Question

An electromagnetic wave going through vacuum is described by \[E = {E_0}\sin \left( {kx - \omega t} \right)\]; \[B = {B_0}\sin \left( {kx - \omega t} \right)\].
Which of the following is true?
A. \[{E_0}k = {B_0}\omega \]
B. \[{E_0}\omega = {B_0}k\]
C. \[{E_0}{B_0} = \omega k\]
D. None of the above

Answer 1

Hint: Recall Maxwell's equation and calculate the curl of the electric field. We have given the solutions of differential equations of electric field and magnetic field. Substitute it into the Maxwell’s equation you obtained after taking the curl.

Formula used:
Maxwell’s equation, \[\nabla \times E = - \dfrac{{\partial B}}{{\partial t}}\], where, E is the electric field and B is the magnetic field.

Complete step by step solution:
To answer this question, we can derive the relation between electric field and magnetic field of the electromagnetic wave using Maxwell’s equation. We will start with one of Maxwell’s equation,
\[\nabla \times E = - \dfrac{{\partial B}}{{\partial t}}\] …… (1)

We assume the electric field is along the y-axis and magnetic field is along the z-axis. Since electric field and magnetic field are only function of distance x and time t, we can write the equation for electric field and magnetic field as follows,
\[\vec E\left( {x,t} \right) = E\left( {x,t} \right)\hat j\] and, \[\vec B\left( {x,t} \right) = B\left( {x,t} \right)\hat k\] …… (2)

We take the curl of electric field as follows,
\[\nabla \times \vec E\left( {x,t} \right) = \left[ {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\
  0&{E\left( {x,t} \right)}&0
\end{array}} \right]\]
\[ \Rightarrow \nabla \times \vec E\left( {x,t} \right) = - \dfrac{{\partial E}}{{\partial x}}\hat k\] …… (3)

From equation (1) and (2), we can write,
\[\dfrac{{\partial E}}{{\partial x}} = - \dfrac{{\partial B}}{{\partial t}}\] …… (4)

We have given the solutions of differential equations of electric field and magnetic field is,
\[E = {E_0}\sin \left( {kx - \omega t} \right)\] …… (5)

And,
\[B = {B_0}\sin \left( {kx - \omega t} \right)\] …… (6)

Substituting equation (5) and (6) in equation (4), we have,
\[\dfrac{\partial }{{\partial x}}\left( {{E_0}\sin \left( {kx - \omega t} \right)} \right) = - \dfrac{\partial }{{\partial t}}\left( {{B_0}\sin \left( {kx - \omega t} \right)} \right)\]
\[ \Rightarrow {E_0}\cos \left( {kx - \omega t} \right)\left( k \right) = - {B_0}\cos \left( {kx - \omega t} \right)\left( { - \omega } \right)\]
\[ \Rightarrow k{E_0} = {B_0}\omega \]

So, the correct answer is “Option A”.

Note:
Another way to answer this question is to remember the relation, \[\dfrac{{{E_0}}}{{{B_0}}} = c\]. We know the relation between speed of light c, wave number k and angular frequency \[\omega \], \[c = \dfrac{\omega }{k}\]. Therefore, we can write, \[\dfrac{{{E_0}}}{{{B_0}}} = \dfrac{\omega }{k}\]. Note that this relation is valid for only electromagnetic waves.