Question

In an electromagnetic wave travelling in free space, the amplitude of magnetic field is \( 6.0 \times 10^{-4} \ \text{T} \). The amplitude of its electric field is:

• A. \( 2 \times 10^4 \ \text{Vm}^{-1} \)
• B. \( 1.5 \times 10^{12} \ \text{Vm}^{-1} \)
• C. \( 1.8 \times 10^5 \ \text{Vm}^{-1} \)
• D. \( 0.3 \times 10^4 \ \text{Vm}^{-1} \)

Hint:

In an electromagnetic wave, the electric and magnetic fields are related by \( E = cB \), where \( c \) is the speed of light. This relationship holds in free space.

Answer 1

The Correct Option is C

The relationship between the electric field \( E \) and the magnetic field \( B \) in an electromagnetic wave is \( E = cB \). Here, \( E \) is the electric field amplitude, \( B \) is the magnetic field amplitude, and \( c \) is the speed of light in vacuum, which is \( 3 \times 10^8 \ \text{m/s} \). Given that \( B = 6.0 \times 10^{-4} \ \text{T} \), we can substitute these values into the equation: \[ E = (3 \times 10^8 \ \text{m/s}) \times (6.0 \times 10^{-4} \ \text{T}) \] \[ E = 1.8 \times 10^5 \ \text{Vm}^{-1} \] Therefore, the electric field amplitude is \( 1.8 \times 10^5 \ \text{Vm}^{-1} \). The correct answer is: \[ \boxed{C} \ 1.8 \times 10^5 \ \text{Vm}^{-1} \]