# NCERT Solutions for Class 12 Physics Chapter 8

## Electromagnetic Waves Class 12

### Exercise : Solutions of Questions on Page Number : 285

Q1 :

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 C2 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 ×109 V/s.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id 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.

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Q2 :

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, Ãâ€° = 300 rad s - 1

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

XC = Capacitive reactance I = V × Ãâ€°C

= 230 × 300 × 100 × 10 - 12

= 6.9 × 10 - 6 A

= 6.9 ÃŽÂ¼A

Hence, the rms value of conduction current is 6.9 ÃŽÂ¼A.

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

(c) Magnetic field is given as:

B Where,

ÃŽÂ¼0 = Free space permeability I0 = 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.

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Q3 :

What physical quantity is the same for X-rays of wavelength 10-10 m, red light of wavelength 6800 Ãƒâ€¦ and radiowaves of wavelength 500 m?

Answer :

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Q4 :

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?

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Q5 :

A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

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Q6 :

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?

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Q7 :

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?

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Q8 :

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ÃŽÂ½ = 50.0 MHz. (a) Determine, B0, Ãâ€°, k, and ÃŽÂ». (b) Find expressions for E and B.

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Q9 :

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hÃŽÂ½ (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?

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Q10 :

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.

(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 x 108 m s-1.]

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Q11 :

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 × 106 rad/s)t]} .

(a) What is the direction of propagation?

(b) What is the wavelength ÃŽÂ»?

(c) What is the frequency ÃŽÂ½?

(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.

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Q12 :

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.

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Q13 :

Use the formula ÃŽÂ»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?

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Q14 :

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 Ãƒâ€¦ - 5896 Ãƒâ€¦ [double lines of sodium]

(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method

(MÃƒÂ¶ssbauer spectroscopy)].

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Q15 :

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?

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Physics : CBSE NCERT Exercise Solutions for Class 12th for Electromagnetic Waves will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

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