Chapter 8 Electromagnetic Waves Class 12 Physics

Question 8.1

Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

(a) Capacitance & dV/dt

Area $A = \pi r^2 = \pi (0.12)^2 \approx 0.045 \text{ m}^2$. Distance $d = 0.05 \text{ m}$.

$$C = \frac{\epsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 0.045}{0.05} \approx 80.1 \times 10^{-12} \text{ F} \approx 80 \text{ pF}$$

Since $q = CV \implies I = C \frac{dV}{dt}$:

$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{80 \times 10^{-12}} = 1.87 \times 10^9 \text{ V/s}$$

(b) Displacement Current

Displacement current $I_d$ is always equal to the conduction current $I_c$ in the connecting wires.

$$I_d = I_c = 0.15 \text{ A}$$

(c) Kirchhoff’s Rule

Yes, if we include both conduction current and displacement current. At the plate, $I_{entering} = I_c$ and $I_{leaving} = I_d$. Since $I_c = I_d$, the junction rule holds.

Result: (a) 80 pF, $1.87 \times 10^9 \text{ V/s}$, (b) 0.15 A, (c) Yes.

Question 8.2

A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius $R = 6.0 \text{ cm}$ has a capacitance $C = 100 \text{ pF}$. The capacitor is connected to a 230 V ac supply with a (angular) frequency of $300 \text{ rad s}^{-1}$.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

(a) RMS Conduction Current

$I_{rms} = V_{rms} / X_C = V_{rms} \times (\omega C)$.

$$I_{rms} = 230 \times 300 \times 100 \times 10^{-12} = 6.9 \times 10^{-6} \text{ A} = 6.9 \mu\text{A}$$

(b) Equality

Yes, conduction current is always equal to displacement current in magnitude.

(c) Amplitude of B

For a point $r < R$ inside the plates, using Ampere-Maxwell Law: $B = \frac{\mu_0 r}{2\pi R^2} I_0$.

Peak current $I_0 = \sqrt{2} I_{rms} = 1.414 \times 6.9 \mu\text{A} \approx 9.76 \mu\text{A}$.

$$B_0 = \frac{4\pi \times 10^{-7} \times 0.03}{2\pi \times (0.06)^2} \times 9.76 \times 10^{-6} \approx 1.63 \times 10^{-11} \text{ T}$$

Result: (a) 6.9 $\mu$A, (b) Yes, (c) $1.63 \times 10^{-11} \text{ T}$

Question 8.3

What is the wavelength of electromagnetic waves in free space whose frequency is $2.0 \times 10^{15} \text{ Hz}$? Which part of the electromagnetic spectrum does this wavelength belong to?

Calculation

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{2 \times 10^{15}} = 1.5 \times 10^{-7} \text{ m} = 1500 \text{ Å}$$

[Image of Electromagnetic Spectrum]

Spectrum Identification

Wavelengths between $10^{-7} \text{ m}$ to $4 \times 10^{-7} \text{ m}$ typically fall in the Ultraviolet (UV) region.

Wavelength: $1.5 \times 10^{-7} \text{ m}$ (UV Region).

Question 8.4

A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Direction

In an EM wave, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and also to the direction of propagation ($\vec{k}$). Since propagation is along Z, $\vec{E}$ and $\vec{B}$ lie in the XY plane. For example, if $E$ is along X, $B$ will be along Y.

Wavelength

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{30 \times 10^6} = \frac{300}{30} = 10 \text{ m}$$

$\vec{E}, \vec{B}$ in XY plane; $\lambda = 10 \text{ m}$.

Question 8.5

A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Calculation

For $\nu_1 = 7.5 \text{ MHz}$:

$$\lambda_1 = \frac{3 \times 10^8}{7.5 \times 10^6} = \frac{300}{7.5} = 40 \text{ m}$$

For $\nu_2 = 12 \text{ MHz}$:

$$\lambda_2 = \frac{3 \times 10^8}{12 \times 10^6} = \frac{300}{12} = 25 \text{ m}$$

Wavelength Band: 40 m to 25 m.

Question 8.6

A charged particle oscillates about its mean equilibrium position with a frequency of $10^9 \text{ Hz}$. What is the frequency of the electromagnetic waves produced by the oscillator?

Concept

An oscillating charge produces electromagnetic waves of the same frequency as the oscillation.

Frequency = $10^9 \text{ Hz}$.

Question 8.7

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 \text{ nT}$. What is the amplitude of the electric field part of the wave?

Formula

$E_0 = c \times B_0$.

Calculation

$$E_0 = 3 \times 10^8 \times 510 \times 10^{-9} = 1530 \times 10^{-1} = 153 \text{ V/m}$$

Amplitude $E_0 = 153 \text{ N/C}$ (or V/m).

Question 8.8

Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 \text{ N/C}$ and that its frequency is $\nu = 50.0 \text{ MHz}$.
(a) Determine, $B_0, \omega, k$, and $\lambda$.
(b) Find expressions for $\vec{E}$ and $\vec{B}$.

(a) Parameters

$$B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 40 \times 10^{-8} = 400 \text{ nT}$$ $$\omega = 2\pi\nu = 2 \times 3.14 \times 50 \times 10^6 = 3.14 \times 10^8 \text{ rad/s}$$ $$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{50 \times 10^6} = 6.0 \text{ m}$$ $$k = \frac{2\pi}{\lambda} = \frac{2\pi}{6} = 1.05 \text{ rad/m}$$

(b) Expressions

Assuming the wave propagates in +x direction with E along Y-axis:

$$\vec{E} = E_0 \sin(kx – \omega t) \hat{j} = 120 \sin(1.05 x – 3.14 \times 10^8 t) \hat{j} \text{ N/C}$$ $$\vec{B} = B_0 \sin(kx – \omega t) \hat{k} = 400 \times 10^{-9} \sin(1.05 x – 3.14 \times 10^8 t) \hat{k} \text{ T}$$

Question 8.9

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h\nu$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Concept

Energy $E = h\nu$. To get eV, divide by $1.6 \times 10^{-19}$.

$$E(\text{eV}) = \frac{6.63 \times 10^{-34} \times \nu}{1.6 \times 10^{-19}} \approx (4.14 \times 10^{-15}) \nu$$

Examples

Radio (1 MHz): $E \approx 10^{-9} \text{ eV}$. Source: Oscillating circuits.
Visible (Yellow, $5 \times 10^{14}$ Hz): $E \approx 2 \text{ eV}$. Source: Electron transitions.
X-rays ($10^{18}$ Hz): $E \approx 4 \text{ keV}$. Source: Inner shell electrons / Deceleration.
Gamma ($10^{20}$ Hz): $E \approx 0.4 \text{ MeV}$. Source: Nuclear transitions.

Relationship: Higher photon energy corresponds to more energetic source mechanisms (Nuclear > Atomic > Electronic Circuits).

Question 8.10

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \text{ Hz}$ and amplitude 48 V/m.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field.

(a) Wavelength

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{2 \times 10^{10}} = 1.5 \times 10^{-2} \text{ m} = 1.5 \text{ cm}$$

(b) Magnetic Amplitude

$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \text{ T}$$

(c) Energy Density Equality

$u_E = \frac{1}{2}\epsilon_0 E_{rms}^2$ and $u_B = \frac{1}{2\mu_0} B_{rms}^2$.

Using $E_{rms} = c B_{rms}$ and $c = 1/\sqrt{\mu_0 \epsilon_0}$:

$$u_E = \frac{1}{2}\epsilon_0 (c B_{rms})^2 = \frac{1}{2}\epsilon_0 \left( \frac{1}{\mu_0 \epsilon_0} \right) B_{rms}^2 = \frac{1}{2\mu_0} B_{rms}^2 = u_B$$

Thus, $u_E = u_B$.

NCERT Solutions