Question

What is meant by displacement current? The amplitude of electric field of an electromagnetic wave is \( E_0 = 120 \, \text{N/C} \) and frequency is \( \nu = 50 \, \text{MHz} \). Find out:
i) Amplitude of the magnetic field \( B_0 \) ii) Wavelength \( \lambda \) of the electromagnetic wave.

Hint:

In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light, and the wavelength is inversely proportional to the frequency of the wave.

Answer 1

Displacement Current:
Displacement current is a term introduced by James Clerk Maxwell to extend Ampère's law to situations involving time-varying electric fields. The displacement current density is given by:
\[ J_D = \varepsilon_0 \frac{\partial E}{\partial t}, \] where:
- \( J_D \) is the displacement current density,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( \frac{\partial E}{\partial t} \) is the rate of change of the electric field with respect to time.
The displacement current plays a crucial role in explaining how electric fields can generate magnetic fields in a vacuum or in materials with varying electric fields, forming the foundation of electromagnetic wave propagation.
i) Amplitude of the Magnetic Field \( B_0 \):
The electric field \( E_0 \) and the magnetic field \( B_0 \) in an electromagnetic wave are related by the speed of light \( c \). The relationship between the electric and magnetic field amplitudes in an electromagnetic wave is:
\[ E_0 = c B_0, \] where:
- \( c \) is the speed of light in vacuum (\( c = 3 \times 10^8 \, \text{m/s} \)),
- \( B_0 \) is the amplitude of the magnetic field.
Rearranging the formula to solve for \( B_0 \):
\[ B_0 = \frac{E_0}{c}. \] Substituting the given values \( E_0 = 120 \, \text{N/C} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ B_0 = \frac{120}{3 \times 10^8} = 4 \times 10^{-7} \, \text{T}. \] So, the amplitude of the magnetic field \( B_0 \) is \( 4 \times 10^{-7} \, \text{T} \).
ii) Wavelength \( \lambda \) of the Electromagnetic Wave:
The wavelength \( \lambda \) of an electromagnetic wave is related to its frequency \( \nu \) by the equation:
\[ \lambda = \frac{c}{\nu}, \] where:
- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \nu \) is the frequency of the wave.
Substituting the given value of frequency \( \nu = 50 \, \text{MHz} = 50 \times 10^6 \, \text{Hz} \):
\[ \lambda = \frac{3 \times 10^8}{50 \times 10^6} = 6 \, \text{m}. \] So, the wavelength \( \lambda \) of the electromagnetic wave is \( 6 \, \text{m} \).