NCERT Solutions: Class 12 Physics Chapter 8 - Electromagnetic Waves

The capacitance is $C=\epsilon_0A/d=\epsilon_0\pi R^2/d=(8.854\times10^{-12})\pi(0.12)^2/0.050=8.0\times10^{-12}\,\text{F}$. Since $I=dQ/dt=C\,dV/dt$, $dV/dt=I/C=0.15/(8.0\times10^{-12})=1.9\times10^{10}\,\text{V s}^{-1}$. The displacement current between the plates equals $\epsilon_0 d\Phi_E/dt=C\,dV/dt$, which is the same as the charging current, $0.15\,\text{A}$. Therefore Kirchhoff's junction rule remains valid when the displacement current is counted as the current continuing through the gap.

For a capacitor in ac, $I_{\text{rms}}=V_{\text{rms}}\omega C=230(300)(100\times10^{-12})=6.9\times10^{-6}\,\text{A}$. The displacement current between the plates equals the conduction current in the leads at every instant. For the field amplitude at radius $r=0.030\,\text{m}$ inside plates of radius $R=0.060\,\text{m}$, use Ampere-Maxwell law: $B_0(2\pi r)=\mu_0 I_0(r^2/R^2)$, where $I_0=\sqrt2 I_{\text{rms}}=9.76\times10^{-6}\,\text{A}$. Thus $B_0=\mu_0 I_0 r/(2\pi R^2)=1.6\times10^{-11}\,\text{T}$.

X-rays, visible light and radio waves are all electromagnetic waves. In vacuum, all electromagnetic waves travel with the same speed $c$, independent of wavelength.

An electromagnetic wave is transverse: $\mathbf{E}\perp\mathbf{B}$ and both are perpendicular to the direction of propagation. For propagation along z, $\mathbf{E}$ and $\mathbf{B}$ may lie along x and y directions. The wavelength is $\lambda=c/\nu=3.0\times10^8/(30\times10^6)=10\,\text{m}$.

Wavelength is $\lambda=c/\nu$. At $7.5\,\text{MHz}$, $\lambda=3.0\times10^8/(7.5\times10^6)=40\,\text{m}$. At $12\,\text{MHz}$, $\lambda=3.0\times10^8/(12\times10^6)=25\,\text{m}$. Thus the wavelength band is $40\,\text{m}$ to $25\,\text{m}$.

An oscillating charge radiates electromagnetic waves at the same frequency as its oscillation. Therefore the emitted electromagnetic waves have frequency $10^9\,\text{Hz}$.

For an electromagnetic wave in vacuum, $E_0=cB_0$. With $B_0=510\,\text{nT}=510\times10^{-9}\,\text{T}$, $E_0=(3.0\times10^8)(510\times10^{-9})=153\,\text{V m}^{-1}$.

The magnetic-field amplitude is $B_0=E_0/c=120/(3.0\times10^8)=4.0\times10^{-7}\,\text{T}$. With $\nu=50.0\times10^6\,\text{Hz}$, $\omega=2\pi\nu=3.14\times10^8\,\text{rad s}^{-1}$ and $\lambda=c/\nu=3.0\times10^8/(50.0\times10^6)=6.0\,\text{m}$. Hence $k=2\pi/\lambda=1.05\,\text{rad m}^{-1}$. The direction of propagation is not specified, so choose propagation along $+x$, electric field along $+y$, and magnetic field along $+z$ so that $\mathbf{E}\times\mathbf{B}$ points along $+x$.

Using $E=h\nu=hc/\lambda$ and $hc\approx1240\,\text{eV nm}$, the photon energy scale follows directly from wavelength. Longer-wavelength waves such as radio and microwaves have very small photon energies and are produced by accelerated charges in circuits or devices. Infrared energies match molecular rotations and vibrations. Visible and ultraviolet energies are comparable to electronic transitions in atoms and molecules. X-ray photons correspond to inner-shell atomic transitions or high-energy electron bombardment, while gamma rays have nuclear energy scales and are produced in nuclear transitions or radioactive decay.

The wavelength is $\lambda=c/\nu=3.0\times10^8/(2.0\times10^{10})=1.5\times10^{-2}\,\text{m}$. The magnetic field amplitude is $B_0=E_0/c=48/(3.0\times10^8)=1.6\times10^{-7}\,\text{T}$. The average electric energy density is $\langle u_E\rangle=\frac12\epsilon_0\langle E^2\rangle=\frac12\epsilon_0(E_0^2/2)=\epsilon_0E_0^2/4$. The average magnetic energy density is $\langle u_B\rangle=\frac{1}{2\mu_0}\langle B^2\rangle=B_0^2/(4\mu_0)=E_0^2/(4\mu_0c^2)=\epsilon_0E_0^2/4$, since $c^2=1/(\mu_0\epsilon_0)$.