EM Waves NCERT Solutions

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.

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.

What physical quantity is the same for X-rays of wavelength10^–10m, red light of wavelength 6800 Å and radio-waves of wavelength 500m?

Ans: Speed of all electromagnetic wave is same 3 × 10⁸ m/s.

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?

Ans: Electric and magnetic field vectors are perpendicular to the direction of wave propagation. Electric and magnetic field vectors in the x-y plane are present in electromagnetic waves travelling in vacuum along the z-direction.

nu = 30MHz = 30 * 10^6 Hz

$lambda = c / nu = {3 * 10^8} / {30 * 10^6} = 10m$

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

Ans: Corresponding wavelength band (range of wavelength) =?
initial frequency nu_1 = 7.5MHz = 7.5 * 10^6 Hz
final frequency nu_2 = 12MHz = 12 * 10^6 Hz

$lambda_1 = c/ nu_1 = {3 * 10^8} / {7.5 * 10^6}$ = 40m

$lambda_2 = c/ nu_1 = {3 * 10^8} / {12 * 10^6}$ = 25m

corresponding wavelength band = lambda_1 - lambda_2 = 40m – 25m

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?

Ans: The frequency of the electromagnetic waves produced by the oscillator is equal to the frequency of oscillating charged particles, 10⁹ Hz.

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?

Ans: B₀ = 510nT = 510 × 10^-9 T, c = 3 ×10⁸ m/s

E₀ = c B₀

E₀ = 3 ×10⁸ × 510 × 10^-9 = 153 N/C

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

Ans: E₀ = 120 N/C , frequency = 50.0 MHz

(a) E₀ = c B₀

[pmath size = 15] B_0 = ? omega =?, k =?, lambda =? [/pmath]

[pmath size = 15] B_0 = E_0 / c = 120 / 3 * 10^8 = 4 * 10^7 [/pmath]

$omega = 2 pi nu = {2 * 3.14 * 50 * 10^6} = 3.14 * 10^8 rad/s$