NCERT Solutions · Class 12 Physics Electromagnetic Waves

(a) Capacitance of a parallel-plate capacitor: \[C=\frac{\epsilon_0 A}{d}=\frac{\epsilon_0 \pi R^2}{d}\] With \(\displaystyle R=0.12\,\mathrm{m}\), \(\displaystyle d=0.05\,\mathrm{m}\): \(\displaystyle A=\pi(0.12)^2=0.04524\,\mathrm{m^2}\) \[C=\frac{(8.854\times10^{-12})(0.04524)}{0.05}=8.01\times10^{-12}\,\mathrm{F}\] Since \(\displaystyle I=C\dfrac{dV}{dt}\) (constant charging current), \[\frac{dV}{dt}=\frac{I}{C}=\frac{0.15}{8.01\times10^{-12}}=1.87\times10^{10}\,\mathrm{V/s}\] C \(\displaystyle \approx\) $\displaystyle 8.01$ pF, dV/dt \(\displaystyle \approx\) $\displaystyle 1.87$\(\displaystyle \times10^{10}\) V/s(b) The displacement current is \(\displaystyle I_d=\epsilon_0 A\dfrac{dE}{dt}=\epsilon_0\dfrac{A}{d}\dfrac{dV}{dt}=C\dfrac{dV}{dt}=I\). So \(\displaystyle I_d = 0.15\,\mathrm{A}\), exactly equal to the conduction current in the wires.(c) Kirchhoff's junction rule, as stated for conduction current alone, fails at a capacitor plate — current flows in through the wire but no conduction current crosses the gap between the plates. However, once the displacement current \(\displaystyle I_d\) between the plates is included (Maxwell's correction to Ampère's law), \(\displaystyle I_d=I\) at every instant, so the sum of conduction current and displacement current is continuous across the plate. The junction rule is valid at each plate only when displacement current is counted along with conduction current.

(a) Capacitive reactance: \(\displaystyle X_C=\dfrac{1}{\omega C}=\dfrac{1}{(300)(100\times10^{-12})}=3.33\times10^{7}\,\Omega\) \[I_{rms}=\frac{V_{rms}}{X_C}=\frac{230}{3.33\times10^{7}}=6.9\times10^{-6}\,\mathrm{A}\] \(\displaystyle I_{rms}\approx 6.9\,\mu\mathrm{A}\)(b) Yes. By charge conservation (Maxwell's equations), the displacement current between the plates equals the conduction current supplying the plates at every instant: \(\displaystyle I_d=I_{conduction}\).(c) Peak current \(\displaystyle I_0=\sqrt2\,I_{rms}=1.414\times6.9\times10^{-6}=9.76\times10^{-6}\,\mathrm{A}\). Inside the plates (\(\displaystyle r<R\)), the Ampère–Maxwell law gives, by symmetry, \[B=\frac{\mu_0 I_0\,r}{2\pi R^2}\] With \(\displaystyle r=0.03\,\mathrm{m}\), \(\displaystyle R=0.06\,\mathrm{m}\): \[B=\frac{(4\pi\times10^{-7})(9.76\times10^{-6})(0.03)}{2\pi(0.06)^2}=1.63\times10^{-11}\,\mathrm{T}\] B \(\displaystyle \approx\) $\displaystyle 1.63$\(\displaystyle \times10^{-11}\) T ($\displaystyle 16.3$ pT)

All electromagnetic waves — X-rays, visible light, and radio waves alike — travel through vacuum at the same speed, the speed of light \(\displaystyle c=3\times10^{8}\,\mathrm{m\,s^{-1}}\), independent of wavelength or frequency. This is the physical quantity common to all three.

For a plane electromagnetic wave in vacuum, \(\displaystyle \vec E\), \(\displaystyle \vec B\), and the direction of propagation form a mutually perpendicular right-handed set (\(\displaystyle \vec E\times\vec B\) points along the propagation direction). Since the wave travels along \(\displaystyle z\), \(\displaystyle \vec E\) and \(\displaystyle \vec B\) both lie in the \(\displaystyle xy\)-plane, perpendicular to \(\displaystyle z\) and perpendicular to each other — e.g. \(\displaystyle E\) along \(\displaystyle x\), \(\displaystyle B\) along \(\displaystyle y\) (or any such perpendicular pair rotated about \(\displaystyle z\)).Wavelength from \(\displaystyle c=\nu\lambda\): \[\lambda=\frac{c}{\nu}=\frac{3\times10^{8}}{30\times10^{6}}=10\,\mathrm{m}\] \(\displaystyle \lambda = 10\,\mathrm{m}\)

Using \(\displaystyle \lambda=c/\nu\): \[\lambda_{min}=\frac{c}{\nu_{max}}=\frac{3\times10^{8}}{12\times10^{6}}=25\,\mathrm{m}\] \[\lambda_{max}=\frac{c}{\nu_{min}}=\frac{3\times10^{8}}{7.5\times10^{6}}=40\,\mathrm{m}\] The wavelength band is $\displaystyle 25$ m to $\displaystyle 40$ m.

An oscillating charge radiates an electromagnetic wave at exactly the frequency of its mechanical oscillation, since the accelerating charge's field oscillates in step with its motion. The electromagnetic wave has frequency \(\displaystyle \nu=10^{9}\,\mathrm{Hz}\) ($\displaystyle 1$ GHz), the same as the oscillator.

For an electromagnetic wave in vacuum, the field amplitudes are related by \(\displaystyle E_0=cB_0\). \[E_0=(3\times10^{8})(510\times10^{-9})=153\,\mathrm{N/C}\] \(\displaystyle E_0 = 153\,\mathrm{N/C}\)

(a) \(\displaystyle B_0=\dfrac{E_0}{c}=\dfrac{120}{3\times10^{8}}=4\times10^{-7}\,\mathrm{T}\)\(\displaystyle \omega=2\pi\nu=2\pi(50.0\times10^{6})=3.14\times10^{8}\,\mathrm{rad/s}\)\(\displaystyle \lambda=\dfrac{c}{\nu}=\dfrac{3\times10^{8}}{50\times10^{6}}=6\,\mathrm{m}\)\(\displaystyle k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{6}=1.05\,\mathrm{rad/m}\)\(\displaystyle B_0=4\times10^{-7}\,\mathrm{T}\), \(\displaystyle \omega=3.14\times10^{8}\,\mathrm{rad/s}\), \(\displaystyle k=1.05\,\mathrm{rad/m}\), \(\displaystyle \lambda=6\,\mathrm{m}\)(b) Taking the wave to propagate along \(\displaystyle x\), with \(\displaystyle E\) along \(\displaystyle y\) and \(\displaystyle B\) along \(\displaystyle z\) (so \(\displaystyle \vec E\times\vec B\) points along \(\displaystyle +x\)): \[E_y=120\cos\!\big(1.05x-3.14\times10^{8}t\big)\,\mathrm{N/C}\] \[B_z=4\times10^{-7}\cos\!\big(1.05x-3.14\times10^{8}t\big)\,\mathrm{T}\]

Using \(\displaystyle E=h\nu\) with \(\displaystyle h=6.63\times10^{-34}\,\mathrm{J\,s}\) and converting to eV (\(\displaystyle 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}\)), taking one representative frequency from each band of the spectrum: The photon energy climbs by roughly $\displaystyle 14$ orders of magnitude from radio to gamma rays, and this scale tracks the source mechanism: radio/microwaves come from oscillating currents in circuits and antennas (low-energy, collective electron motion); infrared from molecular vibrations/rotations; visible/UV from outer electronic transitions in atoms; X-rays from inner-shell electronic transitions or fast-electron deceleration; and gamma rays from nuclear transitions — the more tightly bound and energetic the process, the higher the photon energy and frequency it radiates.

(a) \[\lambda=\frac{c}{\nu}=\frac{3\times10^{8}}{2.0\times10^{10}}=1.5\times10^{-2}\,\mathrm{m}=1.5\,\mathrm{cm}\] \(\displaystyle \lambda = 1.5\,\mathrm{cm}\)(b) \[B_0=\frac{E_0}{c}=\frac{48}{3\times10^{8}}=1.6\times10^{-7}\,\mathrm{T}\] \(\displaystyle B_0 = 1.6\times10^{-7}\,\mathrm{T}\)(c) Average energy density of the electric field (averaging \(\displaystyle \cos^2\) to \(\displaystyle 1/2\)): \[u_E=\left\langle\frac12\epsilon_0E^2\right\rangle=\frac14\epsilon_0E_0^2\] Average energy density of the magnetic field: \[u_B=\left\langle\frac{B^2}{2\mu_0}\right\rangle=\frac{B_0^2}{4\mu_0}\] Substitute \(\displaystyle B_0=E_0/c\) and \(\displaystyle c^2=1/(\mu_0\epsilon_0)\): \[u_B=\frac{E_0^2}{4\mu_0c^2}=\frac{E_0^2}{4\mu_0}\cdot\mu_0\epsilon_0=\frac{\epsilon_0E_0^2}{4}=u_E\] Hence \(\displaystyle u_E=u_B\): the average energy density carried by the electric field equals that carried by the magnetic field, each field carrying half the total energy of the wave.