Electromagnetic Waves — NCERT Solutions

Given radius $r=12\text{ cm}=0.12\text{ m}$ and separation $d=5.0\text{ cm}=0.05\text{ m}$.

(a) Capacitance and rate of change of potential difference

For a parallel-plate capacitor,
$$C=\varepsilon_0\frac{A}{d}$$
where
$$A=\pi r^2=\pi(0.12)^2=0.0452\ \text{m}^2$$
So,
$$C=8.85\times10^{-12}\times\frac{0.0452}{0.05} \approx 8.0\times10^{-12}\ \text{F}$$
Thus, the capacitance is
$$C\approx 8.0\ \text{pF}$$

The charging current is
$$i= C\frac{dV}{dt}$$
Hence,
$$\frac{dV}{dt}=\frac{i}{C}=\frac{0.15}{8.0\times10^{-12}} \approx 1.9\times10^{10}\ \text{V s}^{-1}$$

(b) Displacement current

For a charging capacitor, the displacement current equals the conduction current:
$$i_d=\varepsilon_0\frac{d\Phi_E}{dt}=i=0.15\ \text{A}$$

(c) Kirchhoff's first rule at each plate

Yes, Kirchhoff's junction rule is valid at each plate. The conduction current arriving at a plate does not disappear; it is matched by the displacement current between the plates. Thus charge conservation is satisfied when both conduction current and displacement current are included.

Given $C=100\text{ pF}=100\times10^{-12}\text{ F}$, $V_{\rm rms}=230\text{ V}$, and $\omega=300\text{ rad s}^{-1}$.

(a) rms conduction current

For an AC capacitor circuit,
$$I_{\rm rms}=\omega C V_{\rm rms}$$
So,
$$I_{\rm rms}=300\times 100\times10^{-12}\times 230$$
$$=6.9\times10^{-6}\text{ A}$$
So the rms current is
$$I_{\rm rms}=6.9\ \mu\text{A}$$

(b) Is conduction current equal to displacement current?

In a capacitor circuit, the conduction current in the wires and the displacement current between the plates have the same instantaneous value. So, yes, they are equal in magnitude.

(c) Amplitude of $\mathbf{B}$ at $r=3.0\text{ cm}$

The current amplitude is
$$I_0=\sqrt{2}\,I_{\rm rms}=\sqrt{2}(6.9\times10^{-6}) \approx 9.8\times10^{-6}\text{ A}$$
For a point inside the plates at radius $r=3.0\text{ cm}=0.03\text{ m}$, Ampere-Maxwell law gives
$$B(2\pi r)=\mu_0 I_{\text{enc}}$$
Since the displacement current is uniformly distributed over the plate area,
$$I_{\text{enc}}=I_0\frac{r^2}{R^2}$$
with $R=6.0\text{ cm}=0.06\text{ m}$. Thus,
$$I_{\text{enc}}=9.8\times10^{-6}\times\frac{(0.03)^2}{(0.06)^2} =9.8\times10^{-6}\times\frac14 =2.45\times10^{-6}\text{ A}$$
Now
$$B=\frac{\mu_0 I_{ ext{enc}}}{2\pi r} =\frac{4\pi\times10^{-7}\times2.45\times10^{-6}}{2\pi\times0.03} \approx 1.6\times10^{-11}\text{ T}$$
So the magnetic field amplitude is
$$B_0\approx1.6\times10^{-11}\text{ T}$$

All electromagnetic waves travel in vacuum with the same speed, $c = 3\times10^8\,\text{m s}^{-1}$. Therefore X-rays, red light, and radio waves all have the same speed of light in vacuum.

For a plane electromagnetic wave travelling along the z-direction, the electric field and magnetic field are both perpendicular to the direction of propagation and also perpendicular to each other.

So the fields may be along the x-direction and y-direction respectively, or vice versa.

Given frequency,
$$\nu = 30\text{ MHz} = 30\times10^6\text{ Hz}$$
Wavelength is
$$\lambda = \frac{c}{\nu} = \frac{3\times10^8}{30\times10^6} = 10\text{ m}$$

So the wave has wavelength 10 m.

Use
$$\lambda=\frac{c}{\nu}$$
For $\nu=7.5\text{ MHz}$,
$$\lambda=\frac{3\times10^8}{7.5\times10^6}=40\text{ m}$$
For $\nu=12\text{ MHz}$,
$$\lambda=\frac{3\times10^8}{12\times10^6}=25\text{ m}$$
So the wavelength band is 40 m to 25 m. This exact value is not among the printed options because none were given; the computed answer is used.