Electromagnetic Waves — NCERT Solutions

(a) Capacitance

For a parallel plate capacitor,

$$C=\varepsilon_0\frac{A}{d}$$

Here, radius $r=12\text{ cm}=0.12\text{ m}$, so

$$A=\pi r^2=\pi(0.12)^2=\pi\times 0.0144\approx 0.0452\text{ m}^2$$

and separation

$$d=5.0\text{ cm}=0.050\text{ m}$$

Thus,

$$C=8.85\times 10^{-12}\times \frac{0.0452}{0.050}$$
$$C\approx 8.0\times 10^{-12}\text{ F}$$

So the capacitance is $8.0\times10^{-12}$ F.

The charging current is constant:

$$i=0.15\text{ A}$$

Using

$$i=C\frac{dV}{dt}$$

we get

$$\frac{dV}{dt}=\frac{i}{C}=\frac{0.15}{8.0\times10^{-12}}$$

$$\frac{dV}{dt}\approx 1.9\times10^{10}\text{ V s}^{-1}$$

So the rate of change of potential difference is $1.9\times10^{10}\text{ V s}^{-1}$.

(b) Displacement current

From the chapter, the displacement current is

$$i_d=\varepsilon_0\frac{d\Phi_E}{dt}$$

For a charging capacitor, the displacement current equals the conduction current:

$$i_d=i=0.15\text{ A}$$

So the displacement current is $0.15$ A.

(c) Kirchhoff's first rule

Yes, Kirchhoff's junction rule is valid at each plate of the capacitor if we include displacement current.

At the plates, conduction current enters the capacitor plates, while between the plates there is displacement current of the same magnitude. Therefore, there is no accumulation of charge at the junction in a way that violates current continuity. The generalized Ampere-Maxwell law makes the current the same through any surface bounded by the same loop, so the junction rule remains valid.

Given $C=100\text{ pF}=100\times10^{-12}\text{ F}=1.0\times10^{-10}\text{ F}$, $V_\text{rms}=230\text{ V}$ and $\omega=300\text{ rad s}^{-1}$.

(a) RMS conduction current

For an ac capacitor,

$$i_\text{rms}=\omega C V_\text{rms}$$

So,

$$i_\text{rms}=300\times (1.0\times10^{-10})\times 230$$

$$i_\text{rms}=6.9\times10^{-6}\text{ A}$$

So the rms conduction current is $6.9\,\mu\text{A}$.

(b) Equality of conduction and displacement current

Yes. In a charging or ac capacitor, the displacement current between the plates equals the conduction current in the connecting wires at every instant, so their rms values are also equal.

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

For a point between the plates inside radius $r<R$, the displacement current enclosed is proportional to area:

$$i_d(r)=i_0\frac{r^2}{R^2}$$

First find the current amplitude:

$$i_0=\omega C V_0$$

The given 230 V supply is taken as rms, so

$$V_0=\sqrt{2}\,V_\text{rms}=\sqrt{2}\times 230\approx 325\text{ V}$$

Thus,

$$i_0=300\times 1.0\times10^{-10}\times 325$$

$$i_0=9.75\times10^{-6}\text{ A}$$

Now

$$\frac{r^2}{R^2}=\frac{(0.03)^2}{(0.06)^2}=\frac{1}{4}$$

So enclosed current amplitude is

$$i_d(r)=\frac{9.75\times10^{-6}}{4}=2.44\times10^{-6}\text{ A}$$

Using Ampere-Maxwell law for a circular loop:

$$B(2\pi r)=\mu_0 i_d(r)$$

Hence,

$$B=\frac{\mu_0 i_d(r)}{2\pi r}$$

$$B=\frac{4\pi\times10^{-7}\times 2.44\times10^{-6}}{2\pi\times0.03}$$

$$B=1.6\times10^{-11}\text{ T}$$

So the amplitude of $\mathbf{B}$ is $1.6\times10^{-11}\text{ T}$.

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.