Question

The oscillating magnetic field in a plane electromagnetic wave is given by  B_y = (8 x10^-6)sin[2 x10¹¹t + 300πx]T

(i) Calculate the wavelength of the electromagnetic wave.
(ii) Write down the expression for the oscillating electric field.

Answer 1

Given equation for oscillating magnetic field is,

B_y = (8 × 10^-6) sin [2 × 10¹¹ t + 300 πx]T.

Comparing the given equation with the equation of magnetic field varying sinusoidally with x and t

\(B_y = B_0 \sin\left(\frac{2\pi x}\lambda + \frac{2\pi t}T\right)\)

We get,

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

\(\therefore \lambda = \frac 2{300} = 0.0067 m\)

and \(B_0 = 8 \times 10^{-6}T\)

(i) Wavelength of the electromagnetic wave, \(\lambda = 0.0067m\)

(ii) Magnitude of electric field is calculated by,

\(E_0 = CB_0\)

\(= 3 \times 10^8 \times 8 \times 10^{-6}\)

\(= 24 \times 10^2\)

\(= 2400 \,Vm^{-1}\)

\(\therefore\) The required expression for the oscillating electric field is,

\(E_z = E_0 \sin \left(\frac{2\pi x}\lambda + \frac{2\pi t}T\right)\)

\(= 2400 \sin (300\pi x + 2 \times 10^{11}t)\,V/m\).

Answer 2

(i) Standard equation of magnetic field is

B_y = B₀sin(ωt + kx)T

Comparing this equation with the given equation, we get

B₀ = 8 x 10^-6T, ω = 2 x 10¹¹ rad s^-1, k = 2π/λ = 300π

wavelength  λ  = 2π/300π = 1/150m

(ii) E₀B₀c = 8 x 10^-6 x 3 x 10⁸ = 2.4 x 10³Vm.

According to right hand system of vector E, vector B, vector K, the electric field oscillates along negative Z-axis, so equation is

E_z = - 2.4 x 10³sin(2 x 10¹¹t + 300πx)Vm^-1.