The Electric Side of a Wave

solved ElectromagneticWave

Q. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT $B = 510$ nT. What is the amplitude of the electric field part of the wave, in V/m?

Inputs

magnetic

B

510 nT

Outputs

electric

E

How it's solved

Solved with ElectromagneticWave

Given

$B = 510$ nT

Find $E$ .

Working

$B = 510$ nT $= 5.1 \times 10^{−7}$ T
fields of a light wave(wave) $E = c \cdot B$
$E = 2.99792458 \times 10^{8} \cdot 5.1 \times 10^{−7}$
$E = 152.9$ V/m
$E \approx 153$ V/mto 3 significant figures

Conclusion

In a light wave the electric field is the speed of light times the magnetic field, so a field of a fraction of a microtesla goes with an electric field of a hundred or so volts a metre.

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 :the_electric_side_of_a_wave do4  kind :numerical5  describes "The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = " \6            "%{magnetic}. %{electric}, in V/m?"7  given magnetic: "510 nT"8  asks electric: "What is the amplitude of the electric field part of the wave"9end

solution.rb

1require_relative "../questions/the_electric_side_of_a_wave"2require_relative "../science/electromagnetic_waves/electromagnetic_wave"34solve :the_electric_side_of_a_wave, using: ElectromagneticWave do5  objects :wave67  take magnetic: wave.magnetic_field89  electric = wave.electric_field10  answer electric:, in: "V/m", figures: 31112  conclusion "In a light wave the electric field is the speed of light times the magnetic field, so a field " \13             "of a fraction of a microtesla goes with an electric field of a hundred or so volts a metre."14end

playground.rb

1require_relative "../solutions/the_electric_side_of_a_wave"23play_with :the_electric_side_of_a_wave do4  title "The Electric Side of a Wave"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)