A 50 MHz Wave's Magnetic Field

solved ElectromagneticWave

Q. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C $E = 120$ N/C and that its frequency is ν = 50.0 MHz $ν = 50.0$ MHz. What is the amplitude of its magnetic field, B₀, in teslas?

Inputs

electric

E

120 N/C

frequency

ν

50 MHz

Outputs

magnetic

B

How it's solved

Solved with ElectromagneticWave

Given

$E = 120$ N/C
$ν = 50.0$ MHz

Find $B$ .

Working

$E = 120$ N/C $= 120$ V/m
$ν = 50.0$ MHz $= 5 \times 10^{7}$ Hz
fields of a light wave(wave) $E = c \cdot B$
$120 = 2.99792458 \times 10^{8} \cdot B$
$B = 4.003 \times 10^{−7}$ T
$B \approx 4.00 \times 10^{−7}$ Tto 3 significant figures

Conclusion

The magnetic field of a light wave is its electric field over the speed of light, tiny in teslas whatever the frequency.

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_magnetic_field do4  kind :numerical5  describes "Suppose that the electric field amplitude of an electromagnetic wave is E₀ = %{electric} and that " \6            "its frequency is ν = %{frequency}. %{magnetic}, in teslas?"7  given electric: "120 N/C"8  given frequency: "50.0 MHz"9  asks magnetic: "What is the amplitude of its magnetic field, B₀"10end

solution.rb

1require_relative "../questions/a_fifty_megahertz_wave_s_magnetic_field"2require_relative "../science/electromagnetic_waves/electromagnetic_wave"34solve :a_fifty_megahertz_wave_s_magnetic_field, using: ElectromagneticWave do5  objects :wave67  take electric: wave.electric_field8  take frequency: wave.frequency910  magnetic = wave.magnetic_field11  answer magnetic:, in: "T", figures: 31213  conclusion "The magnetic field of a light wave is its electric field over the speed of light, tiny in teslas " \14             "whatever the frequency."15end

playground.rb

1require_relative "../solutions/a_fifty_megahertz_wave_s_magnetic_field"23play_with :a_fifty_megahertz_wave_s_magnetic_field do4  title "A 50 MHz Wave's Magnetic Field"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)