Question

Suppose that the electric field part of an electromagnetic wave in vacuum is 

${\displaystyle \overrightarrow{\text{E}}=\left\{\left(3.1\frac{\text{N}}{\text{C}}\right)\cos [\left(1.8\frac{\text{rad}}{\text{m}}\right)\text{y}+(5.4\times {10}^8\frac{\text{rad}}{\text{s}})\text{t}]\right\}\hat{\text{i}}}$

(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

(a) From the given electric field vector, it can be inferred that the electric field is directed along the negative x-direction. Hence, the direction of motion is along the negative y-direction i.e., ${\displaystyle -\hat{\text{j}}}$.

(b) It is given that,

${\displaystyle \overrightarrow{\text{E}}=3.1\frac{\text{N}}{\text{C}}\cos [\left(1.8\frac{\text{rad}}{\text{m}}\right)\text{y}+(5.4\times {10}^8\frac{\text{rad}}{\text{s}})\text{t}]\hat{\text{i}}}$ ...........(1)

The general equation for the electric field vector in the positive x direction can be written as:

${\displaystyle \overrightarrow{\text{E}}={\text{E}}_0\sin (\text{kx}-\omega \text{t})\hat{\text{i}}}$ ........(2)

On comparing equations (1) and (2), we get

Electric field amplitude, E₀ = 3.1 N/C

Angular frequency, ω = 5.4 × 10⁸ rad/s

Wave number, k = 1.8 rad/m

Wavelength, ${\displaystyle \lambda =\frac{2\pi }{1.8}}$ = 3.490 m

(c) Frequency of wave is given as:

${\displaystyle \text{v}=\frac{\omega }{2\pi }}$

= ${\displaystyle \frac{5.4\times {10}^8}{2\pi }}$

= 8.6 × 10⁷ Hz

(d) Magnetic field strength is given as:

${\displaystyle {\text{B}}_0=\frac{{\text{E}}_0}{\text{c}}}$

Where,

c = Speed of light = 3 × 10⁸ m/s

∴ ${\displaystyle {\text{B}}_0=\frac{3.1}{3\times {10}^8}}$

= 1.03 × 10⁻⁷ T

(e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:

${\displaystyle \overrightarrow{\text{B}}={\text{B}}_0\cos (\text{ky}+\omega \text{t})\hat{\text{k}}}$

= ${\displaystyle \left\{(1.03\times {10}^{-7} \text{T})\cos [\left(1.8\frac{\text{rad}}{\text{m}}\right)\text{y}+(5.4\times {10}^6\frac{\text{rad}}{\text{s}})\text{t}]\right\}\hat{\text{k}}}$