Question

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is \( E_y = 9.3 \, \text{V/m} \). Then, the magnetic field vector of the wave at that point is:

• A. \( B_z = 9.3 \times 10^{-8} \, \text{T} \)
• B. \( B_z = 1.55 \times 10^{-8} \, \text{T} \)
• C. \( B_z = 6.2 \times 10^{-8} \, \text{T} \)
• D. \( B_z = 3.1 \times 10^{-8} \, \text{T} \)

Hint:

The magnetic field of an electromagnetic wave is related to the electric field by \( B = \frac{E}{c} \), where \( c \) is the speed of light.

Answer 1

The Correct Option is D

For a plane electromagnetic wave, the relationship between the electric field \( E \) and the magnetic field \( B \) is given by: \[ B = \frac{E}{c}, \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light in a vacuum. Given: \[ E = 9.3 \, \text{V/m}, \quad f = 20 \, \text{MHz}. \] The wavelength \( \lambda \) of the wave can be found using the relationship: \[ c = f \lambda \quad \Rightarrow \quad \lambda = \frac{c}{f}. \] Now, using the value \( E = 9.3 \, \text{V/m} \), the magnetic field \( B \) is: \[ B = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \, \text{T}. \]