CBSE Class 12 Physics Chapter 16 Electromagnetic Waves NCERT Solutions

Solution:

Given:

• Radius of each circular plate, $r\; =\; 12\; \backslash text\{\; cm\}\; =\; 0.12\; \backslash text\{\; m\}$
• Distance between the plates, $d\; =\; 5.0\; \backslash text\{\; cm\}\; =\; 0.05\; \backslash text\{\; m\}$
• Charging current, $I\; =\; 0.15\; \backslash text\{\; A\}$
• Permittivity of free space, $\backslash epsilon\_0\; =\; 8.85\; \backslash times\; 10^\{-12\}\; \backslash ,\backslash mathrm\{C\}^2\; \backslash ,\backslash mathrm\{N\}^\{-1\}\; \backslash ,\backslash mathrm\{m\}^\{-2\}$

1. Calculation of Capacitance and Rate of Change of Potential Difference:

The capacitance ($C$) of a parallel plate capacitor is given by the formula:

$$C\; =\; \backslash frac\{\backslash varepsilon\_0\; A\}\{d\}$$

Where $A$ is the area of each plate. Since the plates are circular, $A\; =\; \backslash pi\; r^2$.

Substituting the values:

$$A\; =\; \backslash pi\; (0.12\; \backslash text\{\; m\})^2\; =\; \backslash pi\; (0.0144\; \backslash text\{\; m\}^2)\; \backslash approx\; 0.04524\; \backslash text\{\; m\}^2$$

Now, calculate the capacitance:

$$C\; =\; \backslash frac\{(8.85\; \backslash times\; 10^\{-12\}\; \backslash ,\backslash mathrm\{C\}^2\; \backslash ,\backslash mathrm\{N\}^\{-1\}\; \backslash ,\backslash mathrm\{m\}^\{-2\})\; \backslash times\; (0.04524\; \backslash text\{\; m\}^2)\}\{0.05\; \backslash text\{\; m\}\}$$

$$C\; \backslash approx\; 8.00\; \backslash times\; 10^\{-12\}\; \backslash text\{\; F\}\; =\; 8.00\; \backslash text\{\; pF\}$$

The charge ($q$) on each plate is related to the potential difference ($V$) by $q\; =\; CV$. Differentiating both sides with respect to time ($t$), we get:

$$\backslash frac\{dq\}\{dt\}\; =\; C\; \backslash frac\{dV\}\{dt\}$$

We know that the charging current $I\; =\; \backslash frac\{dq\}\{dt\}$. Therefore:

$$I\; =\; C\; \backslash frac\{dV\}\{dt\}$$

The rate of change of potential difference is:

$$\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{I\}\{C\}\; =\; \backslash frac\{0.15\; \backslash text\{\; A\}\}\{8.00\; \backslash times\; 10^\{-12\}\; \backslash text\{\; F\}\}$$

$$\backslash frac\{dV\}\{dt\}\; \backslash approx\; 1.875\; \backslash times\; 10^\{10\}\; \backslash text\{\; V/s\}$$

So, the rate of charge of potential difference between the plates is approximately $1.875\; \backslash times\; 10^\{10\}\; \backslash text\{\; V/s\}$.

2. Displacement Current:

The displacement current ($I\_d$) between the plates of a capacitor is given by:

$$I\_d\; =\; \backslash epsilon\_0\; \backslash frac\{d\backslash Phi\_E\}\{dt\}$$

Where $\backslash Phi\_E$ is the electric flux. For a parallel plate capacitor, the electric field $E$ between the plates is uniform, and $\backslash Phi\_E\; =\; EA$. Also, $E\; =\; V/d$, so $\backslash Phi\_E\; =\; (V/d)A$.

Then, $\backslash frac\{d\backslash Phi\_E\}\{dt\}\; =\; \backslash frac\{A\}\{d\}\; \backslash frac\{dV\}\{dt\}$.

Substituting this into the formula for $I\_d$:

$$I\_d\; =\; \backslash epsilon\_0\; \backslash frac\{A\}\{d\}\; \backslash frac\{dV\}\{dt\}$$

Since $C\; =\; \backslash frac\{\backslash varepsilon\_0\; A\}\{d\}$, we have $I\_d\; =\; C\; \backslash frac\{dV\}\{dt\}$. From part 1, we know that $C\; \backslash frac\{dV\}\{dt\}\; =\; I$ (the charging current).

Therefore, the displacement current across the plates is equal to the conduction current in the circuit, which is $0.15\; \backslash text\{\; A\}$.

3. Validity of Kirchhoff's First Rule:

Yes, Kirchhoff's first rule (the algebraic sum of currents entering a junction is zero) is valid at each plate of the capacitor, provided that we consider both the conduction current ($I\_c$) flowing into the plate and the displacement current ($I\_d$) leaving the plate (or vice versa). Maxwell's equations show that $I\_c\; =\; I\_d$ in the region between the plates when the capacitor is charging. Thus, the total current is conserved.

Solution:

Given:

• Radius of circular plates, $R\; =\; 6.0\; \backslash text\{\; cm\}\; =\; 0.06\; \backslash text\{\; m\}$
• Capacitance, $C\; =\; 100\; \backslash text\{\; pF\}\; =\; 100\; \backslash times\; 10^\{-12\}\; \backslash text\{\; F\}$
• RMS voltage of the AC supply, $V\_\{rms\}\; =\; 230\; \backslash text\{\; V\}$
• Angular frequency, $\backslash omega\; =\; 300\; \backslash text\{\; rad\; s\}^\{-1\}$
• Distance from the axis, $r\; =\; 3.0\; \backslash text\{\; cm\}\; =\; 0.03\; \backslash text\{\; m\}$

1. RMS Value of Conduction Current:

The capacitive reactance ($X\_C$) is given by:

$$X\_C\; =\; \backslash frac\{1\}\{\backslash omega\; C\}$$

Substituting the values:

$$X\_C\; =\; \backslash frac\{1\}\{(300\; \backslash text\{\; rad\; s\}^\{-1\})\; \backslash times\; (100\; \backslash times\; 10^\{-12\}\; \backslash text\{\; F\})\}\; =\; \backslash frac\{1\}\{3\; \backslash times\; 10^\{-8\}\}\; \backslash Omega$$

$$X\_C\; =\; 3.33\; \backslash times\; 10^\{7\}\; \backslash Omega$$

The rms value of the conduction current ($I\_\{rms\}$) is given by Ohm's law:

$$I\_\{rms\}\; =\; \backslash frac\{V\_\{rms\}\}\{X\_C\}$$

$$I\_\{rms\}\; =\; \backslash frac\{230\; \backslash text\{\; V\}\}\{3.33\; \backslash times\; 10^\{7\}\; \backslash Omega\}\; \backslash approx\; 6.91\; \backslash times\; 10^\{-6\}\; \backslash text\{\; A\}\; =\; 6.91\; \backslash mu\backslash text\{A\}$$

The rms value of the conduction current is approximately $6.91\; \backslash mu\backslash text\{A\}$.

2. Conduction Current vs. Displacement Current:

Yes, the conduction current is equal to the displacement current ($I\_d$) between the plates of the capacitor. The displacement current is given by $I\_d\; =\; \backslash epsilon\_0\; \backslash frac\{d\backslash Phi\_E\}\{dt\}$. For a charging capacitor, the rate of change of electric flux is such that $I\_d\; =\; I\_c$, where $I\_c$ is the conduction current in the external circuit.

3. Amplitude of Magnetic Field ($B$):

The amplitude of the conduction current is $I\_\{0\}\; =\; \backslash sqrt\{2\}\; I\_\{rms\}$.

$$I\_\{0\}\; =\; \backslash sqrt\{2\}\; \backslash times\; 6.91\; \backslash mu\backslash text\{A\}\; \backslash approx\; 9.77\; \backslash mu\backslash text\{A\}$$

The displacement current amplitude is equal to the conduction current amplitude, $I\_\{d0\}\; =\; I\_0\; \backslash approx\; 9.77\; \backslash mu\backslash text\{A\}$.

The magnetic field ($B$) at a distance $r$ from the axis between the plates of a parallel plate capacitor is given by:

$$B\; =\; \backslash frac\{\backslash mu\_0\; I\_d\}\{2\backslash pi\; R^2\}\; r$$

Where $\backslash mu\_0$ is the permeability of free space ($4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m/A\}$), $R$ is the radius of the plates, and $r$ is the distance from the axis.

Substituting the values to find the amplitude of $B$ ($B\_0$):

$$B\_0\; =\; \backslash frac\{\backslash mu\_0\; I\_\{d0\}\}\{2\backslash pi\; R^2\}\; r$$

$$B\_0\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m/A\})\; \backslash times\; (9.77\; \backslash times\; 10^\{-6\}\; \backslash text\{\; A\})\}\{2\backslash pi\; (0.06\; \backslash text\{\; m\})^2\}\; \backslash times\; (0.03\; \backslash text\{\; m\})$$

$$B\_0\; =\; \backslash frac\{(2\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m/A\})\; \backslash times\; (9.77\; \backslash times\; 10^\{-6\}\; \backslash text\{\; A\})\}\{(0.0036\; \backslash text\{\; m\}^2)\}\; \backslash times\; (0.03\; \backslash text\{\; m\})$$

$$B\_0\; \backslash approx\; \backslash frac\{1.954\; \backslash times\; 10^\{-12\}\}\{0.0036\}\; \backslash times\; 0.03\; \backslash text\{\; T\}$$

$$B\_0\; \backslash approx\; 5.43\; \backslash times\; 10^\{-10\}\; \backslash text\{\; T\}$$

The amplitude of the magnetic field at a point 3.0 cm from the axis between the plates is approximately $5.43\; \backslash times\; 10^\{-10\}\; \backslash text\{\; T\}$.