CBSE · Class 12 · Physics · Electromagnetic WavesIn a plane electromagnetic wave, the electric field oscillates sinusoidally with a frequency of $2.0 \times 10^{10} \text{ Hz}$ and an amplitude of $48 \text{ V/m}$. (i) What is the wavelength of the wave? (ii) What is the amplitude of the oscillating magnetic field? (iii) Show that the average energy density of the electric field equals the average energy density of the magnetic field.

Given Data:

• Frequency of wave, $\nu = 2.0 \times 10^{10} \text{ Hz}$
• Amplitude of electric field, $E_0 = 48 \text{ V/m}$
• Speed of light, $c = 3 \times 10^8 \text{ m/s}$

(i) Calculation of Wavelength ($\lambda$):\nThe relation between speed of light, frequency, and wavelength is given by:

$$c = \nu \lambda$$ $$\lambda = \frac{c}{\nu}$|\nSubstitute the given values: $$\lambda = \frac{3 \times 10^8 \text{ m/s}}{2.0 \times 10^{10} \text{ Hz}}$$ $$\lambda = 1.5 \times 10^{-2} \text{ m} = 1.5 \text{ cm}$| Answer (i): The wavelength of the wave is $1.5 \times 10^{-2} \text{ m}$.

(ii) Calculation of Magnetic Field Amplitude ($B_0$):\nThe relation between electric field amplitude and magnetic field amplitude in an electromagnetic wave is:

$$B_0 = \frac{E_0}{c}$$\nSubstitute the values: $$B_0 = \frac{48 \text{ V/m}}{3 \times 10^8 \text{ m/s}}$$ $$B_0 = 1.6 \times 10^{-7} \text{ Tesla} \text{ (or } \text{Wb/m}^2\text{)}$$ Answer (ii): The amplitude of the magnetic field is $1.6 \times 10^{-7} \text{ T}$.

(iii) Proof that Average Energy Densities are Equal:

1. Electric Energy Density ($u_E$): The instantaneous electric energy density is given by: $$u_E = \frac{1}{2} \varepsilon_0 E^2$$ The average electric energy density over a complete cycle is: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 \langle E^2 \rangle$$ Since $\langle E^2 \rangle = \frac{E_0^2}{2}$, we get: $$\langle u_E \rangle = \frac{1}{4} \varepsilon_0 E_0^2$$

2. Magnetic Energy Density ($u_B$): The instantaneous magnetic energy density is given by: $$u_B = \frac{1}{2 \mu_0} B^2$$ The average magnetic energy density over a complete cycle is: $$\langle u_B \rangle = \frac{1}{2 \mu_0} \langle B^2 \rangle$$ Since $\langle B^2 \rangle = \frac{B_0^2}{2}$, we get: $$\langle u_B \rangle = \frac{1}{4 \mu_0} B_0^2$$

3. Comparison: We know that $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$, which implies $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, or $\mu_0 \varepsilon_0 = \frac{1}{c^2}$. Also, $B_0 = \frac{E_0}{c} \implies B_0^2 = \frac{E_0^2}{c^2}$. Substitute $B_0^2$ into the magnetic energy density formula: $$\langle u_B \rangle = \frac{1}{4 \mu_0} \left(\frac{E_0^2}{c^2}\right) = \frac{1}{4 \mu_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right)} E_0^2 = \frac{1}{4} \varepsilon_0 E_0^2$$ Therefore, $$\langle u_E \rangle = \langle u_B \rangle$$ Hence proved.