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
2 answers:
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
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.
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