A 50 MHz Wave's Magnetic Field

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 the amplitude of its magnetic field, B₀, in teslas? expects 4.00 × 10⁻⁷ T

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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?

The playground

Inputs

electric

E

120 N/C

frequency

ν

50 MHz

Outputs

magnetic

B

The solution

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 Ruby

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

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

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

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)