Question

The ratio of E to B in electromagnetic waves is equal to
$\eqalign{
  & {\text{A}}{\text{. }}c \cr
  & {\text{B}}{\text{. }}\dfrac{1}{c} \cr
  & {\text{C}}{\text{. }}4c \cr
  & {\text{D}}{\text{. }}\dfrac{1}{Z} \cr} $

Answer 1

Hint: Electromagnetic waves are produced by accelerated particles and do not require any medium for their propagation. The ratio of electric field and magnetic field can be obtained by first forming the expression for these two quantities individually and then dividing them.

Complete step-by-step answer:
Faraday’s law of electromagnetic induction states that a time-varying magnetic field is a source of changing the electric field. Based on his theoretical studies, Maxwell argued that a time-varying electric field is also a source of the magnetic field.
This means that the change in either field i.e., electric or magnetic, there is a production of the other. Maxwell further showed that these variations in the electric and magnetic field occur at mutually perpendicular directions and have wave-like properties. Hence came forward the idea that a wave of electric and magnetic fields both varying in space and time should exist, one providing the source for another. Such a wave is known as an electromagnetic wave.
An electromagnetic wave is a wave radiated by an accelerated charge and which propagates through space as coupled electric and magnetic fields oscillating perpendicular to each other and the direction of propagation of the wave.
Mathematical expression for the representation of electromagnetic waves:
The following figure shows a plane electromagnetic wave traveling along the X-axis. The electric field oscillates along the Y-axis while the magnetic field oscillates along the Z-axis.

The values of the electric and magnetic fields shown above depend only on x and t.
The electric field vector can be represented mathematically as follows:
$\eqalign{
  & \vec E = {E_y}\hat j = {E_ \circ }\sin \left( {kx - \omega t} \right)\hat j \cr
  & \Rightarrow \vec E = {E_ \circ }\sin \left[ {2\pi \left( {\dfrac{x}{\lambda } - vt} \right)} \right]\hat j \cr
  & \Rightarrow \vec E = {E_ \circ }\sin \left[ {2\pi \left( {\dfrac{x}{\lambda } - \dfrac{t}{T}} \right)} \right]\hat j \cdots \cdots \cdots \left( 1 \right) \cr} $
where $k = 2\pi /\lambda $is the propagation constant of the wave and angular frequency, $\omega = 2\pi v$
Clearly, ${E_x} = {E_y} = 0$
The magnetic field vector can be represented mathematically as follows:
$\eqalign{
  & \vec B = {B_y}\hat k = {B_ \circ }\sin \left( {kx - \omega t} \right)\hat k \cr
  & \Rightarrow \vec B = {B_ \circ }\sin \left[ {2\pi \left( {\dfrac{x}{\lambda } - vt} \right)} \right]\hat k \cr
  & \Rightarrow \vec B = {B_ \circ }\sin \left[ {2\pi \left( {\dfrac{x}{\lambda } - \dfrac{t}{T}} \right)} \right]\hat k \cdots \cdots \cdots \cdots \left( 2 \right) \cr} $
Here ${E_ \circ }{\text{ and }}{B_ \circ }$are the amplitudes of the electric field and magnetic field respectively.
Equation (1) and (2) show that the variation in the electric and magnetic field are in the same phase, i.e. both attain their maximum and minima at the same instant and at the same place (x).
The magnitudes of electric field and magnetic field are related as:
$\dfrac{E}{B} = c{\text{ or }}\dfrac{{{E_ \circ }}}{{{B_ \circ }}} = c$
Therefore, the correct option is A. i.e., the ratio of E to B in electromagnetic waves is equal to c.

Note: Half of the intensity of an electromagnetic wave is provided by its electric field and half by the magnetic field. So the power delivered by the magnetic field of an electromagnetic wave is equal to the power delivered by its electric field, but the magnetic strength is much weaker than the electric field strength.