Question

A plane electromagnetic wave of frequency 20 MHz travels through a space along $ z $-direction. If the electric field vector at a certain point in space is 6 V/m, then what is the magnetic field vector at that point?

• A. \( 2 \times 10^{-8} \, \text{T} \)
• B. \( 3 \times 10^{-5} \, \text{T} \)
• C. \( 2 \, \text{T} \)
• D. \( \frac{1}{2} \, \text{T} \)

Hint:

For electromagnetic waves, the electric field and magnetic field are related by \( E = cB \). This relationship helps in calculating the magnetic field if the electric field is known.

Answer 1

The Correct Option is A

We are given an electromagnetic wave traveling through space and need to find the magnetic field vector \( B \) at a point where the electric field vector \( E = 6 \, \text{V/m} \). The frequency of the wave is \( 20 \, \text{MHz} \).

An electromagnetic wave in free space is characterized by the relationship between the electric field \( E \), magnetic field \( B \), and the speed of light \( c \). The relationship is given by the equation:

\[ B = \frac{E}{c} \]

where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the given electric field \( E = 6 \, \text{V/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \) into the equation, we calculate:

\[ B = \frac{6 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} = 2 \times 10^{-8} \, \text{T} \]

Thus, the magnetic field vector at that point is \( 2 \times 10^{-8} \, \text{T} \).

Answer 2

We are given an electromagnetic wave traveling along the \( z \)-direction with a frequency of 20 MHz, and the electric field vector at a certain point in space is 6 V/m. We need to determine the magnetic field vector at that point.

The relationship between the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \) in an electromagnetic wave is given by:

\[ c = \frac{E}{B} \]

where \( c \) is the speed of light in a vacuum, which is approximately \( 3 \times 10^8 \, \text{m/s} \).

Rearranging the equation to solve for \( B \):

\[ B = \frac{E}{c} \]

Substituting the given values (\( E = 6 \, \text{V/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \)) into the formula:

\[ B = \frac{6}{3 \times 10^8} \]

\[ B = 2 \times 10^{-8} \, \text{T} \]

Therefore, the magnetic field vector at that point is \( 2 \times 10^{-8} \, \text{T} \).