The electric field of a sinusoidal electromagnetic wave obeys the equation E = (360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ]. What is

Question

The electric field of a sinusoidal electromagnetic wave obeys the equation E = (360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ]. What is the amplitude of the magnetic field of this wave? A) 0.06 μT B) 0.23 μT C) 1.10 μT D) 1.20 μT

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Thiên Ân 5 years 2021-08-21T21:50:46+00:00 2 Answers 55 views 0

Answers ( )

  1. thachthao
    0
    2021-08-21T21:51:53+00:00
    Reply

    Answer:

    Option D is correct.

    Explanation:

    Bmax = Emax / c

    The general form for electromagnetic wave equation is

    E = jEmax ×cos(kx-wt)

    We were given

    (360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ].

    So from the equation above

    Emax = 360V/m

    Bmax = 360/(3×10⁸) = 1.2 ×10‐⁶ T.

  2. Helga
    0
    2021-08-21T21:52:39+00:00
    Reply

    Answer

    Option D

    Amplitude of Magnetic field = B = 1.2×10⁻⁶ T

    Explanation:

    The relationship between electric field and magnetic field of an electromagnetic wave is given by

    B = E/c

    Where B is the amplitude of magnetic field and E is the amplitude of electric field and c is the speed of light

    The amplitude of electric field is given as 360 V/m

    B = (360 V/m)/(3×10⁸ m/s)

    B = 1.2×10⁻⁶ V.s/m²

    Since 1 Tesla is equal to 1 V.s/m²

    B = 1.2×10⁻⁶ T

    Therefore, option D is correct

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