| Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th) | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | ||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 11th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 12th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
Chapter 8 Electromagnetic Waves
Introduction
Electricity and magnetism were initially thought to be separate phenomena. However, 19th-century experiments showed that electric currents produce magnetic fields (Oersted, Ampere). Faraday and Henry then demonstrated the converse: a changing magnetic field induces an electric field (electromagnetic induction).
James Clerk Maxwell unified these phenomena with a set of equations (Maxwell's equations) showing that not only electric currents but also a **time-varying electric field** gives rise to a magnetic field. This was a crucial insight, requiring the concept of **displacement current** to ensure consistency in Ampere's circuital law when applied to situations like a charging capacitor.
From Maxwell's equations emerged the prediction of **electromagnetic waves**, which are coupled, time-varying electric and magnetic fields that propagate through space. The speed of these predicted waves in vacuum was remarkably close to the speed of light, leading to the conclusion that **light is an electromagnetic wave**. This unification of electricity, magnetism, and optics was a major scientific achievement.
Heinrich Hertz experimentally verified the existence of electromagnetic waves in 1885, opening the door to their technological use in communication (radio waves by Marconi and others), which has revolutionised the world.
This chapter discusses the necessity of displacement current, the nature and properties of electromagnetic waves, and the broad range of electromagnetic waves that constitute the electromagnetic spectrum.
Displacement Current
To understand the concept of displacement current, consider a charging parallel plate capacitor. A conduction current $i(t)$ flows through the wires connected to the capacitor plates. We can apply Ampere's circuital law ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$) to find the magnetic field at a point outside the capacitor.
If we choose an Amperian loop around the wire [Fig. 8.1(a)], the conduction current $i(t)$ passes through the area bounded by the loop, and Ampere's law gives $B(2\pi r) = \mu_0 i(t)$, implying a magnetic field exists at point P.
However, if we choose a different surface with the same boundary loop, such as a surface shaped like a pot with its bottom between the capacitor plates [Fig. 8.1(b)] or a tiffin box shape [Fig. 8.1(c)], no conduction current passes through these surfaces (as there is a gap between the capacitor plates). According to the original Ampere's law, the integral $\oint \vec{B} \cdot d\vec{l}$ would be zero in these cases, leading to a contradiction in the value of $\vec{B}$ at point P. This indicated an inconsistency in Ampere's law when applied to time-varying situations.
Maxwell resolved this by proposing the existence of an additional term in Ampere's law: the **displacement current ($i_d$)**. He argued that a changing electric field must also act as a source of magnetic field.
Between the capacitor plates, there is a changing electric field $\vec{E}$ as the capacitor charges. The electric flux $\Phi_E$ through the surface S between the plates [Fig. 8.1(c)] is $\Phi_E = EA$. By Gauss's law, $EA = Q/\epsilon_0$, so $Q = \epsilon_0 E A = \epsilon_0 \Phi_E$.
The conduction current $i = dQ/dt = \frac{d}{dt}(\epsilon_0 \Phi_E) = \epsilon_0 \frac{d\Phi_E}{dt}$. This suggested that $\epsilon_0 \frac{d\Phi_E}{dt}$ acts like a current.
Maxwell defined the **displacement current** $i_d$ as:
$\mathbf{i_d = \epsilon_0 \frac{d\Phi_E}{dt}}$
He proposed that the total current ($i_{total}$) contributing to the magnetic field is the sum of the conduction current ($i_c$) and the displacement current ($i_d$):
$\mathbf{i_{total} = i_c + i_d = i_c + \epsilon_0 \frac{d\Phi_E}{dt}}$
The **Ampere-Maxwell law** (generalised Ampere's circuital law) includes both currents:
$\mathbf{\oint_C \vec{B} \cdot d\vec{l} = \mu_0 (i_c + i_d)_{enc} = \mu_0 \left(i_c + \epsilon_0 \frac{d\Phi_E}{dt}\right)_{enc}}$
where $(i_c + i_d)_{enc}$ is the total current (conduction plus displacement) passing through any surface bounded by the loop C.
Using this generalised law, the contradiction is removed. Outside the capacitor, only conduction current $i_c$ is present, $i_d = 0$. Between the plates, $i_c = 0$, and the changing electric field gives rise to a displacement current $i_d = \epsilon_0 \frac{d\Phi_E}{dt}$. For a charging capacitor, the displacement current between the plates is exactly equal to the conduction current in the wires connected to the plates, $i_d = i_c$. Thus, the total current through any surface bounded by the loop is the same, resolving the inconsistency.
Displacement current has the same physical effect as conduction current in producing a magnetic field. Its existence highlights the symmetry between electric and magnetic phenomena: a changing magnetic field produces an electric field (Faraday's law), and a changing electric field produces a magnetic field (due to displacement current).
Maxwell formulated a set of four equations, known as **Maxwell's equations**, which, along with the Lorentz force formula, describe all classical electromagnetic phenomena. These equations are:
1. $\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\epsilon_0$ (Gauss's Law for electricity)
2. $\oint \vec{B} \cdot d\vec{A} = 0$ (Gauss's Law for magnetism, absence of magnetic monopoles)
3. $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$ (Faraday's Law)
4. $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{c\_enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$ (Ampere-Maxwell Law)
The most significant consequence of these equations is the prediction of electromagnetic waves.
Example 8.1. A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 MW across a 2V battery (Fig. 8.3). Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after t = 10–3 s. (The charge on the capacitor at time t is q (t) = CV [1 – exp (–t/t )], where the time constant t is equal to CR.)
Answer:
Given: Circular plates radius $R_{plate} = 1$ m, Capacitance $C = 1 \text{ nF} = 1 \times 10^{-9}$ F. Resistor $R = 1 \text{ M}\Omega = 1 \times 10^6 \Omega$. Battery voltage $V_0 = 2$ V. Time $t = 10^{-3}$ s. Point P is halfway between center and periphery, so at radius $r = R_{plate}/2 = 1/2 = 0.5$ m.
The charge on the capacitor at time $t$ is $q(t) = CV_0 [1 - \exp(-t/\tau)]$, where $\tau = CR$.
Time constant $\tau = CR = (1 \times 10^{-9} \text{ F}) \times (1 \times 10^6 \Omega) = 1 \times 10^{-3}$ s.
At time $t = 10^{-3}$ s, $t/\tau = 10^{-3}/10^{-3} = 1$.
$q(t=10^{-3}\text{s}) = (1 \times 10^{-9} \text{ F}) \times (2 \text{ V}) [1 - \exp(-1)] = 2 \times 10^{-9} [1 - 1/e] \text{ C}$.
Current in the circuit $i_c(t) = \frac{dq}{dt} = \frac{d}{dt}(CV_0 [1 - \exp(-t/\tau)]) = CV_0 \frac{1}{\tau} \exp(-t/\tau) = \frac{V_0}{R} \exp(-t/\tau)$.
At time $t = 10^{-3}$ s, $i_c(t) = \frac{2 \text{ V}}{1 \times 10^6 \Omega} \exp(-1) = 2 \times 10^{-6} \exp(-1) \text{ A}$.
We want to find the magnetic field at point P between the plates. The only current contributing to the magnetic field between the plates is the displacement current $i_d$. The displacement current between the plates is equal to the conduction current in the wires connected to the plates, $i_d = i_c(t)$.
At $t = 10^{-3}$ s, $i_d = i_c(10^{-3}\text{s}) = 2 \times 10^{-6} \exp(-1) \text{ A} \approx 2 \times 10^{-6} \times 0.368 \text{ A} \approx 0.736 \times 10^{-6} \text{ A}$.
To find the magnetic field $B$ at point P (radius $r=0.5$ m), we apply Ampere-Maxwell law. Choose a circular loop of radius $r$ centered on the axis, parallel to the plates and passing through P. By symmetry, $\vec{B}$ is tangential to this loop and has a constant magnitude $B$ around the loop. The line integral $\oint \vec{B} \cdot d\vec{l} = B (2\pi r)$.
The total current enclosed by this loop is the displacement current passing through the smaller circular area of radius $r$. The displacement current is uniformly distributed across the capacitor plates' area. The density of displacement current is $j_d = i_d / A_{plate} = i_d / (\pi R_{plate}^2)$. The displacement current enclosed by the loop of radius $r$ is $i_{d\_enc} = j_d \times (\text{Area of loop}) = j_d \times (\pi r^2) = \frac{i_d}{\pi R_{plate}^2} \times \pi r^2 = i_d \frac{r^2}{R_{plate}^2}$.
At point P, $r = R_{plate}/2 = 0.5$ m. So $i_{d\_enc} = i_d \frac{(0.5)^2}{(1)^2} = 0.25 i_d = \frac{1}{4} i_d$.
Applying Ampere-Maxwell law inside the capacitor (where $i_c=0$): $\oint \vec{B} \cdot d\vec{l} = \mu_0 (i_c + i_d)_{enc} = \mu_0 i_{d\_enc}$.
$B (2\pi r) = \mu_0 i_d \frac{r^2}{R_{plate}^2}$.
$B = \frac{\mu_0 i_d r^2}{2\pi r R_{plate}^2} = \frac{\mu_0 i_d r}{2\pi R_{plate}^2}$.
At point P, $r = 0.5$ m, $R_{plate} = 1$ m. $i_d = 0.736 \times 10^{-6}$ A. $\mu_0 = 4\pi \times 10^{-7}$ T m/A.
$B = \frac{(4\pi \times 10^{-7}) \times (0.736 \times 10^{-6}) \times 0.5}{2\pi \times (1)^2} = \frac{4\pi \times 0.736 \times 0.5}{2\pi} \times 10^{-13} \text{ T} = 2 \times 0.736 \times 0.5 \times 10^{-13} \text{ T} = 0.736 \times 10^{-13}$ T.
The text's answer is $0.74 \times 10^{-13}$ T, which matches.
The magnetic field at point P after $t = 10^{-3}$ s is approximately $0.74 \times 10^{-13}$ T.
Electromagnetic Waves
Maxwell's equations demonstrate that time-dependent electric and magnetic fields can sustain each other and propagate through space as **electromagnetic waves**. This is a consequence of the symmetry between electric and magnetic phenomena (changing $\vec{B}$ creates $\vec{E}$, changing $\vec{E}$ creates $\vec{B}$).
Sources Of Electromagnetic Waves
Electromagnetic waves are produced by **accelerated charges**. Neither stationary charges (produce only static $\vec{E}$) nor charges in uniform motion (produce static $\vec{B}$ and static $\vec{E}$) radiate electromagnetic waves. An oscillating charge is an example of an accelerated charge; it produces oscillating $\vec{E}$ and $\vec{B}$ fields that propagate as a wave.
The frequency of the electromagnetic wave emitted is equal to the frequency of oscillation of the charge. The energy for the wave comes from the accelerated charge.
Hertz experimentally demonstrated electromagnetic waves in 1887 using low frequencies (radio waves). Higher frequencies (like visible light) are not easily produced by simple circuits. J.C. Bose produced shorter wavelength radio waves. Marconi achieved long-distance communication using radio waves, marking the beginning of wireless communication.
Nature Of Electromagnetic Waves
From Maxwell's equations, the electric field $\vec{E}$ and magnetic field $\vec{B}$ in an electromagnetic wave are **perpendicular to each other and to the direction of propagation** of the wave. They are coupled time-varying fields.
A simple example is a plane electromagnetic wave propagating in the z-direction. The electric field may be along the x-direction, oscillating as $E_x = E_0 \sin(kz - \omega t)$. The magnetic field is then along the y-direction, oscillating as $B_y = B_0 \sin(kz - \omega t)$. $\vec{E}$ and $\vec{B}$ oscillate in phase with the same frequency and wavelength.
Here, $k$ is the wave number, related to the wavelength $\lambda$ by $k = 2\pi/\lambda$. $\omega$ is the angular frequency. The speed of propagation is $v = \omega/k$. Maxwell's equations show that in vacuum, the speed is $c = 1/\sqrt{\mu_0\epsilon_0}$. This value is equal to the speed of light, confirming that light is an electromagnetic wave.
The speed of electromagnetic waves in vacuum is constant ($c \approx 3 \times 10^8$ m/s) and independent of wavelength. In a material medium with permittivity $\epsilon$ and permeability $\mu$, the speed is $v = 1/\sqrt{\mu\epsilon}$. The refractive index of a medium is related to this speed.
Electromagnetic waves are **transverse waves** because the oscillating fields are perpendicular to the direction of propagation. They exhibit wave properties like diffraction, refraction, and polarisation.
Electromagnetic waves carry **energy** and **momentum**. Energy density in a region with $\vec{E}$ is $\frac{1}{2}\epsilon_0 E^2$, and with $\vec{B}$ is $\frac{1}{2\mu_0} B^2$. In an electromagnetic wave, energy is shared equally between electric and magnetic fields. Because they carry momentum, they exert pressure called **radiation pressure**. The momentum $p$ delivered to a perfectly absorbing surface with total energy transfer $U$ is $p = U/c$.
The ability of electromagnetic waves to carry energy over large distances is fundamentally important (light from the sun, radio signals for communication).
Example 8.2. A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ˆj V/m. What is B at this point?
Answer:
Given: Frequency $f = 25 \text{ MHz} = 25 \times 10^6 \text{ Hz}$. Wave travels in free space along the x-direction ($\hat{i}$). At a point, electric field $\vec{E} = 6.3 \hat{j}$ V/m. We need to find the magnetic field $\vec{B}$ at this point.
In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of propagation. The direction of propagation is along the x-axis ($\hat{i}$). The electric field $\vec{E}$ is along the y-axis ($\hat{j}$). Therefore, the magnetic field $\vec{B}$ must be perpendicular to both $\hat{i}$ and $\hat{j}$, meaning it is along the z-axis ($\pm \hat{k}$).
The magnitude of the electric and magnetic fields are related by $E_0 = c B_0$. This also applies to the instantaneous magnitudes in phase: $E = cB$. So, the magnitude of $\vec{B}$ is $B = E/c$. In free space, the speed is $c = 3 \times 10^8$ m/s.
$B = \frac{6.3 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 2.1 \times 10^{-8}$ T.
To find the direction of $\vec{B}$, we use the fact that the direction of propagation is given by the direction of $\vec{E} \times \vec{B}$. The direction of propagation is along $\hat{i}$. The direction of $\vec{E}$ is along $\hat{j}$. So, $\hat{i}$ must be in the direction of $\hat{j} \times \vec{B}$. Using the right-hand rule or vector cross products: $\hat{j} \times \hat{k} = \hat{i}$ and $\hat{j} \times (-\hat{k}) = -\hat{i}$. Since the direction of propagation is $+\hat{i}$, $\vec{B}$ must be along $+\hat{k}$ (positive z-axis).
Therefore, $\vec{B} = B \hat{k} = 2.1 \times 10^{-8} \hat{k}$ T.
The magnetic field at this point is $2.1 \times 10^{-8}$ T in the positive z-direction.
Example 8.3. The magnetic field in a plane electromagnetic wave is given by $By = (2 \times 10^{–7})$ T sin ($0.5\times 10^3x+1.5\times 10^{11}t$). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field.
Answer:
The given magnetic field expression is $B_y = B_0 \sin(kx + \omega t)$, with $B_0 = 2 \times 10^{-7}$ T, $k = 0.5 \times 10^3 \text{ rad/m}$, and $\omega = 1.5 \times 10^{11} \text{ rad/s}$. The general form of a sine wave traveling in the x-direction is $\sin(kx \mp \omega t)$. The plus sign here ($kx + \omega t$) indicates the wave is traveling in the **negative x-direction**. The magnetic field is along the y-axis.
(a) Wavelength and frequency:
Wave number $k = 2\pi/\lambda$. $\lambda = 2\pi/k$.
$\lambda = \frac{2\pi \text{ rad}}{0.5 \times 10^3 \text{ rad/m}} = \frac{2\pi}{500} \text{ m} = \frac{3.14159}{250} \text{ m} \approx 0.012566 \text{ m}$.
$\lambda \approx 1.26$ cm (as given in text). Let's use $2\pi/k$ in calculations.
Angular frequency $\omega = 2\pi f$. Frequency $f = \omega/(2\pi)$.
$f = \frac{1.5 \times 10^{11} \text{ rad/s}}{2\pi \text{ rad}} = \frac{1.5 \times 10^{11}}{6.283} \text{ Hz} \approx 0.2387 \times 10^{11} \text{ Hz} = 2.387 \times 10^{10} \text{ Hz}$.
$2.387 \times 10^{10} \text{ Hz} = 23.87 \text{ GHz}$. The text gives 23.9 GHz, which matches.
Wavelength $\lambda \approx 1.26$ cm. Frequency $f \approx 23.9$ GHz.
(b) Expression for the electric field: The wave travels in the negative x-direction. The magnetic field is along the y-axis. The electric field is perpendicular to both direction of propagation ($\hat{i}$) and magnetic field ($\hat{j}$). So it is along the z-axis ($\pm \hat{k}$). The direction of propagation is $\vec{E} \times \vec{B}$. Here propagation is along $-\hat{i}$. $\vec{B}$ is along $\hat{j}$. So $-\hat{i}$ must be in the direction of $\vec{E} \times \hat{j}$. If $\vec{E}$ is along $\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$. This works. If $\vec{E}$ is along $-\hat{k}$, $(-\hat{k}) \times \hat{j} = \hat{i}$. This does not work. So $\vec{E}$ is along the positive z-axis.
The magnitude of the electric field amplitude $E_0 = c B_0$. In free space, $c = 3 \times 10^8$ m/s.
$E_0 = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-7} \text{ T}) = 6 \times 10^1 \text{ V/m} = 60 \text{ V/m}$.
Since $\vec{E}$ and $\vec{B}$ oscillate in phase in free space, the expression for $\vec{E}$ will have the same sinusoidal function as $\vec{B}$.
$\vec{E} = E_z \hat{k}$. $E_z = E_0 \sin(kx + \omega t)$.
$\mathbf{E_z = 60 \sin (0.5 \times 10^3x + 1.5 \times 10^{11}t)}$ V/m.
The expression for the electric field is $\vec{E}(x,t) = [60 \sin (0.5 \times 10^3x + 1.5 \times 10^{11}t)] \hat{k}$ V/m.
Example 8.4. Light with an energy flux of 18 W/cm$^2$ falls on a nonreflecting surface at normal incidence. If the surface has an area of 20 cm$^2$, find the average force exerted on the surface during a 30 minute time span.
Answer:
Given: Energy flux (intensity) $S = 18 \text{ W/cm}^2$. Incidence is normal. Surface area $A = 20 \text{ cm}^2$. Time duration $\Delta t = 30 \text{ minutes} = 30 \times 60 = 1800 \text{ s}$. The surface is non-reflecting (perfect absorber).
Total energy falling on the surface during $\Delta t$ is $U = S \times A \times \Delta t$.
$U = (18 \text{ W/cm}^2) \times (20 \text{ cm}^2) \times (1800 \text{ s}) = 360 \times 1800 \text{ J} = 648000 \text{ J} = 6.48 \times 10^5 \text{ J}$.
For a perfect absorber, the momentum delivered to the surface is $p = U/c$, where $c = 3 \times 10^8$ m/s is the speed of light.
$p = \frac{6.48 \times 10^5 \text{ J}}{3 \times 10^8 \text{ m/s}} = 2.16 \times 10^{-3} \text{ kg m/s}$.
The average force exerted on the surface is the rate of momentum transfer, $F = \frac{\Delta p}{\Delta t}$. The initial momentum of the light absorbed is $p_{initial} = p$. The final momentum of the light is zero. The change in momentum of the light is $p_{final} - p_{initial} = 0 - p = -p$. By momentum conservation, the change in momentum of the surface is $+p$. So $\Delta p_{surface} = p$.
$F_{average} = \frac{\Delta p_{surface}}{\Delta t} = \frac{p}{\Delta t} = \frac{2.16 \times 10^{-3} \text{ kg m/s}}{1800 \text{ s}}$.
$F_{average} = \frac{2.16}{1800} \times 10^{-3} \text{ N} = 0.0012 \times 10^{-3} \text{ N} = 1.2 \times 10^{-6} \text{ N}$.
The average force exerted on the surface is $1.2 \times 10^{-6}$ N.
If the surface were a perfect reflector: The light bounces back, reversing its momentum. The initial momentum is $p_{initial} = p = U/c$. The final momentum is $p_{final} = -p = -U/c$. The change in momentum of the light is $p_{final} - p_{initial} = -p - p = -2p$. By momentum conservation, the change in momentum of the surface is $+2p$. The force is the rate of change of momentum.
$F_{reflector} = \frac{2p}{\Delta t} = 2 \times F_{absorber} = 2 \times 1.2 \times 10^{-6} \text{ N} = 2.4 \times 10^{-6} \text{ N}$.
If the surface were a perfect reflector, the average force would be twice as large, $2.4 \times 10^{-6}$ N.
Example 8.5. Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Answer:
Given: Bulb power output $P_{bulb} = 100$ W. Efficiency $= 2.5\%$. Distance $r = 3$ m. Assume it's a point source.
Power converted to radiation $P_{rad} = P_{bulb} \times \text{Efficiency} = 100 \text{ W} \times 0.025 = 2.5 \text{ W}$.
Since the bulb is a point source, it radiates uniformly in all directions. At distance $r$, the radiation spreads over a sphere of area $A = 4\pi r^2$.
$A = 4\pi (3 \text{ m})^2 = 36\pi \text{ m}^2$.
The intensity (average energy flux) $S_{avg}$ at this distance is the power per unit area:
$S_{avg} = \frac{P_{rad}}{A} = \frac{2.5 \text{ W}}{36\pi \text{ m}^2} \approx \frac{2.5}{113.1} \text{ W/m}^2 \approx 0.0221 \text{ W/m}^2$. The text gives 0.022 W/m$^2$.
In an electromagnetic wave, the energy is shared equally between the electric and magnetic fields. The average energy density is $u_{avg} = u_{E,avg} + u_{B,avg}$. The average energy density of the electric field is $u_{E,avg} = \frac{1}{2} \epsilon_0 E_{rms}^2$. The average energy density of the magnetic field is $u_{B,avg} = \frac{1}{2\mu_0} B_{rms}^2$. And $u_{E,avg} = u_{B,avg}$.
The intensity $S_{avg}$ is related to the average energy density by $S_{avg} = u_{avg} \times c = (u_{E,avg} + u_{B,avg}) c = 2 u_{E,avg} c = 2 \times \frac{1}{2} \epsilon_0 E_{rms}^2 c = \epsilon_0 E_{rms}^2 c$.
$E_{rms}^2 = \frac{S_{avg}}{\epsilon_0 c}$. Using $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$ and $c = 3 \times 10^8 \text{ m/s}$.
$E_{rms}^2 = \frac{0.022}{(8.854 \times 10^{-12}) \times (3 \times 10^8)} = \frac{0.022}{26.562 \times 10^{-4}} = \frac{0.022}{0.0026562} \approx 8.28$ (V/m)$^2$.
$E_{rms} = \sqrt{8.28} \approx 2.877$ V/m. The text gives 2.9 V/m.
The amplitude of the electric field $E_0 = \sqrt{2} E_{rms} = \sqrt{2} \times 2.877 \approx 1.414 \times 2.877 \approx 4.069$ V/m. The text gives 4.07 V/m.
Now calculate the magnetic field. $B_{rms} = E_{rms}/c$.
$B_{rms} = \frac{2.877 \text{ V/m}}{3 \times 10^8 \text{ m/s}} \approx 0.959 \times 10^{-8} \text{ T} = 9.59 \times 10^{-9} \text{ T}$. The text gives $9.6 \times 10^{-9}$ T.
The amplitude of the magnetic field $B_0 = \sqrt{2} B_{rms} = \sqrt{2} \times 9.59 \times 10^{-9} \approx 1.414 \times 9.59 \times 10^{-9} \approx 13.56 \times 10^{-9} \text{ T} = 1.356 \times 10^{-8} \text{ T}$. The text gives $1.4 \times 10^{-8}$ T.
The rms electric field strength is about 2.9 V/m, peak electric field is about 4.07 V/m. The rms magnetic field strength is about $9.6 \times 10^{-9}$ T, peak magnetic field is about $1.4 \times 10^{-8}$ T.
Electromagnetic Spectrum
The **electromagnetic spectrum** is the classification of electromagnetic waves according to their frequency or wavelength. All electromagnetic waves travel at the speed of light $c$ in vacuum, but they have different wavelengths and frequencies ($\lambda = c/f$). Different regions of the spectrum are given names based on how the waves are produced or detected and their typical interactions with matter. There are no sharp boundaries between these regions.
Regions of the electromagnetic spectrum, from longest wavelength (lowest frequency) to shortest wavelength (highest frequency):
Radio Waves
Wavelengths > 0.1 m. Frequencies from ~500 kHz to ~1000 MHz (1 GHz). Produced by accelerated charges in wires (electronic circuits, antennas). Used in radio and television broadcasting (AM, FM, TV bands), cellular communication (UHF). Long-distance transmission via reflection from the ionosphere (short waves) or line-of-sight (TV, FM, cellular).
Microwaves
Wavelengths ~1 mm to 0.1 m. Frequencies ~1 GHz to 300 GHz. Produced by special vacuum tubes (klystrons, magnetrons). Used in radar systems (aircraft navigation, speed guns) due to their short wavelengths allowing good resolution. Also used in microwave ovens (frequency tuned to water molecule resonance for efficient heating).
Infrared Waves
Wavelengths ~700 nm to 1 mm. Frequencies ~4 $\times 10^{14}$ Hz to $4 \times 10^{11}$ Hz. Produced by hot bodies and molecules (vibrations of atoms/molecules). Often called **heat waves** because they are strongly absorbed by many molecules (especially water, CO$_2$) increasing thermal motion. Used in physical therapy (infrared lamps), thermal imaging (night vision, Earth satellites for monitoring temperature, crops), remote controls (LEDs emitting infrared). Play a role in the greenhouse effect (trapping Earth's re-radiated thermal radiation).
Visible Rays
Wavelengths ~400 nm (violet) to 700 nm (red). Frequencies ~7 $\times 10^{14}$ Hz to 4 $\times 10^{14}$ Hz. The part of the spectrum detected by the human eye. Produced by electrons in atoms changing energy levels. Provides visual information about the surroundings. Different organisms may be sensitive to different visible ranges or extend into IR/UV.
Ultraviolet Rays
Wavelengths ~0.6 nm to 400 nm. Frequencies ~5 $\times 10^{16}$ Hz to $7.5 \times 10^{14}$ Hz. Produced by special lamps and hot bodies (like the Sun). Most solar UV is absorbed by the ozone layer in the stratosphere. UV radiation can be harmful; causes tanning (melanin production). Used in LASIK eye surgery (precision beams), water purifiers (killing germs). Ozone layer depletion by CFCs is a major concern.
X-Rays
Wavelengths ~10$^{-4}$ nm to 10 nm. Frequencies ~3 $\times 10^{18}$ Hz to $3 \times 10^{16}$ Hz. Produced by bombarding metal targets with high-energy electrons (in X-ray tubes) or transitions of inner shell electrons in heavy atoms. Used as diagnostic tools (medical imaging) and in cancer treatment (radiotherapy). Can damage living tissues, requiring careful use and shielding.
Gamma Rays
Wavelengths < 10$^{-10}$ m to less than 10$^{-14}$ m. Frequencies > 3 $\times 10^{18}$ Hz. Highest frequency/energy part of the spectrum. Produced in nuclear reactions and emitted by radioactive nuclei. Used in medicine (sterilisation, cancer treatment) and industry (gauging, inspection). Highly penetrating and ionising, requiring heavy shielding.
Table 8.1 summarises the production and detection of these different types of electromagnetic waves. Regions overlap.
| Type | Wavelength range | Production | Detection |
|---|---|---|---|
| Radio | > 0.1 m | Rapid acceleration and decelerations of electrons in aerials | Receiver’s aerials |
| Microwave | 0.1 m to 1 mm | Klystron valve or magnetron valve | Point contact diodes |
| Infra-red | 1 mm to 700 nm | Vibration of atoms and molecules | Thermopiles, Bolometer, Infrared photographic film |
| Light | 700 nm to 400 nm | Electrons in atoms emit light when they move from one energy level to a lower energy level | The eye, Photocells, Photographic film |
| Ultraviolet | 400 nm to 1 nm | Inner shell electrons in atoms moving from one energy level to a lower level | Photocells, Photographic film |
| X-rays | 1 nm to 10$^{-3}$ nm | X-ray tubes or inner shell electrons | Photographic film, Geiger tubes, Ionisation chamber |
| Gamma rays | <10$^{-3}$ nm | Radioactive decay of the nucleus | (Same as X-rays) |
Summary
This chapter introduces electromagnetic waves, their origin, properties, and the electromagnetic spectrum.
- Maxwell's equations unify electricity and magnetism. They show that a changing electric field induces a magnetic field (displacement current) and a changing magnetic field induces an electric field (Faraday's law).
- **Displacement current** ($i_d = \epsilon_0 d\Phi_E/dt$) is due to a time-varying electric field and acts as a source of magnetic field like conduction current. Ampere's law is generalised to Ampere-Maxwell law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (i_c + i_d)_{enc}$.
- **Electromagnetic waves** are coupled, time-varying $\vec{E}$ and $\vec{B}$ fields that propagate through space. Produced by accelerated charges (e.g., oscillating charge). Frequency of wave equals frequency of source.
- $\vec{E}$ and $\vec{B}$ are perpendicular to each other and propagation direction ($\vec{E} \times \vec{B}$). They oscillate in phase.
- Speed in vacuum $c = 1/\sqrt{\mu_0\epsilon_0}$. Speed in medium $v = 1/\sqrt{\mu\epsilon}$. $\lambda = c/f$.
- EM waves are transverse, carry energy and momentum (exert radiation pressure).
- **Electromagnetic spectrum:** Classification by frequency/wavelength. Regions (increasing $\lambda$): gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves. Different interactions with matter.
Exercises
Questions covering the concepts of displacement current, Maxwell's equations, sources and properties of electromagnetic waves, calculations involving E and B field amplitudes, wavelength, frequency, speed, energy, momentum, and radiation pressure of EM waves. Also includes questions on the electromagnetic spectrum and applications of different types of EM waves.
Exercises
Question 8.1. Figure 8.6 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 0.15A.
(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.
Answer:
Question 8.2. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 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.
Answer:
Question 8.3. What physical quantity is the same for X-rays of wavelength $10^{–10}$ m, red light of wavelength 6800 Å and radiowaves of wavelength 500m?
Answer:
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?
Answer:
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?
Answer:
Question 8.6. A charged particle oscillates about its mean equilibrium position with a frequency of $10^9$ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
Question 8.7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0$ = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Question 8.8. Suppose that the electric field amplitude of an electromagnetic wave is $E_0$ = 120 N/C and that its frequency is $\nu$ = 50.0 MHz. (a) Determine, $B_0$, $\omega$, k, and $\lambda$. (b) Find expressions for E and B.
Answer:
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?
Answer:
Question 8.10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}$ Hz and amplitude 48 V m$^{–1}$.
(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. [$c = 3 \times 10^8 \text{ m s}^{–1}$.]
Answer:
ADDITIONAL EXERCISES
Question 8.11. Suppose that the electric field part of an electromagnetic wave in vacuum is $\textbf{E} = \{(3.1 \text{ N/C}) \cos [(1.8 \text{ rad/m}) y \ $$ + (5.4 \times 10^6 \text{ rad/s})t]\}\hat{\textbf{i}}$.
(a) What is the direction of propagation?
(b) What is the wavelength $\lambda$ ?
(c) What is the frequency $\nu$ ?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Answer:
Question 8.12. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:
Question 8.13. Use the formula $\lambda_m T = 0.29$ cmK to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:
Question 8.14. Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å - 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in $^{57}$Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
Answer:
Question 8.15. Answer the following questions:
(a) Long distance radio broadcasts use short-wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?
Answer: