Question

In a plane electromagnetic wave, the electric field of amplitude \( 1 \text{ V m}^{-1} \) varies with time in free space. The average energy density of magnetic field is (in \( \text{Jm}^{-3} \))

• A. \( 8.86 \times 10^{-12} \)
• B. \( 4.43 \times 10^{-12} \)
• C. \( 17.72 \times 10^{-12} \)
• D. \( 2.21 \times 10^{-12} \)
• E. \( 1.11 \times 10^{-12} \)

Hint:

Always remember that for any EM wave in a vacuum, the average energy stored in the electric field is exactly equal to the average energy stored in the magnetic field 120].

Answer 1

The Correct Option is D

Concept: In a plane electromagnetic wave propagating through free space, the energy is distributed equally between the electric and magnetic fields 113].
• Total Average Energy Density (\( u_{avg} \)): This is given by \( u_{avg} = u_E + u_B \) 114].
• Average Electric Energy Density (\( u_E \)): Defined as \( u_E = \frac{1}{4} \epsilon_0 E_0^2 \), where \( E_0 \) is the electric field amplitude 115].
• Average Magnetic Energy Density (\( u_B \)): In free space, \( u_B = u_E \) 116]. Thus, \( u_B = \frac{1}{4} \epsilon_0 E_0^2 \) 117].

Step 1: Identify the values and formula.
Given electric field amplitude \( E_0 = 1 \text{ V m}^{-1} \) and the permittivity of free space \( \epsilon_0 \approx 8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2} \) 118]. The formula for the average magnetic energy density is: \[ u_B = \frac{1}{4} \epsilon_0 E_0^2 \] 119]

Step 2: Perform the calculation.
Substitute the given values: \[ u_B = \frac{1}{4} \times (8.854 \times 10^{-12}) \times (1)^2 \] \[ u_B = 2.2135 \times 10^{-12} \text{ J m}^{-3} \] Rounding to significant figures, we get approximately \( 2.21 \times 10^{-12} \text{ J m}^{-3} \) 119].