Question

If the peak value of the magnetic field of an electromagnetic wave is \( 30 \times 10^{-9} \, \text{T} \), then the peak value of the electric field is

• A. \( 3 \, \text{Vm}^{-1} \)
• B. \( 12 \, \text{Vm}^{-1} \)
• C. \( 6 \, \text{Vm}^{-1} \)
• D. \( 9 \, \text{Vm}^{-1} \) \bigskip

Hint:

The peak values of the electric and magnetic fields in an electromagnetic wave are related by the equation \( E = cB \), where \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, \text{m/s} \)). This relationship helps in finding one when the other is given.

Answer 1

The Correct Option is D

We are given the peak value of the magnetic field of the electromagnetic wave as \( B = 30 \times 10^{-9} \, \text{T} \). The relationship between the electric field \( E \) and magnetic field \( B \) in an electromagnetic wave is given by the equation: \[ E = cB \] where - \( E \) is the peak value of the electric field, - \( c \) is the speed of light in vacuum, \( c = 3 \times 10^8 \, \text{m/s} \), and - \( B \) is the peak value of the magnetic field. 

Step 1: We are provided with the value of \( B = 30 \times 10^{-9} \, \text{T} \) and the speed of light \( c = 3 \times 10^8 \, \text{m/s} \). 

 Step 2: Substitute the given values into the equation \( E = cB \): \[ E = (3 \times 10^8 \, \text{m/s}) \times (30 \times 10^{-9} \, \text{T}) \] 

 Step 3: Perform the multiplication: \[ E = 9 \times 10^0 \, \text{Vm}^{-1} \] 

Step 4: Simplify the result: \[ E = 9 \, \text{Vm}^{-1} \] Thus, the peak value of the electric field is \( \boxed{9 \, \text{Vm}^{-1}} \).