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 Bihar Board Class 12th Physics Solutions Chapter 8 Electromagnetic Waves

Bihar Board Class 12 Physics Electromagnetic Waves Textbook Questions and Answers

Question 1.
Figure 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:
Herd, r = radius of circular plates of the capacitor = 12 cm = 12 x 10-2 m.
d spoling between the plates = 5 cm = 5 x 102m.
I = charging current = 0.15 A.
A = Area of plates = πr2
e0 = 8.854 x 10-12 N-1 C2 nr2
.;. A = Area of plates = πr2
= π x (12 x 102)2 m2

(a) C = Capacitance of capacitor = ?
dV = rate of change of potential difference between the plates = ?
Using the relation, C = ε0 A d, we get

We know that the charge on the plates of the capacitor is given by
q = CV
∴ dqdt=CdVdt (∵ dqdt = I)
or I = C dVdt , or dVdt = IC
∴ dVdt = 0.158.01×10−12 = 1.87 x 1010Vs-1

(b) Id = displacement current across the plates = ?
We know that
Id = dVdt (ε0 øE) = ε0dϕEdt
Where øE = electric flux across the loop = ∮loop E→⋅dS−→
= EA = qε0
Id = ε0ddt(qε0) = ε0ε0⋅dqdt = I
= 0.15 A.
i.e. displacement current is equal to conduction current,
(c) Yes, Kirchhoff’s first rule is valid at each plate of the capacitor if the current is equal to the sum of the conduction current and the displacement current.

Question 2.
A parallel plate capacitor (Fig. below) 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 a.c. 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:
Here R = radius of each circular plate of the capacitor = 6 cm = 6 x 102m.
C = 100 pF = 100 x 10-12 F = 10-10 F.
Er.m.s= r.m.s. value of A.C. supply = 230 V ,
ω = angular frequency of A.C. supply = 300 rad s-1.
(a) Ir.m.s = r.m.s. value of conduction current = ? .
We know that

= 230 x 300 x 10-10 = 69 x 10-7 A = 6.9 x 10-6 A = 6.9 μA.

(b) Yes, I – I0 always for a parallel plate capacitor whether I is steady d.c. or a.c. (oscillating in time) can be proved as follows :

(c) We know that the magnetic’ field at a point between the plates is given by according to Ampere’s Circuital law,

Question 3.
What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 A and radiowaves of wavelength 500 m?
Answer:
X-rays, red light and radio waves all are the electromagnetic waves. They have different wave lengths and frequencies. But the physical quantity which is same for all of these is the velocity of light in vacuum which is denoted by C and is equal to 3 x 108 ms-1.

Question 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:
We know that electromagnetic waves are transverse in nature i.e., the electric field E→ and the magnetic field E→ which constitute the em. waves are always mutually perpendicular to each other as well to the direction of propagation of electromagnetic wave.

Here the electromagnetic wave is travelling along Z-axis. Electric and magnetic field vectors E→ and B→ are normal to each other travelling along Y and X axis respectively (as per figure). Thus we conclude that with E→ and B→ are in XY plane and are perpendicular to each other.

Here v = frequency of e.m. wave 30 MHz = 30 x 106 Hz
C = Velocity of e.m. wave = 3 x 108 ms-1.
λ = ?
Weknow that
C = vλ .
∴ λ = Cv = 3×10830×106 = 10m

Question 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:
Here, frequency band of stations, v = 7.5 MHz to 12 MHz.
= 7.5 x 106 Hz to 12 x 106 Hz.
C = Velocity of radio waves = 3 x 108 ms-1.
λ = Wavelength band of stations = ?
We know that C = vλ .
∴ λ = Cλ

(i) When v = 7.5 MHz = 7.5 x 106 Hz.
Then
3×1087.5×106 = 40 m

(ii) When v = 12 MHz = 12 x 106 Hz.
Then
3×1087.5×106 = 25 m’
∴ Wavelength band of stations is 25 m to 40 m.

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
We know that an accelerated particle radiates electromagnetic waves. An oscillating charge (example accelerated charge) oscillating with certain frequency produces oscillating electric field which produces oscillating magnetic field. These two oscillating fields regenerate each other. The frequency of electro-magnetic waves is equal to the frequency of oscillation of charge particle which is 109Hz here.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Here B0 = amplitude of magnetic field part of e.m. wave in vaccum = 510.nT = 510 x 10-9 T.
C = Velocity of e.m. waves in vacuum = 3 x 108 ms-1.
E0 = amplitude of electric field part of e.m. wave in vacuum = ?
We know that for electromagnetic waves,

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, B0 ω, k, and λ (b) Find expression for E and B.
Answer:
Here, E0 = amplitude of electric field = 120 NC-1
v = frequency of e.m. wave = 50 MHz = 50 x 106 Hz.
C = velocity of em. wave = 3 x 108 ms-1.
(a) (i) B0 = ?,
(ii) k = ?
We know that

(iii) λ is given by
λ = Cυ
= 3×10850×106 = 6m
(iv) ω = 27iv = 2 x 3.14 x 50 x 106 = 3.14 x 108 rad s-1.
Expressions for E→ and B→ are along Y and Z axis respectively.
.’. Equation for E→ is.
Ey−→ = E0 sin (kx – ωt) j^
= 120 sin (1.05 x – 3.14 x 108 t) j^ (NC-1)

Equation of B is :
Bz−→ = B0 sin (kx – ωt) k^ (T)
= 4 x 10-7 sin (1.05 x – 3.14 x 108 1) k (T)

Aliter:
Equation for E→ may be written as

Question 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?
Answer:
Energy of photon is given by the relation, ‘
E = hv
E = hCλ In SI
h = 6.62 x 10-34 Js = Plank’s constant.
C = 3 x 108 ms-1
= Velocity of e.m. wave in free space.

If λ, in metre, E in J. To convert E in electron volt we divide it by 1.6 x 10-19

Different parts of e.m. Spectrum are :
1. y – rays : λ ranging from 10-10 m to less than 10-14 m.

If λ = 10-14 m, Energy = 108 eV
Energy of y rays ranges between 104 to 108 eV.

2. x – rays: X from 10-8 m down to 10-13 m
For λ = 10-8 m

Energy of γ-rays lies 10° to 103 eV.

4. Visible Radiation : γ from 4000 Å to 7000 Å
For λ = 4000 Å = 4 x 10-7m.
As proved in ultra violet Radiation Energy = 3.1 eV = 100 eV

5. ‘Infra red radiations –
λ ranging from 7 x 10-7 m to 7 x 10-4 m.
As in visible radiations for λ = 7 x 10-7 m energy is of the order of
100 eV and λ = 7 x 10-4m engry is 11000 times that is of the order of 10-3eV
Energy of the order of 108 to 10-10 eV

6. Micro waves –
A has range From 1 mm to 0.3 m
For λ = 1 mm or 10-3 m energy is equal to

For A = 0.3 menergy = 4.1 x 10-6eV
Energy is of the order of 10-6 eV.

7. Radio waves : A, has range 1 m to few km.

≈ 10-6 eV and for λ of the order of few km.
Energy of a photon that a source produces indicates the spacing of relevant energy levels of that source. e.g. X = 10-12 m corresponds to photo energy = 1.24 x 106 eV = 1.24 MeV. This indicates that nuclear energy levels (transition between which causes y-ray emission) are typically spaced by 1 MeV or so. Similarly a visible wavelength X = 5 x 10-7 m corresponds to photo energy = 2.5 eV. It means that energy levels i.e., transitions between which give visible radiation are typically spaced by a few eV.

Question 10.
In a plane electromagnetic wave the electric field oscillates sinusoidally at a frequency of 2.0 x 1010 Hz and amplitude 48 V m-1 J
(a) What is the wavelength of a 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 = 3x 108 ms-1].
Answer:
Here, v = frequency of E→ = 2.0 x 1010 Hz ,
E0 = amplitude of electric field = 48 Vm-1
C = 3 x 108 ms-1.
(a) λ = wavelength of the wave = ?
We know that, λ Cv = 3×1082×1010 =1.5 x 10-2 m.

(b) B0 = amplitude of oscillating magnetic field = ?
We know that

(c) Let uE and uB be the energy density of E→ field and B→ field respectively.
To prove uE = uB.
We know that

∴From (i) and (ii), we get thai μE = μBHence proved:

Question 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 x 108rad/s)} i^
(a) What is the direction of propagation?
(b) What is the wavelength X?
(c) What is the frequency v ?
(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:
Here, E→ = (3.1 NC-1) cos [(1.8 rad m-1) y + (5.4 x 108 rad s-1) t] i^ ……..(1)
The given equation is of the type :
E→ = E0 cos (ky + ωt) i^ ……(2)
Comparing (1) and (2), we get
E0 = 3.1 NC-1
k = 1.8 rad m-1
ω = 5.4 x 108 rad s-1.

(a) The electric field propagates along the direction of i^ i.e., along
x – axis and the e.m. wave propagates along Y axis i.e. along i^.

(b) λ = wavelength = ?
We know that co = 2πv = 2πCλ

(c) v = frequency =?

(d) B0 = amplitude of magnetic field part of the wave =?
We know that C = E0 B0

(e) Expression for B = ?
It is given by
B = B0 cos (ky + ωt)
= (1.03 x 10-8T)
cos [(1.8 rad m-1) + (5.4 x 108 rad s-1) t]

Direction of B :
We know that
C→ = E→ x B→
i.e.,C→ acts perpendicular to the plane of E→ and B→ Now C→ acts along Y axis i.e., j^, then B→ will act along Z axis i.e., along j^
B→ = (1.03 x 10-8 T) cos [(1.8 rad m-1) + (5.4 x 108 rad s-1) t] k

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiatidn.
(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:
Here, P = power of bulb = 100 W.
P’ = Power converted into visible radiation = 5% of P
= 5100 x 100 = 5 Js-1

(a) r = 1m, I = average intensity of visible radiation = ?
The bulb as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding sphere is:
A = 4nr2
We know that

Question 13.
Use the formula λmT = 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:
We know that every body at a given temperature T emits radiations of all wavelengths in certain region, thus we can say that it produces a continuous spectrum of wavelengths. For a black body, the wave length corresponding to maximum intensity of radiation at a given temperature T is given by the relation according to Wein’s displacement law as :

Temperatures for other wavelengths can be found similarly. These numbers tell us the tempera hire rangps required for obtaining radiations in different parts of the e.m. spectrum. Thus to obtain visible radiations say having λm = 5000 A = 5000 x 10~10 m = 5 x 10~5 cm, the source should have a temperature, T given by

It should be noted that a body at lower temperature will also produce this wavelength but not with maximum intensity.

Question 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 A – 5896 A [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)].
Answer:
(a) Here, λ = 21 cm. This wavelength of the electromagnetic waves corresponds to radio waves (which have short wavelength or high frequency end).
(b) Here v = 1057 MHz; so the corresponding wavelength is given by

Thus this wavelength also corresponds to short wavelength end i. e., radiowaves.
(c) Here T = 2.7 K
∴ According to wein’s displacement law,

This wavelength corresponds to microwave region of the electromagnetic spectrum.
(d) λ – 5890 Å – 5896 Å which is doublet of sodium light and it lies in the visible region (yellow) of the electromagnetic spectrum.
(e) Here, E = energy = 14.4 KeV = 14.4 x 103 x 1.6 x 10-19 J (∴ 1eV = 1.6x 10-19J)
Also we know that

which corresponds to X-ray region (i.e., soft y-ray region) of the electromagnetic spectrum.

Question 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 it 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 serve ‘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 short-wave bands because ionosphere reflects the waves lying in the frequency range of short wave band i.e., the waves in these bands.

(b) Yes, it is necessary to use satellites for long distance T.V. transmission because T V- signals being of high frequency are not reflected by the ionosphere. Therefore”to reflect the signals back to earth’s surface for long distance transmission, satellites are used.

(c) Optical and radio waves can penetrate the atmosphere where— as X-rays being of much smaller wavelength are absorbed by the earth’s atmosphere which is transparent to visible light and radiowaves and opaque to X-rays. That is why we can work with optical and radio telescope on earth’s surface but X-ray astronomy is possible only from the sakllìtes orbiting above the earth’s atmosphere.

(d) The small ozone layer present on the top of the stratosphereabsorbs the ultraviolet radiations, X-rays, y-rays from the sun along with the cosmic rays which are dangerous and cause genetic damage to the Livthgells and tissues preventing them from reaching the earth’s surface and thus helps in the survival of the life.

(e) Due to green house effect, the temperature of earth’s surface is raised in the presence of the atmosphere. In the absence of the atmosphere, the heat received by the earth during day time will be completely lost during night. Hence the temperature of earth would be lowered because of the absence of green House effect.

(f) The clouds produced by a global nuclear war would perhaps cover most parts of the sky preventing the solar light from reaching many parts of the globe and thus it may cause a severe nuclear winter.