NCERT Class 12 Physics — Electromagnetic Waves, Chapter 8 Exercises. All 10 questions solved.

The exercises at the end of Chapter 8 fall into two groups. Questions 8.1 and 8.2 are about the displacement current — the idea Maxwell added to Ampere’s law so that current is continuous through the gap of a charging capacitor. Questions 8.3 to 8.10 are about the waves that idea predicts: their speed, the directions of their fields, how the electric and magnetic amplitudes are related, and the energy carried by a single photon in each part of the spectrum. Almost all the arithmetic runs on three results:

• Displacement current — $i_d = \varepsilon_0\,\dfrac{\text{d}\Phi_E}{\text{d}t}$, where $\Phi_E$ is the electric flux; it enters the Ampere–Maxwell law alongside the conduction current, $\oint\vec B\cdot\text{d}\vec l = \mu_0(i_c + i_d)$.
• Waves in vacuum — every electromagnetic wave travels at $c = 1/\sqrt{\mu_0\varepsilon_0} = 3\times10^8\ \text{m s}^{-1}$, so $c = \nu\lambda$, and $\omega = 2\pi\nu$, $k = 2\pi/\lambda$.
• Field amplitudes — $E_0 = cB_0$, with $\vec E$ and $\vec B$ perpendicular to each other and to the direction of travel.

Key insight. Two facts carry the whole exercise. A changing electric field acts exactly like a current, so the current that stops at one plate of a capacitor carries on across the gap as displacement current (8.1, 8.2). And in vacuum every electromagnetic wave, from X-rays to radio waves, travels at the same speed $c$; what distinguishes them is only frequency, and frequency fixes the wavelength, the ratio of the fields and the energy of each photon (8.3 to 8.10).

Question 8.1

Figure 8.5 shows a capacitor made of two circular plates each of radius $12\ \text{cm}$, and separated by $5.0\ \text{cm}$. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15\ \text{A}$. (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.

Question 8.1 (a)

Solution. The plates are large compared with their separation, so the capacitor is treated as an ideal parallel-plate capacitor. The plate area is $A = \pi(0.12)^2 = 4.52\times10^{-2}\ \text{m}^2$, and

$$C = \frac{\varepsilon_0A}{d} = \frac{(8.854\times10^{-12})(4.52\times10^{-2})}{0.050} = 8.0\times10^{-12}\ \text{F} = 8.0\ \text{pF}$$

The charge on the plates is $Q = CV$, and the current is the rate at which charge arrives, so $i = \text{d}Q/\text{d}t = C\,\text{d}V/\text{d}t$. With the current constant:

$$\frac{\text{d}V}{\text{d}t} = \frac{i}{C} = \frac{0.15}{8.0\times10^{-12}} = 1.87\times10^{10}\ \text{V s}^{-1}$$

Question 8.1 (b)

Solution. No charge crosses the gap, but the electric field in the gap grows as the plates charge up, and a changing electric flux is a displacement current, $i_d = \varepsilon_0\,\text{d}\Phi_E/\text{d}t$. Ignoring the fringing at the edges, the field is confined between the plates, so $\Phi_E = EA$ with $E = Q/\varepsilon_0A$. Then

$$i_d = \varepsilon_0A\,\frac{\text{d}E}{\text{d}t} = \varepsilon_0A\cdot\frac{1}{\varepsilon_0A}\frac{\text{d}Q}{\text{d}t} = \frac{\text{d}Q}{\text{d}t} = i$$

The displacement current across the gap is exactly equal to the conduction current in the wires.

Question 8.1 (c)

Solution. Yes, provided “current” means the total of conduction and displacement currents. A conduction current of $0.15\ \text{A}$ flows into one plate and none flows out of it across the gap, so for conduction current alone the junction rule fails: charge piles up on the plate. But a displacement current of $0.15\ \text{A}$ leaves the plate across the gap, as part (b) shows. Counting both, what flows into each plate equals what flows out.

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 $230\ \text{V}$ 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 $\vec B$ at a point $3.0\ \text{cm}$ from the axis between the plates.

Question 8.2 (a)

Solution. The capacitor’s reactance is $X_C = 1/\omega C$, so the rms current is

$$I = \frac{V}{X_C} = V\omega C = (230)(300)(100\times10^{-12}) = 6.9\times10^{-6}\ \text{A}$$

Question 8.2 (b)

Solution. Yes. The argument in Question 8.1(b), $i_d = \varepsilon_0A\,\text{d}E/\text{d}t = \text{d}Q/\text{d}t = i$, holds at every instant; nowhere does it need the current to be steady. So when the conduction current oscillates, the displacement current oscillates with it, equal to it at every moment.

Question 8.2 (c)

Solution. Apply the Ampere–Maxwell law to a circle of radius $r = 3.0\ \text{cm}$, centred on the axis and lying between the plates. No conduction current passes through it, only displacement current. The field between the plates is uniform, so the displacement current is spread evenly over the plate area $\pi R^2$, and the circle encloses the fraction $\pi r^2/\pi R^2$ of it:

$$B\,(2\pi r) = \mu_0\,i_d\,\frac{r^2}{R^2} \quad\Longrightarrow\quad B = \frac{\mu_0}{2\pi}\,\frac{r}{R^2}\,i_d$$

This holds at every instant, so $B$ oscillates in phase with $i_d$, and the amplitude of $B$ follows from the amplitude of the current. Since $i_d = i$, that amplitude is $i_0 = \sqrt2\times6.9\ \mu\text{A} = 9.76\ \mu\text{A}$:

$$B_0 = (2\times10^{-7})\,\frac{0.030}{(0.060)^2}\,(9.76\times10^{-6}) = 1.63\times10^{-11}\ \text{T}$$

Question 8.3

What physical quantity is the same for X-rays of wavelength $10^{-10}\ \text{m}$, red light of wavelength $6800\ \text{Å}$ and radiowaves of wavelength $500\ \text{m}$?

Solution. All three are electromagnetic waves, and in vacuum every electromagnetic wave travels at the same speed, $c = 1/\sqrt{\mu_0\varepsilon_0}$, whatever its wavelength. Their frequencies are very different — $\nu = c/\lambda$ gives $3\times10^{18}\ \text{Hz}$, $4.4\times10^{14}\ \text{Hz}$ and $6\times10^{5}\ \text{Hz}$ — but the product $\nu\lambda$ is the same for all three.

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 $30\ \text{MHz}$, what is its wavelength?

Solution. Electromagnetic waves are transverse: $\vec E$ and $\vec B$ are both perpendicular to the direction of travel, and to each other. With the wave along $z$, both fields lie in the $x$-$y$ plane, at right angles to one another, and $\vec E\times\vec B$ points along $+z$. The wavelength is

$$\lambda = \frac{c}{\nu} = \frac{3\times10^8}{30\times10^6} = 10\ \text{m}$$

Question 8.5

A radio can tune in to any station in the $7.5\ \text{MHz}$ to $12\ \text{MHz}$ band. What is the corresponding wavelength band?

Solution. Wavelength and frequency are inversely related, $\lambda = c/\nu$, so each end of the frequency band gives one end of the wavelength band, with the order reversed:

$$\lambda_1 = \frac{3\times10^8}{7.5\times10^6} = 40\ \text{m}, \qquad \lambda_2 = \frac{3\times10^8}{12\times10^6} = 25\ \text{m}$$

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?

Solution. An accelerating charge radiates, and an oscillating charge sets up electric and magnetic fields that change at exactly the rate it oscillates. The wave it sends out therefore has the same frequency as the oscillation.

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?

Solution. In an electromagnetic wave in vacuum the field amplitudes are in the fixed ratio $E_0/B_0 = c$:

$$E_0 = cB_0 = (3\times10^8)(510\times10^{-9}) = 153\ \text{N/C}$$

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 $\vec B$.

Question 8.8 (a)

Solution. Each quantity follows from one defining relation:

$$B_0 = \frac{E_0}{c} = \frac{120}{3\times10^8} = 4.00\times10^{-7}\ \text{T} = 400\ \text{nT}$$

$$\omega = 2\pi\nu = 2\pi(50.0\times10^6) = 3.14\times10^8\ \text{rad s}^{-1}$$

$$\lambda = \frac{c}{\nu} = \frac{3\times10^8}{50.0\times10^6} = 6.00\ \text{m}, \qquad k = \frac{2\pi}{\lambda} = \frac{2\pi}{6.00} = 1.05\ \text{rad m}^{-1}$$

Question 8.8 (b)

Solution. The question does not fix the directions, so choose them: let the wave travel along $+x$ with $\vec E$ along $y$. Then $\vec B$ must be along $z$, so that $\vec E\times\vec B$ points along $+x$ ($\hat{\jmath}\times\hat k = \hat{\imath}$). The two fields oscillate in phase, each as $\sin(kx – \omega t)$:

$$\vec E = (120\ \text{N/C})\,\sin\!\big[(1.05\ \text{rad/m})\,x – (3.14\times10^8\ \text{rad/s})\,t\big]\,\hat{\jmath}$$

$$\vec B = (400\ \text{nT})\,\sin\!\big[(1.05\ \text{rad/m})\,x – (3.14\times10^8\ \text{rad/s})\,t\big]\,\hat k$$

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?

Solution. With $\nu = c/\lambda$, the photon energy is $E = hc/\lambda$. Dividing by $e = 1.6\times10^{-19}\ \text{C}$ converts joules to electronvolts, and with $h = 6.63\times10^{-34}\ \text{J s}$:

$$E = \frac{hc}{e\lambda} = \frac{(6.63\times10^{-34})(3\times10^8)}{(1.6\times10^{-19})\,\lambda} = \frac{1.24\times10^{-6}}{\lambda}\ \text{eV}\quad(\lambda\text{ in metres})$$

So a $1\ \text{m}$ wave carries photons of $1.24\times10^{-6}\ \text{eV}$, and every factor of ten shorter in wavelength is a factor of ten more in energy. Applying this to the wavelength ranges of Table 8.1 in the chapter:

A source emits a photon when it drops from one energy level to a lower one, so the photon energy tells us how far apart the source’s energy levels are. Gamma rays, with $\lambda = 10^{-12}\ \text{m}$ giving $1.24\ \text{MeV}$, come from nuclei, whose energy levels are about an MeV apart. X-rays, at keV energies, come from the tightly bound inner-shell electrons of heavy atoms. Visible light, for example $\lambda = 5\times10^{-7}\ \text{m}$ giving $2.5\ \text{eV}$, and ultraviolet come from the outer electrons of atoms, whose levels are a few eV apart. Infra-red, from about $10^{-3}\ \text{eV}$ up to an eV or so, matches the smaller energies of vibrating atoms and molecules. At radio energies the steps are so tiny that the radiation is simply produced by electrons accelerated in an aerial, with no quantum transition involved.

Question 8.10

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times10^{10}\ \text{Hz}$ and amplitude $48\ \text{V m}^{-1}$. (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 $\vec E$ field equals the average energy density of the $\vec B$ field. [$c = 3\times10^8\ \text{m s}^{-1}$.]

Question 8.10 (a)

Solution. The wave travels at $c$, so

$$\lambda = \frac{c}{\nu} = \frac{3\times10^8}{2.0\times10^{10}} = 1.5\times10^{-2}\ \text{m}$$

Question 8.10 (b)

Solution. The magnetic amplitude is fixed by the electric one:

$$B_0 = \frac{E_0}{c} = \frac{48}{3\times10^8} = 1.6\times10^{-7}\ \text{T}$$

Question 8.10 (c)

Solution. The energy densities of the two fields are $u_E = \tfrac12\varepsilon_0E^2$ and $u_B = B^2/2\mu_0$. Both fields vary as $\sin(kx – \omega t)$ and the average of $\sin^2$ over a cycle is $\tfrac12$, so the averages are

$$\bar u_E = \tfrac14\varepsilon_0E_0^2, \qquad \bar u_B = \frac{B_0^2}{4\mu_0}$$

The two are equal because the amplitudes are tied together. Put $B_0 = E_0/c$ into $\bar u_B$ and use $c^2 = 1/\mu_0\varepsilon_0$:

$$\bar u_B = \frac{E_0^2}{4\mu_0c^2} = \frac{E_0^2\,\mu_0\varepsilon_0}{4\mu_0} = \tfrac14\varepsilon_0E_0^2 = \bar u_E$$

For this wave each is $\tfrac14(8.854\times10^{-12})(48)^2 = 5.1\times10^{-9}\ \text{J m}^{-3}$.

Common mistakes

• Question 8.1(a): putting the radius in centimetres. The area is $\pi(0.12)^2\ \text{m}^2$. Leaving $r$ and $d$ in centimetres gives a capacitance a hundred times too large, since $A/d$ has the dimension of length.
• Question 8.1(c): saying the junction rule fails. It fails only if conduction current is counted alone. The current that stops at a plate carries on across the gap as displacement current, and the total is continuous.
• Question 8.2(c): using the whole displacement current. A circle of radius $3\ \text{cm}$ inside plates of radius $6\ \text{cm}$ encloses only a quarter of the plate area, so only a quarter of $i_d$. Using all of it makes $B$ four times too large.
• Question 8.2(c): using the rms current for the amplitude. The amplitude of $B$ comes from the amplitude of the current, $\sqrt2\times6.9\ \mu\text{A}$, not the rms value.
• Question 8.3: answering “frequency” or “energy”. Those differ enormously between X-rays and radio waves. The speed in vacuum is the only one of these quantities they share.
• Question 8.4: placing $\vec E$ or $\vec B$ along $z$. Electromagnetic waves are transverse, so neither field has a component along the direction of travel.
• Question 8.8(a): writing $k = 1/\lambda$. The wave number carries a $2\pi$, $k = 2\pi/\lambda$, just as $\omega = 2\pi\nu$.
• Question 8.10(c): forgetting the average. The instantaneous densities are $\tfrac12\varepsilon_0E^2$ and $B^2/2\mu_0$. Averaging introduces the same factor of $\tfrac12$ into both, so it does not change the equality, but it does change the value of each.

Practise next

• Chapter 9, Ray Optics and Optical Instruments — the visible part of this spectrum, followed as rays through mirrors, lenses and instruments.
• Chapter 7, Alternating Current — the capacitive reactance behind Question 8.2(a).