The Magnetic Side of a Wave

solved ElectromagneticWave

Q. A plane electromagnetic wave of frequency 25 MHz $ν = 25$ MHz travels in free space along the x-direction. At a particular point in space and time, its electric field is 6.3 V/m $E = 6.3$ V/m along y. What is the magnetic field at this point, in teslas?

Inputs

frequency

ν

25 MHz

electric

E

6.3 V/m

Outputs

magnetic

B

How it's solved

Solved with ElectromagneticWave

Given

$ν = 25$ MHz
$E = 6.3$ V/m

Find $B$ .

Working

$ν = 25$ MHz $= 2.5 \times 10^{7}$ Hz
fields of a light wave(wave) $E = c \cdot B$
$6.3 = 2.99792458 \times 10^{8} \cdot B$
$B = 2.101 \times 10^{−8}$ T
$B \approx 2.1 \times 10^{−8}$ Tto 2 significant figures

Conclusion

In a light wave the electric field is always the speed of light times the magnetic field, whatever the frequency; the magnetic field stands across both the electric field and the motion.

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_magnetic_side_of_a_wave do4  kind :numerical5  describes "A plane electromagnetic wave of frequency %{frequency} travels in free space along the x-direction. At a " \6            "particular point in space and time, its electric field is %{electric} along y. %{magnetic}, in teslas?"7  given frequency: "25 MHz"8  given electric: "6.3 V/m"9  asks magnetic: "What is the magnetic field at this point"10end

solution.rb

1require_relative "../questions/the_magnetic_side_of_a_wave"2require_relative "../science/electromagnetic_waves/electromagnetic_wave"34solve :the_magnetic_side_of_a_wave, using: ElectromagneticWave do5  objects :wave67  take frequency: wave.frequency8  take electric: wave.electric_field910  magnetic = wave.magnetic_field11  answer magnetic:, in: "T", figures: 21213  conclusion "In a light wave the electric field is always the speed of light times the magnetic field, " \14             "whatever the frequency; the magnetic field stands across both the electric field and the motion."15end

playground.rb

1require_relative "../solutions/the_magnetic_side_of_a_wave"23play_with :the_magnetic_side_of_a_wave do4  title "The Magnetic 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)