CBSE Class 12 Physics Electromagnetic Waves NCERT Solutions

Given:

Radius of each circular plate, r = 12 \text{ cm} = 0.12 \text{ m}

Distance between the plates, d = 5.0 \text{ cm} = 0.05 \text{ m}

Charging current, I = 0.15 \text{ A}

Permittivity of free space, \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{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 = \frac{\epsilon_0 A}{d}, where A is the area of each plate and d is the distance between the plates.

The area of each circular plate is A = \pi r^2.

Substituting the values:

A = \pi (0.12 \text{ m})^2 = 0.0452 \text{ m}^2

Now, calculate the capacitance:

C = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (0.0452 \text{ m}^2)}{0.05 \text{ m}} \approx 8.00 \times 10^{-12} \text{ F}

So, the capacitance is approximately 8.00 \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 \frac{dq}{dt} = C \frac{dV}{dt}.

We know that the charging current I = \frac{dq}{dt}. Therefore,

I = C \frac{dV}{dt}

The rate of change of potential difference is \frac{dV}{dt} = \frac{I}{C}.

Substituting the values:

\frac{dV}{dt} = \frac{0.15 \text{ A}}{8.00 \times 10^{-12} \text{ F}} \approx 1.875 \times 10^9 \text{ V/s}

Thus, the rate of change of potential difference between the plates is approximately 1.87 \times 10^9 \text{ V/s}.

2. Displacement Current Across the Plates:

According to Maxwell's equations, the displacement current (I_d) between the plates of a capacitor is equal to the conduction current (I) flowing into the plates, provided the capacitor is being charged or discharged.

Therefore, the displacement current across the plates is I_d = I = 0.15 \text{ A}.

3. Validity of Kirchhoff's First Rule:

Yes, Kirchhoff's first rule (junction rule) is valid at each plate of the capacitor, but only if we consider both the conduction current entering the plate and the displacement current leaving the plate (or vice versa). The rule states that the algebraic sum of currents entering a junction must equal the algebraic sum of currents leaving it. In the case of a capacitor plate during charging, the conduction current flows towards the plate, and the displacement current effectively flows away from the plate, ensuring the rule holds.

Given:

Radius of circular plates, R = 6.0 \text{ cm} = 0.06 \text{ m}

Capacitance, C = 100 \text{ pF} = 100 \times 10^{-12} \text{ F}

Supply voltage (rms), V_{rms} = 230 \text{ V}

Angular frequency, \omega = 300 \text{ rad s}^{-1}

1. RMS Value of Conduction Current:

The capacitive reactance (X_C) of the capacitor is given by X_C = \frac{1}{\omega C}.

Substituting the values:

X_C = \frac{1}{(300 \text{ rad s}^{-1}) \times (100 \times 10^{-12} \text{ F})} = \frac{1}{3 \times 10^{-8}} \Omega = 3.33 \times 10^7 \Omega

The rms value of the conduction current (I_{rms}) is given by Ohm's law applied to the capacitive reactance:

I_{rms} = \frac{V_{rms}}{X_C}

Substituting the values:

I_{rms} = \frac{230 \text{ V}}{3.33 \times 10^7 \Omega} \approx 6.91 \times 10^{-6} \text{ A} = 6.91 \mu \text{A}

Therefore, the rms value of the conduction current is approximately 6.91 \mu \text{A}.

2. Conduction Current vs. Displacement Current:

In an AC circuit with a capacitor, the conduction current flows through the wires to the plates, and the displacement current exists between the plates due to the changing electric field. According to Maxwell's equations, the conduction current entering the plates is equal to the displacement current between the plates at any instant. So, yes, the magnitude of the conduction current is equal to the magnitude of the displacement current.

3. Amplitude of B at a Point 3.0 cm from the Axis:

The magnetic field (B) between the plates of a parallel plate capacitor carrying a time-varying current can be found using Ampere's law modified by Maxwell. For a point at a radial distance r from the axis, inside the plates (where r < R), the magnetic field is given by B = \frac{\mu_0 I_d}{2 \pi R^2} r, where I_d is the instantaneous displacement current and R is the radius of the plates.

The instantaneous displacement current I_d is related to the conduction current I by I_d = I. The amplitude of the conduction current I_0 is related to the rms current by I_{rms} = \frac{I_0}{\sqrt{2}}, so I_0 = I_{rms} \sqrt{2}.

The amplitude of the displacement current is I_{d0} = I_0 = (6.91 \times 10^{-6} \text{ A}) \times \sqrt{2} \approx 9.77 \times 10^{-6} \text{ A}.

We need to find the amplitude of B at r = 3.0 \text{ cm} = 0.03 \text{ m}.

Substituting the values into the formula for B (amplitude B_0):

B_0 = \frac{\mu_0 I_{d0}}{2 \pi R^2} r = \frac{(4 \pi \times 10^{-7} \text{ T m/A}) \times (9.77 \times 10^{-6} \text{ A})}{2 \pi (0.06 \text{ m})^2} \times (0.03 \text{ m})

B_0 = \frac{(2 \times 10^{-7} \text{ T m/A}) \times (9.77 \times 10^{-6} \text{ A})}{(0.0036 \text{ m}^2)} \times (0.03 \text{ m})

B_0 = \frac{1.954 \times 10^{-12}}{0.0036} \times 0.03 \text{ T} \approx 5.43 \times 10^{-10} \times 0.03 \text{ T}

B_0 \approx 1.63 \times 10^{-11} \text{ T}

Therefore, the amplitude of the magnetic field B at a point 3.0 cm from the axis between the plates is approximately 1.63 \times 10^{-11} \text{ T}.