Electromagnetic Waves

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New answer posted•a year ago
S
a year ago

8.9 Energy of a photon is given as:

E = h ν = hcλ , where

h = Planck’s constant = 6.6 ×10-34 Js

c = Speed of light = 3 ×108 m/s

λ = wavelength of radiation

Hence, E = 6.6×10-34×3×108λ J = 1.98×10-25λ J = 1.98×10-25λ×1.602×10-19 eV = 1.236×10-6λ eV

The following table lists the photon energies for

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New answer posted•a year ago
S
a year ago

8.8 Electric field amplitude, E0 = 120 N/C

Frequency of source, ν = 50 MHz = 50 ×106 Hz

Speed of light, c = 3 ×108 m/s

Magnitude of magnetic field strength is given as

B0= E0c = 1203×108 = 4 ×10-7 T = 400 ×10-9 T = 400 nT

Angularfrequencyofthesourceisgivenas ω = 2 πν = 2 ×π×&n

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New answer posted•a year ago
S
a year ago

8.7 The amplitude of magnetic field of the electromagnetic wave in a vacuum,

B0 = 510 nT = 510 *10-9 T

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

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

E = c B0 = 3 *108* 510 *10-9 N/C = 153 N

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New answer posted•a year ago
S
a year ago

8.6 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. 109 Hz.

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New answer posted•a year ago
S
a year ago

8.5 The lowest tuning frequency ν1 = 7.5 MHz = 7.5 ×106 Hz

The highest tuning frequency ν2 = 12 MHz = 12 ×106 Hz

Speed of light, c = 3 ×108 m/s

The wavelength for lowest tuning frequency, λ1 = cν1 = 3×1087.5×106 = 40 m

The wavelength for highest tuning freq

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New answer posted•a year ago
S
a year ago

8.4 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 ×106 /s

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

Wavelength

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New answer posted•a year ago
S
a year ago

8.3 The speed of light (3 *108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

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New answer posted•a year ago
S
a year ago

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

Capacitance of parallel capacitor, C = 100 pF = 100 ×10-12 F

Supply voltage, V = 230 V

Angular frequency, ω = 300 rad/s

rms value of the conduction current, I = VXc , where X c = c a p a c i t i v e r e a c t a n c e = 1 ω C

Hence, I = V ×ω×C = 230 ×300× 100 ×10-12 = 6

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New answer posted•a year ago
S
a year ago

8.1 Radius of the each circular plate, r = 12 cm = 0.12

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

Charging current, I = 0.15 A

Permittivity of free space, ? 0 = 8.85 × 10 - 12 C 2 N - 1 m - 2

Capacitance between two plates is given by the relation,

C = ? 0 A d  , where A = Area of each plate = ? r 2 = ? × 0.12 2

C = 8.85 × 10 - 12 × ? × 0.12 2 0.05  = 8.007 × 10 - 12

F

Charge

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New answer posted•a year ago
S
a year ago

Electromagnetic waves do not require a medium to propagate because they generate self-sustaining oscillations of magnetic and electric fields. Hence, it can travel through the vacuum of space. A changing magnetic field generates an electric field and vice versa. This property allows the wave to carr

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