# NCERT Solutions For Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves: Maxwell introduced the term displacement current to remove inconsistency in Ampers law. The solutions of NCERT class 12 physics chapter 8 electromagnetic waves starts with the question related to the concept of displacement current. Other topics covered in the chapter are Maxwell’s equations and electromagnetic waves and their properties and applications, the equation of electromagnetic wave and the relation between the magnetic field, electric field, and speed of light. Problems related to the electromagnetic spectrum are also discussed in CBSE NCERT solutions for class 12 physics chapter 8 electromagnetic waves. The solutions of NCERT helps in testing the knowledge about the concepts studied in the chapter. Different electromagnetic waves and their wavelength are listed in the below table. Questions based on this table are also discussed in the NCERT solutions for class 12 physics chapter 8 electromagnetic waves.

## Range of wavelength:

 Electromagnetic wave type Wavelength range Radio wave Greater than 0.1 m Microwave 0.1m to 1 mm Infra-red wave 1mm to 700 nm Light 700 nm to 400 nm Ultraviolet 400 nm to 1nm X-rays 1nm to 0.001nm Gamma rays Less than 0.001nm

## NCERT solutions for class 12 physics chapter 8 electromagnetic waves exercise:

Calculate the capacitance and the rate of change of potential
difference between the plates.

Radius of the discs(r) = 12cm

Area of the discs(A) = $\pi r^{2}=\pi (0.12)^{2}=.045m^{2}$

Permittivity, $\epsilon _{0}$=$8.85\times 10^{-12}C^{2}N^{-1}m^{2}$

Distance between the two discs = 5cm=0.05m

$Capacitance= \frac{\varepsilon _{0}A}{d}$

=$\frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}$

=$8.003\times 10^{-12}F$

=8.003 pF

$Rate\ of \ change\ of \ potential =\frac{dv}{dt}$

But

$V =\frac{Q}{C}$

Therefore,

$Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C}$

$=\frac{0.15A}{8.003pF}$

=$1.87\times 10^{10}Vs^{-1}$

## 1(b). 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.

Obtain the displacement current across the plates.

The value of the displacement current will be the same as that of the conduction current which is given to be 0.15A. Therefore displacement current is also 0.15A.

Is Kirchhoff’s first rule (junction rule) valid at each plate of the
capacitor? Explain.

Yes, Kirchoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first but once we take into consideration both the conduction and displacement current the Kirchoff's first rule will hold good.

## 2.(a) 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 $300 rad s ^{-1}$.

What is the rms value of the conduction current?

Capacitance(C) of the parallel plate capacitor = 100 pF

Voltage(V) = 230 V

Angular Frequency (ω) = $300\ rad\ s^{-1}$

$Rms\ Current(I) =\frac{Voltage}{Capacitive\ Reactance}$

$Capacitive \ Reactance(X_{c})= \frac{1}{C\omega }=\frac{1}{100\ pF\times 300\ rad\ s^{-1}}$

$X_{c}= 3.33\times 10^{7} \Omega$

$I=\frac{V}{X_{c}}=\frac{230V}{3.33\times 10^{7}\Omega }= 6.9\times 10^{-6}A$

RMS value of conduction current is $6.9\mu A$

Is the conduction current equal to the displacement current?

Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere – Maxwell Law.

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

We know the Ampere - Maxwells Law,

$\oint B\cdot \vec{dl} = \mu _{0}(i_{c}+\varepsilon _{0}\frac{d\phi _{E}}{dt})$

Between the plates conduction current $i_{c}=0$.

For a loop of radius r smaller than the radius of the discs,

$\mu _{0}\varepsilon _{0}\frac{\mathrm{d} \phi _{E}}{\mathrm{d} t}=\mu _{0}i_{d}\frac{\pi r^{2}}{\pi R^{2}}=\mu _{0}i_{d}\frac{ r^{2}}{ R^{2}}$

$B(2\pi r)=\mu _{0}i_{d}\frac{r^{2}}{R^{2}}$

$B=\mu _{0}i_{d}\frac{r}{2\pi R^{2}}$

Since we have to find the amplitude of the magnetic field we won't use the RMS value but the maximum value of current.

$\\i_{max}=\sqrt{2}\times i_{rms}\\ i_{max}=\sqrt{2}\times6.9\mu A\\ i_{max}=9.76\mu A$

$\\B_{amp}=\mu _{0}i_{max}\frac{r}{2\pi R^{2}}\\ B_{amp}=\frac{4\pi\times 10^{-7}\times 9.33\times 10^{-6}\times (.03) }{2\pi\times (0.06)^{2} }\\ B_{amp}=1.63\times 10^{-11}T$

The amplitude of B at a point 3.0 cm from the axis between the plates is  $1.63\times 10^{-11}T$.

The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to $3\times 10^{8}ms^{-1}$ (speed of light in vacuum).

Since the electromagnetic wave travels along the z-direction its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.

Frequency of wave = 30 MHz

Wavelength=

$\frac{Speed\ of\ light}{Frequency}=\frac{3\times 10^{8}}{30\times 10^{6}}$

$=10m$.

Frequency range = 7.5 MHz to 12 MHz

Speed of light = $3\times 10^{8}ms^{-1}$

Wavelength corresponding to the frequency of 7.5 MHz =

$\frac{3\times 10^{8}ms^{-1}}{7.5\times 10^{6}Hz}$

$=40m$

Wavelength corresponding to the frequency of 12 MHz =

$\frac{3\times 10^{8}ms^{-1}}{12\times 10^{6}Hz}$

$=25m$

The corresponding wavelength band is 25m to 40m.

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore electromagnetic waves produced will have a frequency of  109 Hz.

## 7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 n T$. What is the amplitude of the electric field part of the wave?

Magnetic Field (B0)=510 nT = 510 $\times$ 10-9 T

Speed of light(c) = 3 $\times$ 108 ms-1

Electric Field = B $\times$ c

= 510 $\times$ 10-9 T   $\times$ 3 $\times$ 108 ms-1

= 153 NC-1

E0 = 120 NC-1

$\nu =50.0\ MHz$

(a)

$Magnetic\ Field \ amplitude(B0) =\frac{E_{0}}{c}$

=$\frac{120}{3\times 10^{8}}$

$=400 nT$

Angular frequency ($\omega$)  = 2$\pi \nu$

$=2$$\times \pi \times 50$$\times$$10^{6}$

=3.14$\times$10rad s-1

Propagation constant(k)

$\frac{2\pi }{\lambda }$

=$\frac{2\pi \nu }{\lambda\nu }$

=$\frac{\omega }{c}$

=$\frac{3.14\times 10^{8} }{3\times 10^{8}}$

Wavelength($\lambda$) = $\frac{c}{\nu }$

=$\frac{3\times 10^{8}}{50\times 10^{6}}$$= 6 m$

Assuming the electromagnetic wave propagates in the positive z-direction the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction as they are mutually perpendicular and $E_{0}\times B_{0}$  gives the direction of propagation of the wave.

$\vec{E}=E_{0}sin(kx-\omega t)\hat{i}$

$= 120sin(1.05x-3.14\times 10^{8} t)NC^{-1}\hat{i}$

$\vec{B}=B_{0}sin(kx-\omega t)\hat{j}$

$= 400sin(1.05x-3.14\times 10^{8} t)\ nT\hat{j}$

## 9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (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?

$E=h\nu=\frac{hc}{\lambda}$

$E=h\nu=\frac{6.6\times 10^{-34}\times3\times10^8}{\lambda \times1.6\times10^{-19}}eV=\frac{12.375\times 10^{-7}}{\lambda}eV$

Now substitute the different range of wavelength in the electromagnetic spectrum to obtain the energy

 EM wave one wavelength is taken from the range Energy in eV Radio 1 m $1.2375\times10^{-6}$ Microwave 1 mm $1.2375\times10^{-3}$ Infra-red 1000 nm 1.2375 Light 500 nm 2.475 Ultraviolet 1nm 1237.5 X-rays 0.01 nm 123750 Gamma rays 0.0001 nm 12375000

What is the wavelength of the wave?

Frequency($\nu$) =20$\times$1010 Hz

E0 = 48 Vm-1

$Wavelength(\lambda) =\frac{Speed\ of\ light(c)}{Frequency (\nu )}$

=$\frac{3\times 10^{8}}{20\times 10^{10}}$

$=1.5 mm$

What is the amplitude of the oscillating magnetic field?

The amplitude of the oscillating magnetic field(B0) = $\frac{E_{0}}{c}$

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

$=160 nT$

Show that the average energy density of the E field equals the average energy density of the B field. $[c = 3 \times 10^8 m s^{-1}. ]$

The average energy density of the Electric field(UE

$\frac{1}{2}\epsilon E^{2}$

=$\frac{1}{2}\epsilon (Bc)^{2}$                 (as E=Bc)

=$\frac{1}{2}\epsilon \frac{B^{2}}{\mu \epsilon }$                      $(c=\frac{1}{\sqrt{\mu \epsilon }})$

=U

Therefore the average energy density of the electric field is equal to the average energy density of the Magnetic field.

$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$

$E = \left \{ ( 3.1 N/C )\sin [ (\frac{\pi }{2}-(( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t ))] \right \} \hat i$

The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.

$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$

What is the wavelength?

From the equation of the wave given we can infer k = 1.8 rad m-1

$Wavelength(\lambda )=\frac{2\pi }{k}$

$=\frac{2\pi }{1.8}$

$=3.49 m$

## 11 (c) Suppose that the electric field part of an electromagnetic wave in vacuum is

$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$

What is the frequency n?

From the given equation of the electric field we can infer angular frequency($\omega$) = 5.4$\times$10rad s-1

Frequency($\nu$) = $\frac{\omega }{2\pi }$

=$\frac{5.4\times10^{8}}{2\pi }$

=8.6$\times$10Hz

=86 MHz

$E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s t )] \right \} \hat i$
What is the amplitude of the magnetic field part of the wave?

From the given equation of the electric field, we can infer Electric field amplitude (E0) =3.1 NC-1

$Magnetic field amplitude (B_0) =\frac{E_{0}}{c}$

$=\frac{3.1}{3\times 10^{8}}$

=1.03$\times$10-7 T

## 11(e)  Suppose that the electric field part of an electromagnetic wave in vacuum is  $E = \left \{ ( 3.1 N/C )\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat i$

Write an expression for the magnetic field part of the wave.

As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction the magnetic field vector must be directed along the negative z-direction. ($-\hat{i}\times -\hat{k}=-\hat{j}$)

Therefore, $\vec{B}= \left \{ B_{0}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \} \hat k$

$\vec{B}= \left \{ 1.03\times 10^{-7}\cos [ ( 1.8 rad /m)y + ( 5.4 \times 10 ^6 rad /s )t] \right \}T \hat k$

Assume that the radiation is emitted isotropically and neglect reflection.

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 1m

$=\frac{5}{4\pi (1)^{2}}$

=0.398 Wm-2

Total power which is converted into visible radiation = 5% of 100W = 5W

The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.

Intensity for a sphere of radius 10 m

$=\frac{5}{4\pi (10)^{2}}$

=3.98$\times$10-3 Wm-2

 EM wave one wavelength is taken from the range Temperature $T=\frac{0.29}{\lambda}$ Radio 100 cm $2.9\times 10^{-3}K$ Microwave 0.1cm 2.9 K Infra-red 100000ncm 2900K Light 50000 ncm 5800K Ultraviolet 100ncm $2.9\times10^{6}K$ X-rays 1 ncm $2.9\times10^8K$ Gamma rays 0.01 nm $2.9\times10^{10}K$

These numbers indicate the temperature ranges required for obtaining radiations in different parts of the spectrum

## 14 b) 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.

1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).

Frequency($\nu$)=1057 MHz

Wavelength($\lambda$$=\frac{c}{\nu }$

$=\frac{3\times 10^{8}}{1057\times 10^{6} }$

=0.283 m

=28.3 cm

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].

Using formula, $\lambda _{m}T=0.29\ cmK$

$\lambda =\frac{0.29}{2.7}$

= 0.107cm

Microwaves.

## 14 d) 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.

$5890 \dot{A} - 5896 \dot{A}$   [double lines of sodium]

Visible light.

14.4 keV [energy of a particular transition in $^{57} Fe$ nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].

E=14.4 keV

Wavelength($\lambda$) = $\frac{hc}{E}$

=$\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{14.4\times 10^{3}\times 1.6\times 10^{-19}}$

= 0.85 A0

X-rays

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

As TV signals are of high frequencies they are not reflected by the ionosphere and therefore satellites are to be used to reflect them.

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiations coming from the sun which are very harmful to humans.

If the earth did not have an atmosphere its average surface temperature be lower than what it is now as in the absence of atmosphere there would be no greenhouse effect.

The use of nuclear weapons would cause the formation of smoke clouds preventing the light from the sun reaching earth surface and it will also deplete the atmosphere and therefore stopping the greenhouse effect and thus doubling the cooling effect.

## NCERT solutions for class 12 physics chapter wise:

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Significance of NCERT solutions for class 12 physics chapter 8 electromagnetic induction:

• For CBSE board exam 4% questions are expected from this chapter.
• CBSE NCERT solutions for class 12 physics chapter 8 electromagnetic waves will give a clear idea about how to use the formulas studied in the chapter.
• For exams like NEET and JEE main one or two questions are asked from the chapter.
• Questions based on wave equation are expected from the chapter. The relation between electric field, magnetic field and speed of light is also important. There are questions based on these concepts in the NCERT solutions for class 12 physics chapter 8 electromagnetic waves.