A 50 MHz Wave's Wave Number

solved ElectromagneticWave

Q. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz $ν = 50.0$ MHz. What is its wave number, k, in rad/m?

Inputs

frequency

ν

50 MHz

Outputs

number

k

How it's solved

Solved with ElectromagneticWave

Given

$ν = 50.0$ MHz

Find $k$ .

Working

$ν = 50.0$ MHz $= 5 \times 10^{7}$ Hz
light travels at c(wave) $c = ν \cdot λ$
$2.99792458 \times 10^{8} = 5 \times 10^{7} \cdot λ$
$λ = 5.99584916$ m
wave number and wavelength(wave) $k = \frac{2 \cdot π}{λ}$
$k = \frac{2 \cdot 3.14159265}{5.99584916}$
$k = 1.048$ rad/m
$k = 1.04792251$ m⁻¹ $= 1.04792251$ rad/m
$k \approx 1.05$ rad/mto 3 significant figures

Conclusion

The wave number counts the radians of phase in each metre, two pi over the wavelength.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_fifty_megahertz_wave_s_wave_number do4  kind :numerical5  describes "Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its " \6            "frequency is ν = %{frequency}. %{number}, in rad/m?"7  given frequency: "50.0 MHz"8  asks number: "What is its wave number, k"9end

solution.rb

1require_relative "../questions/a_fifty_megahertz_wave_s_wave_number"2require_relative "../science/electromagnetic_waves/electromagnetic_wave"34solve :a_fifty_megahertz_wave_s_wave_number, using: ElectromagneticWave do5  objects :wave67  take frequency: wave.frequency89  number = wave.wave_number10  answer number:, in: "rad/m", figures: 31112  conclusion "The wave number counts the radians of phase in each metre, two pi over the wavelength."13end

playground.rb

1require_relative "../solutions/a_fifty_megahertz_wave_s_wave_number"23play_with :a_fifty_megahertz_wave_s_wave_number do4  title "A 50 MHz Wave's Wave Number"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	ElectromagneticWave
speed of light	$c$	m/s	—	ElectromagneticWave
electric field	$E$	V/m	—	ElectromagneticWave
magnetic field	$B$	T	—	ElectromagneticWave
wave number	$k$	m⁻¹	—	ElectromagneticWave
angular frequency	$ω$	rad/s	—	ElectromagneticWave
wavelength	$λ$	m	0 … ∞	ElectromagneticWave
frequency	$ν$	Hz	0 … ∞	ElectromagneticWave
electric field of ground	$E_{g}$	V/m	—
magnetic field of ground	$B_{g}$	T	—
electric field of wave	$E$	V/m	—
magnetic field of wave	$B$	T	—
wave number of ground	$k_{g}$	m⁻¹	—
wavelength of ground	$λ_{g}$	m	0 … ∞
wave number of wave	$k$	m⁻¹	—
wavelength of wave	$λ$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of wave	$ω$	rad/s	—
frequency of wave	$ν$	Hz	0 … ∞

E_{w} = c \cdot B_{w}

fields of a light wave

k_{w} = \frac{2 \cdot π}{λ_{w}}

wave number and wavelength

ω_{w} = 2 \cdot π \cdot ν_{w}

angular frequency and frequency

c = ν_{w} \cdot λ_{w}

light travels at c

E_{g} = c \cdot B_{g}

fields of a light wave(ground)

E = c \cdot B

fields of a light wave(wave)

k_{g} = \frac{2 \cdot π}{λ_{g}}

wave number and wavelength(ground)

k = \frac{2 \cdot π}{λ}

wave number and wavelength(wave)

ω_{g} = 2 \cdot π \cdot ν_{g}

angular frequency and frequency(ground)

ω = 2 \cdot π \cdot ν

angular frequency and frequency(wave)

c = ν_{g} \cdot λ_{g}

light travels at c(ground)

c = ν \cdot λ

light travels at c(wave)