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Question: Average energy density of electromagnetic wave $E_y = E_0 \sin(kx - \omega t)$ and $B_z = B_0 \sin (...

Average energy density of electromagnetic wave Ey=E0sin⁡(kx−ωt)E_y = E_0 \sin(kx - \omega t) and Bz=B0sin⁡(kx−ωt)B_z = B_0 \sin (kx - \omega t) is given by

A

12ϵ0E02\frac{1}{2} \epsilon_0 E_0^2

B

B022μ0\frac{B_0^2}{2\mu_0}

C

ϵ0E024+B024μ0\frac{\epsilon_0 E_0^2}{4} + \frac{B_0^2}{4\mu_0}

D

All of these

Answer

All of these

Explanation

Solution

The electric field of the electromagnetic wave is given by Ey=E0sin⁡(kx−ωt)E_y = E_0 \sin(kx - \omega t), and the magnetic field by Bz=B0sin⁡(kx−ωt)B_z = B_0 \sin (kx - \omega t).

The instantaneous energy density due to the electric field is: ue=12ϵ0E2=12ϵ0(E0sin⁡(kx−ωt))2=12ϵ0E02sin⁡2(kx−ωt)u_e = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 (E_0 \sin(kx - \omega t))^2 = \frac{1}{2} \epsilon_0 E_0^2 \sin^2(kx - \omega t)

The instantaneous energy density due to the magnetic field is: um=12μ0B2=12μ0(B0sin⁡(kx−ωt))2=12μ0B02sin⁡2(kx−ωt)u_m = \frac{1}{2\mu_0} B^2 = \frac{1}{2\mu_0} (B_0 \sin(kx - \omega t))^2 = \frac{1}{2\mu_0} B_0^2 \sin^2(kx - \omega t)

To find the average energy density, we need to average the instantaneous energy densities over one full cycle. The average value of sin⁡2(θ)\sin^2(\theta) over a full cycle is 12\frac{1}{2}.

So, the average electric energy density is: ⟨ue⟩=12ϵ0E02⟨sin⁡2(kx−ωt)⟩=12ϵ0E02×12=14ϵ0E02\langle u_e \rangle = \frac{1}{2} \epsilon_0 E_0^2 \langle \sin^2(kx - \omega t) \rangle = \frac{1}{2} \epsilon_0 E_0^2 \times \frac{1}{2} = \frac{1}{4} \epsilon_0 E_0^2

The average magnetic energy density is: ⟨um⟩=12μ0B02⟨sin⁡2(kx−ωt)⟩=12μ0B02×12=14μ0B02\langle u_m \rangle = \frac{1}{2\mu_0} B_0^2 \langle \sin^2(kx - \omega t) \rangle = \frac{1}{2\mu_0} B_0^2 \times \frac{1}{2} = \frac{1}{4\mu_0} B_0^2

The total average energy density is the sum of the average electric and magnetic energy densities: ⟨u⟩=⟨ue⟩+⟨um⟩=14ϵ0E02+14μ0B02\langle u \rangle = \langle u_e \rangle + \langle u_m \rangle = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4\mu_0} B_0^2 This matches option C.

For an electromagnetic wave in vacuum, the magnitudes of the electric and magnetic fields are related by E0=cB0E_0 = c B_0, where cc is the speed of light in vacuum. The speed of light is also given by c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}. Therefore, E02=c2B02=(1μ0ϵ0)2B02=1μ0ϵ0B02E_0^2 = c^2 B_0^2 = \left(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\right)^2 B_0^2 = \frac{1}{\mu_0 \epsilon_0} B_0^2. From this relation, we can write ϵ0E02=B02μ0\epsilon_0 E_0^2 = \frac{B_0^2}{\mu_0}.

Now, let's substitute this relationship into the expression for ⟨ue⟩\langle u_e \rangle: ⟨ue⟩=14ϵ0E02=14(B02μ0)=B024μ0\langle u_e \rangle = \frac{1}{4} \epsilon_0 E_0^2 = \frac{1}{4} \left(\frac{B_0^2}{\mu_0}\right) = \frac{B_0^2}{4\mu_0} This shows that ⟨ue⟩=⟨um⟩\langle u_e \rangle = \langle u_m \rangle. The average energy density is equally divided between the electric and magnetic fields.

Using this equality, we can express the total average energy density in alternative forms:

  1. In terms of E0E_0 only: ⟨u⟩=⟨ue⟩+⟨um⟩=⟨ue⟩+⟨ue⟩=2⟨ue⟩=2(14ϵ0E02)=12ϵ0E02\langle u \rangle = \langle u_e \rangle + \langle u_m \rangle = \langle u_e \rangle + \langle u_e \rangle = 2 \langle u_e \rangle = 2 \left(\frac{1}{4} \epsilon_0 E_0^2\right) = \frac{1}{2} \epsilon_0 E_0^2 This matches option A.

  2. In terms of B0B_0 only: ⟨u⟩=⟨ue⟩+⟨um⟩=⟨um⟩+⟨um⟩=2⟨um⟩=2(14μ0B02)=B022μ0\langle u \rangle = \langle u_e \rangle + \langle u_m \rangle = \langle u_m \rangle + \langle u_m \rangle = 2 \langle u_m \rangle = 2 \left(\frac{1}{4\mu_0} B_0^2\right) = \frac{B_0^2}{2\mu_0} This matches option B.

Since options A, B, and C are all correct expressions for the average energy density of an electromagnetic wave, the correct choice is "All of these".