Electromagnetic Waves Class 12th Physics Part I CBSE Solution

Class 12^th Physics Part I CBSE Solution

Exercises

1. Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and…
2. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm…
3. What physical quantity is the same for X-rays of wavelength 10-10 m, red light of…
4. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about…
5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the…
6. A charged particle oscillates about its mean equilibrium position with a frequency of 10^9…
7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is…
8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and…
9. The terminology of different parts of the electromagnetic spectrum is given in the text.…
10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency…

Additional Exercises

1. Suppose that the electric field part of an electromagnetic wave in vacuum is (a) What is…
2. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the…
3. Use the formula λmT = 0.29 cm K to obtain the characteristic temperature ranges for…
4. Given below are some famous numbers associated with electromagnetic radiations in…
5. Long distance radio broadcasts use short-wave bands. Why? Answer the following question:…
6. It is necessary to use satellites for long distance TV transmission. Why? Answer the…
7. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only…
8. The small ozone layer on top of the stratosphere is crucial for human survival. Why?…
9. If the earth did not have an atmosphere, would its average surface temperature be higher…
10. Some scientists have predicted that a global nuclear war on the earth would be followed by…

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Exercises

Question 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:

Given: Radius of circular plates, r = 12 cm = 0.12 m

Distance between the circular plates, d = 5.0 cm = 0.05 m

Current, I = 0.15 A

(a) Capacitance between the circular plates is given by the following expression:

Where A is the area of plate and is given by the relation: A = π r²

And Absolute permittivity ϵ₀ = 8.854 ×10^-12 C²N^-1m^-2

Now putting the values in above relation

⇒ 

⇒ C = 8.0032 ×10^-12 F

Rate of change of potential difference between the plates can be calculated as follows:

We know that charge on each plate can be find out using the following expression:

q = CV .... (i)

where C is the capacitance between the two plates

V is the potential difference between the plates

Capacitance between the two plates C = 8.0032 ×10^-12 F

Converting in pF, we get C = 80.032 pF

Now differentiating equation (i) w.r.t time

Since 

⇒ 

⇒ 

(b) The displacement current will be equal to the conduction current across the plates. Therefore, displacement current I_d = I_c = 0.15 A.

(c) If we take the total current as sum of conduction current and displacement current then the Kirchhoff’s first rule is valid at each plate of the capacitor.

Question 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:

Given: Radius of circular plates R = 6.0 cm = 0.06 m

Capacitance C = 100 pF = 100 × 10^-12 F

Voltage V = 230 V

Angular frequency, ω = 300 rad s^-1

The figure is:

(a) The root mean square value of the conduction current is given by the following relation:

I_rms = V/X_c

Where I_rms is the root mean square value of the conduction current

V is the voltage

X_c is the capacitive reactance and is given by X_c = 1/ω_c

⇒ I_{r. m. s} = V × ω_C

Substituting the values we get

I_rms = 230 V×300 rad s^-1×100×10^-12F

I_rms = 6.9 × 10^-6A= 6.9 μ A

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

(c) Amplitude of magnetic field B at a point 3.0 cm from the axis between the plates is given by the following expression:

Distance between the two plates = 3.0 cm = 0.03 m

μ₀ is the free space permeability and is equal to 4π×10^-7NA^-2

I₀ is equal to the maximum value of current i.e. √2I

On calculating, we get 162.63

⇒ B = 166 × 10^-11 T

Question 3.

What physical quantity is the same for X-rays of wavelength 10^–10 m, red light of wavelength 6800 Å and radio waves of wavelength 500m?

Answer:

The physical quantity which is same for X-rays of wavelength 10^–10 m, red light of wavelength 6800 Å and radio waves of wavelength 500 m is the speed of light in vacuum which is equal to 3×10⁸ m/s. The speed of light in vacuum remains same for all wavelengths. This is because speed of light is independent of wavelength in the vacuum.

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:

Given: The electromagnetic wave travels in vacuum along z-direction

The direction of electric field vector and magnetic field vector will be in XY plane.

Frequency of the wave = 30 MHz

Converting into Hz, v = 30× 10⁶ Hz.

Wavelength of the wave can be calculated using the following relation:

λ = c/v

⇒ 

⇒λ = 10 m

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:

Given: Minimum frequency v₁ = 7.5 MHz

In Hz, v₁ = 7.5 × 10⁶ Hz

Maximum frequency v₂ = 12 MHz

In Hz, v₂ = 12 ×10⁶ Hz

Speed of light c = 3 × 10⁸ m/s

Now wavelength for frequency v₁ = 7.5 MHz is given by the following relation:

λ ₁ = c/v₁

⇒ λ₁ = 40 m

The wavelength for frequency v₂ = 12 MHz is given by the following relation:

λ₂ = c/v₂

λ ₂ = 25m

∴ the range is 20 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:

The frequency of the electromagnetic waves produced by the oscillator is 10⁹ Hz. This is because the frequency of the electromagnetic waves produced by the oscillator is equal to the frequency of the electromagnetic waves produced by the oscillator.

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:

Given: Amplitude of magnetic field part of harmonic electromagnetic wave, B₀ = 510 nT

Converting into Tesla, B₀ = 510 × 10^-9 T

Speed of light in vacuum, c = 3 × 10⁸ m/s

Amplitude of the electric field part of the wave can be calculated by the following expression:

E = c B₀

E = 3× 10⁸ms^-1 ×510 × 10^-9 T

⇒ E = 153 N/C

Question 8.

Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b) Find expressions for E and B.

Answer:

Given: Electric field amplitude of electromagnetic wave, E₀ = 120 N/C

Frequency of electromagnetic wave, v = 50 MHz

(a) Magnitude of magnetic field is given by the following expression:

B₀ = E₀/c

⇒ B₀ = 40 × 10^-8 T

Converting it into nT, we get B₀ = 400 nT

Angular frequency can be calculated as follows:

ω = 2πv

where ν is the linear frequency.

ω = 2×3.14×50×10⁶Hz

ω = 314 × 10⁶ rad/s

ω = 3.14 × 10⁸ rad/s

Propagation constant k is given by the following expression:

k = ω /c

K = 1.05 rad/m

Wavelength can be calculated by the following expression:

λ = c/v

⇒ λ = 6 m

(b) The expression for electric field and magnetic field can be written as follows:

Let a wave is propagating in positive x- direction.

Since all the three vectors are mutually perpendicular to each other. The electric field vector will be in positive y- direction and magnetic field vector will be in positive z-direction.

Electric field vector can be expressed in the form of equation as given below:

⇒ 

Substituting values in above equation we get

⇒ 

Magnetic field vector can be expressed in form of following equation:

⇒ 

Substituting values in above equation we get

⇒ 

Question 9.

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

The energy of a photon E = hv

Where v is the frequency of the wave.

And, h is planck’s constant = 6.626×10^-34J-sec

v is given by the relation v = c/λ,

where c is the velocity of light = 3×10⁸ m/sec

Putting in above formula

E = hc/λ

In electron volts,

 eV

 eV

⇒ eV

The different scales of photon energies obtained are related to the sources of electromagnetic radiation as the different photon energies of the spectrum shows the spacing between the energy levels of the source.

Question 10.

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ 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 × 10⁸ m s^–1.]

Answer:

Given: Frequency of electromagnetic wave f = 2.0×10¹⁰ Hz

Amplitude of the electromagnetic wave i.e. E₀ = 48 V m^–1

(a) Wavelength of the electromagnetic wave can be calculated as follows:

v = c / λ

also, λ = c/v

where c is the speed of light in vacuum

v is the frequency of electromagnetic wave

and λ is the wavelength of electromagnetic wave

Substituting the values

On calculating, we get

⇒ λ = 0.015 m

(b) Amplitude of the electromagnetic wave i.e. E₀ = 48 V m^–1

Amplitude of the oscillating magnetic field is given by the following expression:

B₀ = E₀/c

On calculating we get:

⇒ B₀ = 1.6 × 10^-7 Tesla

(c) The average energy density of the E field equals the average energy density of the B field can be proved as follows:

The energy density of electric field  (i)

Where is the permittivity of free space

The energy density of magnetic field is given as follows:

_{......................}(ii)

Where is the permeability of free space

The relation between Electric field and magnetic field i.e. E and B is as follows:

On squaring both the sides, we get following relation

Or, 

From equation (i) and (ii), we get

∴ U_E = U_B

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Additional Exercises

Question 1.

Suppose that the electric field part of an electromagnetic wave in vacuum is 

(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?

(a) Write an expression for the magnetic field part of the wave.

Answer:

(a) Given: E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]}

_{As it is clear from the above electric field vector, that electric field is in the negative x direction. Therefore, the direction of propagation of the vector will be in negative y-direction i.e. –j.}

(b) As we know that the general equation for electric field vector in the positive x-direction is:

..........(i)

The electric field part of an electromagnetic wave in vacuum is

E = {(3.1 N/C) cos [(1.8 rad /m) y + (5.4 × 10⁶ rad/s)t_{..............(ii)}

Comparing equation (i) and (ii), we get

E₀ = 3.1 N

(c) Angular frequency ω = 5.4 × 10⁶ rad/s

Wave number i.e. k = 1.8 rad/m

Wavelength of the wave can be calculated as follows:

λ = 2π/k

(d) Angular frequency of the wave can be expressed as follows:

ω = 2π v

⇒ 5.4 × 10⁶ = 2 × 3.14 × v

⇒ 

⇒ 

⇒ V = 0.86 × 10⁶ Hz

(e) Magnetic field strength can be calculated as follows:

B₀ = E₀/c

⇒ 

⇒ B₀ = 1.03 × 10^-7 T

(f) Since magnetic field vector is in the negative z-direction, the general equation for magnetic field vector can be written as follows:

The expression for the magnetic field part of the wave can be written as follows:

B = B₀ cos (ky + wt) k

B = {1.03 × 10^-7 cos [(1.8 rad/m)y + (5.4 × 10⁶ rad/s)t]}k

Question 2.

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:

(a) Given : Power of the bulb = 100 W

Since, 5% of the power of a 100 W light bulb is converted to visible radiation, therefore the power of visible radiation can be calculated as follows:

P’ = 5W

The average intensity of visible radiation at a distance of 1m from the bulb can be calculated by using the following formula:

⇒ 

⇒ 

⇒ I = 0.398 W/m²

(b) The average intensity of visible radiation at a distance of 10 m from the bulb can be calculated by using the following formula:

I = P’/4π d²

I = 

I = 5/1256

I = 0.00398 W/m²

Question 3.

Use the formula λ_mT = 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:

According to the planck’s law of radiation:

λ_m T = 0.29 cm K

or λ_m = 0.29/T cm K

where T is the temperature and λ_m is the maximum wavelength of the wave

for different values of maximum wavelengths such as λ_m = 10^-4 cm, 5 × 10^-5 cm and 10^-6 cm, the values for temperatures can be calculated as follows:

T = 0.29/λ_m cm K

When, λ_m = 10^-4 cm

T = 0.29/10^-4 cm K

T = 2900 K

When, λ_m = 5× 10^-5 cm

T =  cm K

T = 5800 K

When, λ_m = 10^-6 cm

T = 0.29/10^-6 cm K

T = 290000 K

The above data shows that temperature is required to get radiations of different parts of electromagnetic spectrum and the temperature increases with decrease in the wavelength.

Question 4.

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 ⁵⁷Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].

Answer:

(a) The wavelength emitted by atomic hydrogen in interstellar space i.e. 21 cm is the radio waves as radio waves are shortest in wavelength of the electromagnetic spectrum.

(b) The frequency of radiation arising from the two close energy levels in hydrogen known as Lamb shift i.e. 1057 MHz is radio waves as it belongs to the short wavelength end of the electromagnetic spectrum.

(c) Given: Temperature = 2.7 K

λ_m = 0.29/T cm K

putting the value of temperature in above equation

λ_m = 0.29/2.7 cm K

λ_m = 0.11 cm

The above wavelength belongs to the microwaves of the electromagnetic spectrum.

(d) 5890 Å - 5896 Å [double lines of sodium] belongs to the yellow light of the visible spectrum of electromagnetic waves.

(e) Given: Energy E = 14.4 keV

Energy is given by the relation E = hv

h = 6.6 × 10^-34 Js

v = E/h

⇒ 4 × 10¹⁸ Hz

the electromagnetic spectrum belongs to X-rays.

Question 5.

Answer the following question:

Long distance radio broadcasts use short-wave bands. Why?

Answer:

Long distance radio broadcasts use short-wave bands because ionosphere refracts only shortwave bands and long wave bands are not refracted by the ionosphere.

Question 6.

Answer the following question:

It is necessary to use satellites for long distance TV transmission. Why?

Answer:

Satellites are used for long distance T.V. transmission because due to the high frequencies of the T.V. signals and high energy of the TV signals, they can’t be reflected by the ionosphere. Due to which satellites are required for reflecting the T.V. signals by the ionosphere.

Question 7.

Answer the following question:

Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?

Answer:

Optical and radio-telescopes are built on the ground because atmosphere can absorb the X-rays but visible and radio-waves are not absorbed by the atmosphere and they can penetrate through it. That’s why Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth.

Question 8.

Answer the following question:

The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Answer:

The small ozone layer on top of the stratosphere is crucial for human survival because ozone layer present on top of the stratosphere helps in preventing the harmful UV radiation from reaching the Earth’s surface and absorbs it. If it reaches the Earth’s surface it can cause many serious harms to the people.

Question 9.

Answer the following question:

If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

Answer:

When there will be no atmosphere on surface of Earth, then there will be no green house effect which will led to decrease in the temperature. The decrease in temperature will result in making the environment cool and it will be difficult to survive by human.

Question 10.

Answer the following question:

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:

The environmental condition post nuclear war will be very worst. The war will result in formation of clouds in the sky all over resulting in the extreme winter conditions. It will prevent sunlight to reach the Earth’s surface and the Ozone layer will also gets depleted.