Question

A plane electromagnetic wave travels in free space along $\mathrm{x}-$direction. If the value of $\overrightarrow{\mathrm{B}}$ (in tesla) at a particular point in space and time is $1.2\times {10}^{-8}\hat{\mathrm{k}}$, the value of $\overrightarrow{\mathrm{E}}$ (in $\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.$) at that point is,

• A. $3.6\hat{\mathrm{k}}$
• B. $1.2\hat{\mathrm{k}}$
• C. $1.2\hat{\mathrm{j}}$
• D. $3.6\hat{\mathrm{j}}$

Answer 1

The correct option is D

$3.6\hat{\mathrm{j}}$

Step 1. Given data

It is given that the magnetic field of the plane electromagnetic wave is, $1.2\times {10}^{-8}\hat{\mathrm{k}}\mathrm{T}$.

We have to find the value of $\overrightarrow{\mathrm{E}}$.

Step 2. Formula to be used.

As we know that, the magnitude of the electric field is simply defined as the force per charge on the test charge.

For an electromagnetic wave, ratio of magnitudes of electric and magnetic field is,

$\frac{\mathrm{E}}{\mathrm{B}}=C$

Here, $\mathrm{C}$ is the speed of electromagnetic wave in vacuum.

Step 3. Find the value of $\overrightarrow{\mathrm{E}}$.

The direction of propagation of the electromagnetic wave is along positive $\mathrm{x}-$ direction.

The magnitude of $\overrightarrow{\mathrm{E}}$ is,

$\mathrm{E}=\mathrm{B}\times \mathrm{c}$

$=\left(1.2\times {10}^{-8}\right)\times \left(3\times {10}^{-8}\right)$

$=3.6\mathrm{V}/\mathrm{m}$

So, the magnetic field is positive $\mathrm{z}-$ direction and the wave propagates along positive $\mathrm{x}-$ direction.

Step 4. Calculate the magnitude of the electric field.

Thus, $\overrightarrow{\mathrm{E}}$should be in a direction perpendicular to both $\mathrm{x}$ and $\mathrm{z}$ axes.

That means along $\mathrm{y}-$ axis $\to \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{E}}=3.6\hat{\mathrm{j}}\mathrm{V}/\mathrm{m}$

Thus, the value of $\overrightarrow{\mathrm{E}}$ is $3.6\hat{\mathrm{j}}\mathrm{V}/\mathrm{m}$

Hence, option $\mathrm{D}$ is the correct answer.