A 50 MHz Wave's Wavelength

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. What is its wavelength, λ, in metres? expects 6.00 m

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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 wavelength, λ, in metres?

The playground

Inputs

frequency

ν

50 MHz

Outputs

wavelength

λ

The solution

Solved with ElectromagneticWave

Given

$ν = 50.0$ MHz

Find $λ$ .

Working

$ν = 50.0$ MHz $= 5 \times 10^{7}$ Hz
light travels at c(wave) $c = ν \cdot λ$
$2.99792458 \times 10^{8} = 5 \times 10^{7} \cdot λ$
$λ = 6.00 m$
$λ \approx 6.00$ mto 3 significant figures

Conclusion

Every electromagnetic wave in vacuum travels at the speed of light, so its wavelength is that speed over its frequency.

The Ruby

1require_relative "../physics"23question :a_fifty_megahertz_wave_s_wavelength 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}. %{wavelength}, in metres?"7  given frequency: "50.0 MHz"8  asks wavelength: "What is its wavelength, λ"9end

1require_relative "../questions/a_fifty_megahertz_wave_s_wavelength"2require_relative "../science/electromagnetic_waves/electromagnetic_wave"34solve :a_fifty_megahertz_wave_s_wavelength, using: ElectromagneticWave do5  objects :wave67  take frequency: wave.frequency89  wavelength = wave.wavelength10  answer wavelength:, in: "m", figures: 31112  conclusion "Every electromagnetic wave in vacuum travels at the speed of light, so its wavelength is that speed " \13             "over its frequency."14end

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

The laws

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)