $\nu_2 = 12\text{ MHz}$:
$$\lambda_2 = \frac{3 \times 10^8}{12 \times 10^6} = 25\text{ m}$$

$$\boxed{\text{Corresponding wavelength band: } 25\text{ m to } 40\text{ m}}$$

Concept: An oscillating charged particle acts as an oscillating electric dipole and produces electromagnetic waves. The frequency of the electromagnetic waves produced equals the frequency of oscillation of the charged particle.

Given: Frequency of oscillation of charged particle $= 10^9\text{ Hz}$

$$\boxed{\text{Frequency of electromagnetic waves produced} = 10^9\text{ Hz} = 1\text{ GHz}}$$

This lies in the microwave region of the electromagnetic spectrum.

Given:
- Amplitude of magnetic field: $B_0 = 510\text{ nT} = 510 \times 10^{-9}\text{ T}$
- Speed of light in vacuum: $c = 3 \times 10^8\text{ m s}^{-1}$

Formula: In an electromagnetic wave, the amplitudes of electric and magnetic fields are related by:
$$\frac{E_0}{B_0} = c$$

$$E_0 = c \times B_0 = 3 \times 10^8 \times 510 \times 10^{-9}$$

$$E_0 = 3 \times 10^8 \times 5.10 \times 10^{-7}$$

$$E_0 = 15.3 \times 10^{1} = 153\text{ N/C}$$

$$\boxed{E_0 = 153\text{ N C}^{-1}}$$

Given:
- $E_0 = 120\text{ N/C}$
- $\nu = 50.0\text{ MHz} = 5.0 \times 10^7\text{ Hz}$
- $c = 3 \times 10^8\text{ m s}^{-1}$

---

(a) Determination of $B_0$, $\omega$, $k$, and $\lambda$:

Amplitude of magnetic field $B_0$:
$$B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 4 \times 10^{-7}\text{ T} = 400\text{ nT}$$

Angular frequency $\omega$:
$$\omega = 2\pi\nu = 2\pi \times 5.0 \times 10^7$$
$$\omega = 3.14 \times 10^8\text{ rad s}^{-1}$$

Wavelength $\lambda$:
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5.0 \times 10^7} = 6.0\text{ m}$$

Wave number $k$:
$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{6.0} = \frac{\pi}{3} \approx 1.05\text{ rad m}^{-1}$$

Summary:
$$\boxed{B_0 = 4 \times 10^{-7}\text{ T},\quad \omega = 3.14 \times 10^8\text{ rad s}^{-1},\quad k \approx 1.05\text{ rad m}^{-1},\quad \lambda = 6.0\text{ m}}$$

---

(b) Expressions for $\mathbf{E}$ and $\mathbf{B}$:

Assuming the wave propagates along the $z$-direction, with $\mathbf{E}$ along the $x$-direction and $\mathbf{B}$ along the $y$-direction:

$$\mathbf{E} = E_0 \sin(kz - \omega t)\,\hat{x}$$
$$\mathbf{E} = 120\sin(1.05\,z - 3.14 \times 10^8\,t)\,\hat{x}\text{ N/C}$$

$$\mathbf{B} = B_0 \sin(kz - \omega t)\,\hat{y}$$
$$\mathbf{B} = 4 \times 10^{-7}\sin(1.05\,z - 3.14 \times 10^8\,t)\,\hat{y}\text{ T}$$

Given:
- Planck's constant: $h = 6.626 \times 10^{-34}\text{ J s}$
- $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$
- Speed of light: $c = 3 \times 10^8\text{ m s}^{-1}$
- Formula: $E = h\nu = \dfrac{hc}{\lambda}$

Photon energies for different regions of the EM spectrum:

Using $E(\text{eV}) = \dfrac{hc}{\lambda \times 1.6 \times 10^{-19}}$:

| Region | Wavelength range | Frequency range | Photon Energy (approx.) |
|---|---|---|---|
| Radio waves | $\sim 10^6\text{ m}$ to $0.1\text{ m}$ | $\sim 10^2$ to $3\times10^9\text{ Hz}$ | $\sim 10^{-9}$ to $10^{-5}\text{ eV}$ |
| Microwaves | $\sim 0.1\text{ m}$ to $1\text{ mm}$ | $\sim 3\times10^9$ to $3\times10^{11}\text{ Hz}$ | $\sim 10^{-5}$ to $10^{-3}\text{ eV}$ |
| Infrared | $\sim 1\text{ mm}$ to $700\text{ nm}$ | $\sim 3\times10^{11}$ to $4\times10^{14}\text{ Hz}$ | $\sim 10^{-3}$ to $2\text{ eV}$ |
| Visible | $\sim 700\text{ nm}$ to $400\text{ nm}$ | $\sim 4\times10^{14}$ to $7\times10^{14}\text{ Hz}$ | $\sim 2$ to $3\text{ eV}$ |
| Ultraviolet | $\sim 400\text{ nm}$ to $1\text{ nm}$ | $\sim 7\times10^{14}$ to $3\times10^{17}\text{ Hz}$ | $\sim 3$ to $10^3\text{ eV}$ |
| X-rays | $\sim 1\text{ nm}$ to $10^{-3}\text{ nm}$ | $\sim 3\times10^{17}$ to $3\times10^{20}\text{ Hz}$ | $\sim 10^3$ to $10^6\text{ eV}$ |
| Gamma rays | < 10^{-3}\text{ nm} | > 3\times10^{20}\text{ Hz} | > 10^6\text{ eV} |

Relation to sources:

- Radio waves (very low photon energy $\sim 10^{-9}$ to $10^{-5}$ eV): produced by oscillating electrons in antennas/circuits — energy transitions are very small.
- Microwaves ($\sim 10^{-3}$ eV): produced by special vacuum tubes; correspond to molecular rotational energy transitions.
- Infrared ($\sim 10^{-3}$ to $2$ eV): produced by hot bodies and molecules; correspond to molecular vibrational transitions.
- Visible light ($\sim 2$ to $3$ eV): produced by atomic electron transitions in the outer shells of atoms.
- Ultraviolet ($\sim 3$ to $10^3$ eV): produced by atoms with high-energy electron transitions (e.g., the sun, arc lamps).
- X-rays ($\sim 10^3$ to $10^6$ eV): produced by inner-shell electron transitions in heavy atoms or by deceleration of fast electrons.
- Gamma rays (> 10^6 eV): produced by nuclear transitions — the highest energy photons, originating from atomic nuclei.

Conclusion: The photon energy scale directly reflects the energy scale of the source. Higher energy sources (nuclear transitions, inner-shell atomic transitions) produce higher frequency, shorter wavelength, higher photon energy radiation.

Given:
- Frequency: $\nu = 2.0 \times 10^{10}\text{ Hz}$
- Amplitude of electric field: $E_0 = 48\text{ V m}^{-1}$
- $c = 3 \times 10^8\text{ m s}^{-1}$

---

(a) Wavelength of the wave:

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{2.0 \times 10^{10}}$$

$$\boxed{\lambda = 1.5 \times 10^{-2}\text{ m} = 0.015\text{ m} = 1.5\text{ cm}}$$

This wave lies in the microwave region.

---

(b) Amplitude of the oscillating magnetic field:

Using the relation $\dfrac{E_0}{B_0} = c$:

$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8}$$

$$\boxed{B_0 = 1.6 \times 10^{-7}\text{ T} = 160\text{ nT}}$$

---

(c) Show that average energy density of $\mathbf{E}$ field equals that of $\mathbf{B}$ field:

Average energy density of the electric field:
$$u_E = \frac{1}{2}\varepsilon_0 E^2$$

For a sinusoidally varying field, the time-average of $E^2$ is $\dfrac{E_0^2}{2}$:
$$\langle u_E \rangle = \frac{1}{2}\varepsilon_0 \cdot \frac{E_0^2}{2} = \frac{\varepsilon_0 E_0^2}{4}$$

Average energy density of the magnetic field:
$$u_B = \frac{B^2}{2\mu_0}$$

Similarly:
$$\langle u_B \rangle = \frac{1}{2\mu_0} \cdot \frac{B_0^2}{2} = \frac{B_0^2}{4\mu_0}$$

Now, using $E_0 = cB_0$ and $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$, so $c^2 = \dfrac{1}{\mu_0\varepsilon_0}$:

$$\langle u_B \rangle = \frac{B_0^2}{4\mu_0} = \frac{1}{4\mu_0} \cdot \frac{E_0^2}{c^2} = \frac{E_0^2}{4\mu_0} \cdot \mu_0\varepsilon_0 = \frac{\varepsilon_0 E_0^2}{4}$$

Therefore:
$$\langle u_E \rangle = \frac{\varepsilon_0 E_0^2}{4} = \langle u_B \rangle$$

$$\boxed{\langle u_E \rangle = \langle u_B \rangle}$$

Hence, the average energy density of the electric field equals the average energy density of the magnetic field in an electromagnetic wave. Proved.