Question

Suppose that the electric field part of an electromagnetic wave in vacuum is $E=\left(3.1N/C\right)cos\left[\right(1.8rad/m)y+(5.4\times {10}^{6}rad/s\left)t\right]\hat{i}$(a) What is the direction of propagation?
(b) What is the wavelength $\lambda$ ?
(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

(a)

Electric field is directed along negative x-direction.

hence the direction of motion is along the negative y-direction.

(b)

From the given equation, $k=1.8rad/m$

$\lambda =2\pi /k=3.49$

(c)

Frequency of wave is $\nu =\omega /2\pi =8.6\times {10}^7=86Mhz$

(d)

For the given equation, $E_o=3.1N/C$

$B_o=E_o/c=1.03\times {10}^{-7}=103nT$

(e)

Magnetic wave is directed along negative z-direction.

Thus, $B=B_{o}cos(ky+\omega t)\hat{k}$

$\Rightarrow B=1.03\times {10}^{-7}cos(1.8y+5.4\times {10}^{6}t)\hat{k}$