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An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the –z direction. the electric field has an amplitude of 2.70×10−3v/m and is parallel to the x axis. what are the frequency and the magnetic-field amplitude of this wave?

Respuesta :

skyluke89
(a) frequency of the wave

The electromagnetic wave travels at the speed of light: [tex]c=3 \cdot 10^8 m/s[/tex]. Since its wavelength is [tex]\lambda=435 nm=435 \cdot 10^{-9}m[/tex], we can find the frequency of the wave by using the relationship between f, [tex]\lambda [/tex] and c:
[tex]f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{435 \cdot 10^{-9}m}=6.9 \cdot 10^{14} Hz [/tex]

(b) magnetic field amplitude

The relationship between the electric field amplitude E, the magnetic field amplitude B and the speed of light c for an electromagnetic wave is
[tex]E=cB[/tex]
And since we know E, by re-arranging the equation we can find the magnitude of B:
[tex]B= \frac{E}{c}= \frac{2.70 \cdot 10^{-3}V/m}{3 \cdot 10^8 m/s}=9\cdot 10^{-12} T [/tex]
johanrusli

A. the frequency of the wave is about 6.90 × 10¹⁴ Hz

B. the magnetic field strength is 9 × 10⁻¹² T

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Further explanation

Let's recall the speed of wave formula in vacuum as follows:

[tex]\boxed {c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}}[/tex]

[tex]\boxed {c = \frac{E}{B}}[/tex]

where:

c = speed of wave in vacuum ( m/s )

μo = the permeability constant of vacuum

εo =  the permittivity of free space

E = electric field strength ( V/m )

B = magnetic field strength ( T )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

wavelength of electromagnetic wave = λ = 435 nm = 4.35 × 10⁻⁷ m

electric field strength = E = 2.70 × 10⁻³ V/m

speed of electromagnetic wave in vacuum = c = 3 × 10⁸ m/s

Asked:

A. frequency = f = ?

B. magnetic field strength = B = ?

Solution:

Part A:

[tex]c = \lambda f[/tex]

[tex]f = c \div \lambda[/tex]

[tex]f = (3 \times 10^8) \div ( 4.35 \times 10^{-7} )[/tex]

[tex]\boxed{f \approx 6.90 \times 10^{14} \texttt{ Hz}}[/tex]

[tex]\texttt{ }[/tex]

Part B:

[tex]c = E \div B[/tex]

[tex]B = E \div c[/tex]

[tex]B = ( 2.70 \times 10^{-3} ) \div ( 3 \times 10^8 )[/tex]

[tex]\boxed{B = 9 \times 10^{-12} \texttt{ T}}[/tex]

[tex]\texttt{ }[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Light

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