Question

A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular space and time, \(\overrightarrow E =6.6 \;\hat jV/m  .\)What is \(\overrightarrow B\) at this point ?

• A. \(-2.2 \times 10^{-}\hat i\;T \)
• B. \(-2.2 \times 10^{-8}\hat k\;T\)
• C. \(2.2 \times 10^{-8}\hat i\;T\)
• D. \(2.2 \times 10^{-8}\hat k\;T\)

Answer 1

The Correct Option is D

Given: 

• Electric field magnitude (\( |\vec{E}| \)) = \( 6.6 \, \hat{j} \, \text{V/m} \)
• Frequency (\( f \)) = \( 20 \, \text{MHz} \)
• Wave propagation direction (\( \vec{C} \)) = \( \hat{i} \, \text{(x-direction)} \)

Step 1: Calculate the Magnitude of \( \vec{B} \)

The magnitude of the magnetic field (\( |\vec{B}| \)) is related to the electric field magnitude (\( |\vec{E}| \)) by:

\[ |\vec{B}| = \frac{|\vec{E}|}{c}, \]

where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light in free space. Substituting the given values:

\[ |\vec{B}| = \frac{6.6}{3 \times 10^8} = 2.2 \times 10^{-8} \, \text{T}. \]

Step 2: Determine the Direction of \( \vec{B} \)

In an electromagnetic wave, the electric field (\( \vec{E} \)), magnetic field (\( \vec{B} \)), and propagation direction (\( \vec{C} \)) are mutually perpendicular and follow the right-hand rule:

\[ \vec{E} \times \vec{B} = \vec{C}. \]

• \( \vec{C} \) is in the \( \hat{i} \)-direction (x-axis).
• \( \vec{E} \) is in the \( \hat{j} \)-direction (y-axis).
• Using the right-hand rule, \( \vec{B} \) must be in the \( \hat{k} \)-direction (z-axis), as \( \hat{j} \times \hat{k} = \hat{i} \).

Step 3: Combine Magnitude and Direction

Combining the magnitude and direction, the magnetic field is:

\[ \vec{B} = (2.2 \times 10^{-8}) \, \hat{k} \, \text{T}. \]

Final Answer:

\( \vec{B} = 2.2 \times 10^{-8} \, \hat{k} \, \text{T} \).