Question

The oscillating electric field of an electromagnetic wave is given by: 

E_y =30sin [2 x 10¹¹t + 300πx]Vm^-1 

(a) Obtain the value of the wavelength of the electromagnetic wave. 

(b) Write down the expression for the oscillating magnetic field

Answer 1

(a) Given equation is \(E_Y = 30 \sin [2\times 10^{11}t + 300\pi x] \text{Vm}^{-1}\)

Comparing the given equation with the equation of

\(E_Y = E_0 \sin \left[\frac{2\pi t}T + \frac{2\pi x}\lambda\right]\),

We get,

\(\frac{2\pi}\lambda = 300\pi \)

⇒ \(\frac 1 \lambda = 150\)

⇒ \(\lambda = \frac 1{150} \) m

⇒ \(\lambda = 6.67 \times 10^{-3}\) m

(b) Oscillating electric field,

\(E_Y = 30 \sin [2\times 10^{11}t + 300\pi x] \text{Vm}^{-1}\)

Here, \(E_0 = 30\)

Hence, \(B_0 = \frac{E_0}c\)

\(= \frac{30}{3 \times 10^ 8}\)

\(= 10^{-7}\)

\(\therefore \) Oscillating magnetic field is given by,

\(B_Z = (10^{-7}) \sin [2\times 10^{11}t + 300\pi x] T\)

Answer 2

(a) Given equation is 

E_y =30sin [2 x 10¹¹t + 300πx]Vm^-1 

Comparing with standard equation 

E_y = E₀sin(ωt + kx)Vm^-1, we get

(b) The wave is propagating along X-axis, electric field is oscillating along Y-axis, so according to right hand system of (vector E, vector B, vector K), the magnetic field must oscillate along Z-axis.