Question:

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

Answer:

a) B0 = (E0/(c)) = (120 N/C)/(3.00 x 10^8 m/s) = 4.00 x 10^-7 T

ω = 2πν = 2π(50.0 x 10^6 Hz) = 3.14 x 10^8 rad/s

k = ω/c = (3.14 x 10^8 rad/s)/(3.00 x 10^8 m/s) = 1.05 x 10^7 m^-1

λ = c/ν = (3.00 x 10^8 m/s)/(50.0 x 10^6 Hz) = 6.00 m

b) E = E0cos(ωt - kx)

B = B0cos(ωt - kx)

Question:

Use the formula λm​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?

Answer:

1. First, identify the formula λm​T = 0.29 cm K. This formula is used to calculate the characteristic temperature ranges for different parts of the electromagnetic spectrum.

2. Next, use the formula to calculate the characteristic temperature ranges. For example, if the wavelength of a certain part of the electromagnetic spectrum is 0.1 cm, then the characteristic temperature range is 0.29 cm x 0.1 cm = 0.029 K.

3. The numbers obtained from the formula tell us the characteristic temperature range for different parts of the electromagnetic spectrum. For example, a wavelength of 0.1 cm indicates a temperature range of 0.029 K, while a wavelength of 1 cm indicates a temperature range of 0.29 K.

Question:

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

Answer:

Step 1: Convert the frequency range from 7.5 MHz to 12 MHz to hertz (Hz).

7.5 MHz = 7,500,000 Hz 12 MHz = 12,000,000 Hz

Step 2: Calculate the wavelength using the formula λ = c/f where c is the speed of light (3 x 10^8 m/s) and f is the frequency.

λ = 3 x 10^8 m/s / 7,500,000 Hz = 0.4 m

λ = 3 x 10^8 m/s / 12,000,000 Hz = 0.25 m

Step 3: The corresponding wavelength band is 0.25 m to 0.4 m.

Question:

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:

Answer:

1. The electric and magnetic field vectors of the plane electromagnetic wave travel perpendicular to each other and perpendicular to the direction of the wave (the z-direction).
2. The wavelength of the wave is equal to the speed of light (c) divided by the frequency (f) of the wave, or λ = c/f. In this case, the wavelength is equal to c/30 MHz, or 0.01 meters.

Question:

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0×10^10Hz and amplitude 48Vm^−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 ×108ms^−1]

Answer:

(a) Wavelength = Speed of light/Frequency = (3 x 10^8 m/s)/(2 x 10^10 Hz) = 1.5 x 10^-2 m

(b) Magnetic Field Amplitude = Electric Field Amplitude/Speed of Light = 48 Vm^-1 / (3 x 10^8 m/s) = 1.6 x 10^-7 T

(c) Average Energy Density of Electric Field = 1/2 x Electric Field Amplitude^2 = 1/2 x (48 Vm^-1)^2 = 0.00576 J/m^3

Average Energy Density of Magnetic Field = 1/2 x Magnetic Field Amplitude^2 = 1/2 x (1.6 x 10^-7 T)^2 = 1.024 x 10^-14 J/m^3

Therefore, Average Energy Density of Electric Field = Average Energy Density of Magnetic Field

Question:

A charged particle oscillates about its mean (equilibrium) position with a frequency of 109Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:

Answer: The frequency of the electromagnetic waves produced by the oscillator is also 109Hz.

Question:

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:

1. The terminology of different parts of the electromagnetic spectrum is given in the text.

2. 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.

3. The photon energy in units of eV for different parts of the electromagnetic spectrum is calculated using the formula E=hν, where E is the energy of a quantum of radiation (photon), h is Planck’s constant (6.626 x 10^-34 Js) and ν is the frequency of the radiation.

4. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

The different scales of photon energies obtained are related to the sources of electromagnetic radiation in that different sources emit different frequencies of radiation, and thus different energies of photons. For example, gamma rays have a higher frequency and thus higher energy than visible light.

Question:

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0​=510nT. What is the amplitude of the electric field part of the wave?

Answer:

Answer: The amplitude of the electric field part of the harmonic electromagnetic wave in vacuum is E0 = (1/c) × B0, where c is the speed of light in vacuum.

Therefore, E0 = (1/c) × 510nT = (1/3 x 10^8 m/s) × 510nT = 1.7 x 10^-3 V/m

Question:

What physical quantity is the same for X-rays of wavelength 10−10m, red light of wavelength 6800 A˚ and radiowaves of wavelength 500m?

Answer:

Answer: The physical quantity that is the same for X-rays of wavelength 10−10m, red light of wavelength 6800 A˚ and radiowaves of wavelength 500m is the frequency.

Question:

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 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?

Answer:

(a) Short-wave bands are used for long distance radio broadcasts because they can travel much further than other radio frequencies due to their higher frequency and shorter wavelength.

(b) Satellites are necessary for long distance TV transmission because they can provide a much wider coverage area than terrestrial transmitters and can reach remote areas that would otherwise be difficult to reach.

(c) Optical and radio telescopes are built on the ground because they require a stable platform for precise measurements, whereas X-ray astronomy is possible from satellites orbiting the earth because X-rays have very short wavelengths and cannot penetrate the atmosphere.

(d) The small ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet radiation from the sun, which can be harmful to humans and other life forms.

(e) If the earth did not have an atmosphere, its average surface temperature would be lower than what it is now because the atmosphere acts as an insulator, trapping heat and keeping the temperature relatively stable.

(f) The prediction of a nuclear winter following a global nuclear war is based on the fact that the smoke and dust from the explosions would block out the sun, causing a drop in global temperatures and leading to a decrease in food production and other devastating effects.

Question:

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) 5890A˚−5896A˚ [double lines of sodium] (e) 14.4 ke V [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mo¨ssbauer spectroscopy)].

Answer:

(a) Radio waves (b) Radio waves (c) Microwaves (d) Visible light (e) Gamma rays

Question:

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) Intensity at a distance of 1 m from the bulb = 100 W x 5% = 5 W

(b) Intensity at a distance of 10 m from the bulb = (5 W) / (10 m)^2 = 0.005 W/m^2

Question:

Suppose that the electric field part of an electromagnetic wave in vacuum is E=(3.1 N/C)cos[(1.8rad/m)y+(5.4×10^6rad/s)t]i^(a) What is the direction of propagation? (b) What is the wavelength λ ? (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:

a) The direction of propagation is in the +x direction.

b) The wavelength is λ = 1/1.8 m = 0.556 m.

c) The frequency is v = 5.4×10^6 rad/s.

d) The amplitude of the magnetic field part of the wave is B = (3.1 N/C)/(2πv).

e) The magnetic field part of the wave is B = (3.1 N/C)cos[(1.8rad/m)y+(5.4×10^6rad/s)t]j^(a).