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

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

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.