The Magnetic Side of a Wave

Q. A plane electromagnetic wave of frequency 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 along y. What is the magnetic field at this point, in teslas? expects 2.1 × 10⁻⁸ T

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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?

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Inputs

frequency

ν

25 MHz

electric

E

6.3 V/m

Outputs

magnetic

B

The solution

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.

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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

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

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

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)