Question

Suppose that the electric field part of an electromagnetic wave in vacuum is
E= {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 x 10⁶ rad/s)t]} \(\hat i\)

(a) What is the direction of propagation?

(b) What is the wavelength λ?

(c) What is the frequency V?

(d) What is the amplitude of the magnetic field part of the wave?

(e) Write an expression for the magnetic field part of the wave.

Answer 1

The electric field part of the electromagnetic wave is given by,
\(\vec E\) = (3.1 NC^-1 ) cos [(1,8 rad m^-1) y + (5.4 x 10⁶ rad s¹) t]\(\hat i\) it follows that
E₀ = 3.1 NC^-1
k =1.8 rad m^-1
ω = 5.4 × 10⁶ rad s^-1
(a) Since the argument of since in the expression for electric field is of type (ky + cot), the direction of propagation of electro magnetic wave is along negative Y-axis.

(b) The wavelength of the wave,

(e) The electric field and the direction of propagation of the e.m. wave are along negative X-axis and negative Y-axis respectively. Therefore, the magnetic field is along negative Z-axis and expression for it is given by
B = (10.33nT) cos [(1.8 rad m^-1) y + (5.4 x 10⁶rad s^-1)t] \(\hat k\)