Question

A plane electromagnetic wave of frequency $50MHz$ travels in free space along the positive x-direction. At a particular point in space and time, $\vec E = 6.3\hat jV/m$. The corresponding magnetic field at that point will be:
(A) $18.9 \times {10^{ - 8}}\hat kT$
(B) $6.3 \times {10^{ - 8}}\hat kT$
(C) $2.1 \times {10^{ - 8}}\hat kT$
(D) $18.9 \times {10^8}\hat kT$

Answer 1

Hint
The magnitude of the magnetic field will be given by the magnitude of the electric field divided by the speed of light. From the direction of the electromagnetic wave and the electric field we can find the direction of the magnetic field as they are mutually perpendicular to each other.
In this solution, we will be using the following formula,
$\Rightarrow \dfrac{{\left| {\vec E} \right|}}{{\left| {\vec B} \right|}} = c$
Where $\left| {\vec E} \right|$ is magnitude of electric field, $\left| {\vec B} \right|$ is magnitude of magnetic field and $c$ is the speed of light.

Complete step by step answer
The electromagnetic wave travels in a direction which is perpendicular to the direction of the magnetic field and the electric field. In the question we are given the electric field as, $\vec E = 6.3\hat jV/m$. From here the magnitude of the electric field is $\left| {\vec E} \right| = 6.3V/m$.
Now the speed of light is given by, $c = 3 \times {10^8}m/s$. So from the electric field we can find the value of the magnetic field as,
$\Rightarrow \dfrac{{\left| {\vec E} \right|}}{{\left| {\vec B} \right|}} = c$
Where we can write the magnetic field as,
$\Rightarrow \left| {\vec B} \right| = \dfrac{{\left| {\vec E} \right|}}{c}$
So substituting the values we get
$\Rightarrow \left| {\vec B} \right| = \dfrac{{6.3}}{{3 \times {{10}^8}}}$
This gives on calculation the magnitude of the magnetic field as,
$\left| {\vec B} \right| = 2.1 \times {10^{ - 8}}T$
Now since the electromagnetic field, electric and magnetic fields are mutually perpendicular, so from there we can find the direction of the magnetic field. Now if we take the cross product of the direction of the electric and magnetic field, we get the electromagnetic field.
In the question it is given that the electromagnetic field moves in X-direction, that is, $\hat i$ and electric field moves in Y-direction, that is $\hat j$. So, we can take the direction of magnetic field as $\hat A$ and write,
$\Rightarrow \hat j \times \hat A = \hat i$
So from here we can write the $\hat A$ is given by $\hat k$, since $\hat j \times \hat k = \hat i$.
So the magnetic field is $\vec B = 2.1 \times {10^{ - 8}}\hat kT$
Hence, the correct answer is option (C).

Note
The electromagnetic waves are formed as a result of the vibrations of the electric and magnetic field perpendicular to each other and also to the direction of propagation. The electromagnetic field travels with constant speed, that is $c = 3 \times {10^8}m/s$, in vacuum and is deflected by neither electric nor magnetic field.