#### Page No 285:

#### 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 0.15 A.

**(a) ** Calculate
the capacitance and the rate of charge 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:

Radius
of each circular plate, *r* = 12 cm = 0.12 m

Distance
between the plates, *d* = 5 cm = 0.05 m

Charging
current, *I* = 0.15 A

Permittivity
of free space, =
8.85 × 10^{−12} C^{2} N^{−1} m^{−2}

**(a)** Capacitance
between the two plates is given by the relation,

*C*

Where,

*A* = Area of each plate

Charge
on each plate, *q* = *CV*

Where,

V = Potential difference across the plates

Differentiation
on both sides with respect to time (*t*) gives:

Therefore,
the change in potential difference between the plates is 1.87 ×10^{9} V/s.

**(b)** The displacement current across the plates is the same as
the conduction current. Hence, the displacement current, *i*_{d} is 0.15 A.

**(c) **Yes

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

#### Page No 286:

#### 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:

Radius
of each circular plate, *R* = 6.0 cm = 0.06 m

Capacitance
of a parallel plate capacitor, *C* = 100 pF = 100 × 10^{−12} F

Supply voltage, *V* = 230 V

Angular frequency, *ω* = 300 rad s^{−1}

**(a)** Rms value of conduction current, *I*

Where,

*X*_{C} = Capacitive reactance

∴ *I* = *V* × *ωC*

= 230 × 300 × 100 × 10^{−12}

= 6.9 × 10^{−6} A

= 6.9 μA

Hence, the rms value of conduction current is 6.9 μA.

**(b)** Yes,
conduction current is equal to displacement current.

**(c)** Magnetic
field is given as:

*B*

Where,

μ_{0} = Free space permeability

*I*_{0} = Maximum value of current =

*r* = Distance between the plates from the axis = 3.0 cm = 0.03 m

∴*B*

= 1.63 × 10^{−11} T

Hence,
the magnetic field at that point is 1.63 × 10^{−11} T.

#### 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 500 m?

#### Answer:

The
speed of light (3 × 10^{8} m/s) in a vacuum is the same for all wavelengths. It is independent
of the wavelength in the vacuum.

#### 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:

The
electromagnetic wave travels in a vacuum along the z-direction. The
electric field (*E*)
and the magnetic field (*H*)
are in the *x*–*y* plane. They are mutually perpendicular.

Frequency
of the wave, ν
= 30 MHz = 30 × 10^{6} s^{−1}

Speed of light in a
vacuum, *c* = 3 × 10^{8} m/s

Wavelength of a wave is given as:

#### 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:

A
radio can tune to minimum frequency, *ν*_{1} = 7.5 MHz= 7.5 × 10^{6} Hz

Maximum
frequency, *ν*_{2} = 12 MHz = 12 × 10^{6} Hz

Speed
of light, *c* = 3 × 10^{8} m/s

Corresponding
wavelength for *ν*_{1} can be calculated as:

Corresponding
wavelength for *ν*_{2} can be calculated as:

Thus, the wavelength band of the radio is 40 m to 25 m.

#### 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:

The
frequency of an electromagnetic wave produced by the oscillator is
the same as that of a charged particle oscillating about its mean
position i.e., 10^{9} Hz.

#### 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:

Amplitude of magnetic field of an electromagnetic wave in a vacuum,

*B*_{0} = 510 nT = 510 × 10^{−9} T

Speed of light in a
vacuum, *c* = 3 × 10^{8} m/s

Amplitude of electric field of the electromagnetic wave is given by the relation,

*E* = *cB*_{0}

=
3 × 10^{8 }× 510 × 10^{−9 }=
153 N/C

Therefore, the electric field part of the wave is 153 N/C.

#### Question 8.8:

Suppose
that the electric field amplitude of an electromagnetic wave is *E*_{0} = 120 N/C and that its frequency is *ν* = 50.0 MHz. (a) Determine, *B*_{0}*, *ω, *k, *and λ. (b) Find expressions
for **E **and **B**.

#### Answer:

Electric field
amplitude, *E*_{0} = 120 N/C

Frequency of source, *ν* = 50.0 MHz = 50 × 10^{6} Hz

Speed of light, *c* = 3 × 10^{8 }m/s

**(a)** Magnitude of magnetic field strength is given as:

Angular frequency of source is given as:

*ω* = 2π*ν*

= 2π × 50 × 10^{6}

= 3.14 × 10^{8} rad/s

Propagation constant is given as:

Wavelength of wave is given as:

**(b)** Suppose the wave is propagating in the positive *x* direction. Then, the electric field vector will be in the positive *y* direction and the magnetic field vector will be in the positive *z* direction. This is because all three vectors are mutually
perpendicular.

Equation of electric field vector is given as:

And, magnetic field vector is given as:

#### Question 8.9:

The
terminology of different parts of the electromagnetic spectrum is
given in the text. Use the formula *E *= *h**ν* (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:

Energy of a photon is given as:

Where,

*h* = Planck’s
constant = 6.6 × 10^{−34} Js

*c* = Speed of
light = 3 × 10^{8} m/s

λ = Wavelength of radiation

The given table lists
the photon energies for different parts of an electromagnetic
spectrum for different*λ*.

λ (m) 10 ^{3}1 10 ^{−3}10 ^{−6}10 ^{−8}10 ^{−10}10 ^{−12}*E*(eV)12.375 × 10 ^{−10}12.375 × 10 ^{−7}12.375 × 10 ^{−4}12.375 × 10 ^{−1}12.375 × 10 ^{1}12.375 × 10 ^{3}12.375 × 10 ^{5}

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

#### Question 8.10:

In
a plane electromagnetic wave, the electric field oscillates
sinusoidally at a frequency of 2.0 × 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 × 10^{8} m s^{−1}.]

#### Answer:

Frequency of the
electromagnetic wave, *ν* = 2.0 × 10^{10} Hz

Electric field
amplitude, *E*_{0} = 48 V m^{−1}

Speed of light, *c* = 3 × 10^{8} m/s

**(a)** Wavelength
of a wave is given as:

**(b)** Magnetic
field strength is given as:

**(c)** Energy
density of the electric field is given as:

And, energy density of the magnetic field is given as:

Where,

∈_{0} = Permittivity of free space

μ_{0} = Permeability of free space

We
have the relation connecting *E* and *B* as:

*E* = *cB* … (1)

Where,

… (2)

Putting equation (2) in equation (1), we get

Squaring both sides, we get

#### Page No 287:

#### Question 8.11:

Suppose
that the electric field part of an electromagnetic wave in vacuum is **E **=
{(3.1 N/C) cos [(1.8 rad/m) *y *+
(5.4 × 10^{6} rad/s)*t*]} .

**(a) ** What is the direction of
propagation?

**(b) ** What is the wavelength λ?

**(c) ** What is the frequency *ν*?

**(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:

**(a)** From the given electric field vector, it can be inferred
that the electric field is directed along the negative *x* direction. Hence, the direction of motion is along the negative *y *direction i.e., .

**(b)** It is given that,

The
general equation for the electric field vector in the positive *x* direction can be written as:

On comparing equations (1) and (2), we get

Electric
field amplitude, *E*_{0} = 3.1 N/C

Angular
frequency, *ω* = 5.4 × 10^{8} rad/s

Wave
number, *k* = 1.8 rad/m

Wavelength, = 3.490 m

**(c)** Frequency of wave is given as:

**(d)** Magnetic field strength is given as:

Where,

*c* = Speed of light = 3 × 10^{8} m/s

**(e)** On observing the given vector field, it can be observed
that the magnetic field vector is directed along the negative z
direction. Hence, the general equation for the magnetic field vector
is written as:

#### 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 1 m from the bulb?

**(b) ** at a distance of 10 m?

Assume that the radiation is emitted isotropically and neglect reflection.

#### Answer:

Power rating of bulb, *P* = 100 W

It is given that about 5% of its power is converted into visible radiation.

Power of visible radiation,

Hence, the power of visible radiation is 5W.

**(a)** Distance of
a point from the bulb, *d* = 1 m

Hence, intensity of radiation at that point is given as:

**(b)** Distance of
a point from the bulb, *d*_{1} = 10 m

Hence, intensity of radiation at that point is given as:

#### Question 8.13:

Use the formula λ_{m} *T*=
0.29 cm K to obtain the characteristic temperature ranges for
different parts of the electromagnetic spectrum. What do the numbers
that you obtain tell you?

#### Answer:

A body at a particular temperature produces a continous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,

Where,

λ_{m} = maximum wavelength

*T* = temperature

Thus, the temperature for different wavelengths can be obtained as:

For *λ*_{m} = 10^{−4} cm;

For *λ*_{m} = 5 ×10^{−5} cm;

For *λ*_{m} = 10^{−6} cm; and so on.

The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.

#### 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:

**(a)** Radio waves; it belongs to the short wavelength end of the
electromagnetic spectrum.

**(b)** Radio waves; it belongs to the short wavelength end.

**(c)** Temperature, *T* = 2.7 °K

λ_{m }is given by Planck’s law as:

This wavelength corresponds to microwaves.

**(d)** This is the yellow light of the visible spectrum.

**(e)** Transition energy is given by the relation,

*E* = *h**ν*

Where,

*h *= Planck’s constant* = *6.6 × 10^{−34} Js

ν = Frequency of radiation

Energy, *E *= 14.4 K eV

This corresponds to X-rays.

#### 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 radio telescopes 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:

**(a) **Long distance radio broadcasts use shortwave bands because
only these bands can be refracted by the ionosphere.

**(b) **It is necessary to use satellites for long distance TV
transmissions because television signals are of high frequencies and
high energies. Thus, these signals are not reflected by the
ionosphere. Hence, satellites are helpful in reflecting TV signals.
Also, they help in long distance TV transmissions.

**(c) **With reference to X-ray astronomy, X-rays are absorbed by
the atmosphere. However, visible and radio waves can penetrate it.
Hence, optical and radio telescopes are built on the ground, while
X-ray astronomy is possible only with the help of satellites orbiting
the Earth.

**(d)** The small ozone layer on the top of the atmosphere is
crucial for human survival because it absorbs harmful ultraviolet
radiations present in sunlight and prevents it from reaching the
Earth’s surface.

**(e) **In theabsenceof an atmosphere, there would
be no greenhouse effect on the surface of the Earth. As a result, the
temperature of the Earth would decrease rapidly, making it chilly and
difficult for human survival.

**(f) **A global nuclear war on the surface of the Earth would
have disastrous consequences. Post-nuclear war, the Earth will
experience severe winter as the war will produce clouds of smoke that
would cover maximum parts of the sky, thereby preventing solar light
form reaching the atmosphere. Also, it will lead to the depletion of
the ozone layer.

**NCERT Solutions for Class 12 Physics Chapters**

**Physics Part I**

- Chapter 1 – Electric Charges And Fields
- Chapter 2 – Electrostatic Potential And Capacitance
- Chapter 3 – Current Electricity
- Chapter 4 – Moving Charges And Magnetism
- Chapter 5 – Magnetism And Matter
- Chapter 6 – Electromagnetic Induction
- Chapter 7 – Alternating Current
- Chapter 8 – Electromagnetic Waves

**Physics Part II**

- Chapter 1 – Ray Optics And Optical Instruments
- Chapter 2 – Wave Optics
- Chapter 3 – Dual Nature Of Radiation And Matter
- Chapter 4 – Atoms
- Chapter 5 – Nuclei
- Chapter 6 – Semiconductor Electronics: Materials Devices and Simple Circuits
- Chapter 7 – Communication Systems

**NCERT Solutions for Class 12-science:**