Question

Write an equation for the magnetic field of an electromagnetic wave traveling in vacuum if the amplitude is 0.0028 T and the frequency is 2.0 × 10⁸ Hz. Find the magnetic field when x = 32 cm and t = 6.8 ns.

Answer 1

An equation for the magnetic field of an electromagnetic wave traveling in vacuum which will be given as -

B = B₀ sin (k x - t)

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Wavelength of a wave is given by -

= c / f

where, c = speed of light in vacuum = 3 x 10⁸ m/s

f = frequency of a wave = 2 x 10⁸ Hz

then, we get

= [(3 x 10⁸ m/s) / (2 x 10⁸ Hz)]

= 1.5 m

Angular frequency of a wave is given by -

= 2f

= [(6.28 rad) (2 x 10⁸ Hz)]

= 1.256 x 10⁹ rad/s

Propagation constant of a wave is given by -

k = / c

k = [(1.256 x 10⁹ rad/s) / (3 x 10⁸ m/s)]

k = 4.186 rad/m

The magnetic field of an electromagnetic wave which will be given by -

B = B₀ sin (k x - t)

where, B₀ = amplitude of magnetic field = 0.0028 T

x = distance traveled by a wave = 0.32 m

t = time taken by a wave = 6.8 x 10^-9 s

then, we get

B = (0.0028 T) sin {[(4.186 rad/m) (0.32 m)] - [(1.256 x 10⁹ rad/s) (6.8 x 10^-9 s)]}

B = (0.0028 T) sin [(1.33952 rad) - (8.5408 rad)]

B = (0.0028 T) sin (-7.20128 rad)

B = [(0.0028 T) (-0.1253 rad)]

B = - 0.00035084 T

B - 3.51 x 10^-4 T