Question

A plane electromagnetic wave of frequency 50MHz travels in free space along the x-direction. At a particular point in the space and time, $\vec E = 6 \cdot 3\widehat j{\text{V}}{{\text{m}}^{ - 1}}$. At this point $\vec B$ is equal to,
A) $8 \cdot 33 \times {10^{ - 8}}\hat kT$
B) $18 \cdot 9 \times {10^{ - 8}}\hat kT$
C) $4 \cdot 1 \times {10^{ - 8}}\hat kT$
D) $2 \cdot 1 \times {10^{ - 8}}\hat kT$

Answer 1

The characteristics of the wave can help us find the direction of the magnetic field. Let suppose the wave has the direction of motion in the x-direction, and then the direction of the magnetic field will be perpendicular to the direction of motion of the wave and the direction of the electric field.

Formula used:
${{\text{B}}_{\text{o}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{o}}}}}{c}$ where ${{\text{B}}_{\text{o}}}$ is the magnetic field ${{\text{E}}_{\text{o}}}$ is the electric field and c is the speed of light, the value of speed of light is given by $c = 3 \times {10^8}$.

Complete step by step answer:
As it is given that the frequency of the wave is $50MHz$ and is in the x-direction and the electric field is given as $\vec E = 6 \cdot 3\widehat j{\text{V}}{{\text{m}}^{ - 1}}$.It is given that the wave is moving towards the x-direction and the direction of the electric field is in the y-direction.
Now let us calculate the value of the magnetic field
As the magnetic field is given by,
$\Rightarrow {{\text{B}}_{\text{o}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{o}}}}}{c}$
Replace the value of the electric field and the speed of light in the above relation. The value of electric field is $E = 6 \cdot 3$ and the value of speed of light is equal to $c = 3 \times {10^8}m{s^{ - 1}}$.
$\Rightarrow {{\text{B}}_{\text{o}}}{\text{ = }}\dfrac{{6 \cdot 3}}{{3 \times {{10}^8}}}$
$\Rightarrow {{\text{B}}_{\text{o}}}{\text{ = 2}} \cdot {\text{1}} \times {\text{1}}{{\text{0}}^{ - 8}}T$
The direction of the magnetic field will be equal to$\vec E \times \vec B$. So the direction of the magnetic field is $\hat k$.

Therefore, the magnetic field of the wave is given by ${{\text{B}}_{\text{o}}}{\text{ = 2}} \cdot {\text{1}} \times {\text{1}}{{\text{0}}^{ - 8}}\hat kT$. So, the correct answer is option D.

Note:
The direction of the magnetic field will be perpendicular to the direction of the motion of the wave and the electric field also, so the cross product is how you calculate the direction of the magnetic field of the given wave.