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3 September, 12:11

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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Answers (2)
  1. 3 September, 12:50
    0
    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
  2. 3 September, 12:51
    0
    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.
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