Q:1. Fig. shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by and external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule valid at each plate of the capacitor? Explain.

(Given ε_{0} = 8.85 × 10^{–}12 C^{2} N^{-1} m^{-2})

Sol: capacitance of a parallel plate capacitor is

$\large C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \pi r^2 }{d} $

$\large C = \frac{8.85 \times 10^{-12} \times 3.14 \times (12\times 10^{-2})^2}{5 \times 10^{-3}} $

= 80.1 pF

q = C V

Differentiating w.r.t time

$\large \frac{dq}{dt} = C \frac{dV}{dt} $

$\large I = C \frac{dV}{dt} $

$\large \frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{-12}} $

= 1.87 × 10^{9} Vs^{-1}

(b) Displacement current = conduction current = 0.15 A

(c) Yes, Kirchhoff’s first rule is valid at each plate of the capacitor provided we take the current to be the sum of the conduction and displacement currents.

Q:2. A parallel plate capacitor made of circular plates each of radius R = 6.0 cm. Has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c. supply with an angular frequency of 300 rad. s^{-1}.

(a) What is the r.m.s value of the conduction current ?

(b) Is the conduction current equal to the displacement current ?

(c) Determine the magnitude of B at a point 3.0 cm from the axis between the plates.

Sol:(a) $\large X_C = \frac{1}{\omega C}$

$\large I_v = \frac{E_v}{X_C} = \frac{E_v}{\frac{1}{\omega C}}$

$\large I_v = E_v \times \omega C = 230 \times 300 \times 100 \times 10^{-12}$

= 6.9 × 10^{-6} A

(b) Yes , conduction current is equal to the displacement current .

$\large I_d = \epsilon_0 \frac{d\phi}{dt} = \epsilon_0 \frac{d(E A)}{dt}$

$\large I_d = \epsilon_0 \frac{d(E A)}{dt} =\epsilon_0 A\frac{dE}{dt} $

$\large I_d = \epsilon_0 A\frac{d}{dt}(\frac{\sigma}{\epsilon_0}) $

$\large I_d = \epsilon_0 A\frac{d}{dt}(\frac{q}{A \epsilon_0}) = \frac{dq}{dt} = I $

(c) $\large B = \frac{\mu_0 (\sqrt{2} I_v) r}{2 \pi R^2}$ ; (As I_{o} = √2 I_{v})

$\large B = \frac{4 \pi \times 10^{-7} (\sqrt{2} \times 6.9 \times 10^{-6}) 0.03}{2 \pi (0.06)^2}$

= 1.63 × 10^{-11} T

Q:3. What physical quantity is the same for X – rays of wavelength 10^{-10} m, red light of wavelength 6800 A° and radiowaves of wavelength 500 m?

Sol:

The speed in vacuum is same for all the given wavelengths, which is 3 × 10^{8} m/s , but wavelength changes .

Q:4. A plane electromagnetic wave travels in vacuum along Z – direction. What can you say about the directions of electric and magnetic field vectors? If the frequency of the wave is 30 MHz. what is wavelength ?

Sol. E and B vectors must be in X and Y – directions.

Direction of Wave Propagation $ || (\vec{E} \times \vec{B}) $

As we know, $\large \lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{30 \times 10^6} $

= 10 m

Q:5. A radio can tune to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band ?

Sol: $\large \lambda_1 = \frac{c}{\nu_1} = \frac{3 \times 10^8}{7.5 \times 10^6}$ = 40 m

$\large \lambda_2 = \frac{c}{\nu_2} = \frac{3 \times 10^8}{12 \times 10^6}$ = 25 m

Wavelength band : 40 m to 25 m

Q:6. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of electromagnetic waves produced by the oscillator ?

Sol: The frequency of electromagnetic wave is the same as that of oscillating charged particle about its equilibrium position; which is 10^{9} Hz.

Q:7.The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B_{0} = 510 nT. What is the amplitude of the electric field part of the wave ?

Sol: $\large c = \frac{E_0}{B_0} \Rightarrow E_0 = c B_0 $

$\large E_0 = 3 \times 10^8 \times 510 \times 10^{-9} $

= 153 N/C

Q:8. Suppose that the electric field amplitude of and electromagnetic wave is E_{0} = 120 N/C and its frequency is v = 50.0 MHz. (a) Determine B_{0} , ω , k and λ . (b) Find expressions for $\vec{E}$ and $\vec{B}$ .

Sol: $\large B_0 = \frac{E_0}{c} $

$\large B_0 = \frac{120}{3 \times 10^8} $

= 4 × 10^{-7} T

$\large \omega = 2 \pi \nu = 2 \times 3.14 \times 50 \times 10^6 $

= 3.14 x 10^{8} rad/s

$\large k = \frac{\omega}{c} = \frac{3.14 \times 10^8}{3 \times 10^8} $

k = 1.05 rad/m

$\large \lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{50 \times 10^6}$

(b) E = E_{0} sin(ω t – k x)

E = 120 sin(3.14 × 10^{8} t – 1.05 x )

B = 4 × 10^{-7} sin(3.14 × 10^{8} t – 1.05 x )

Q:10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10^{10} Hz and amplitude 48 V/m, (a) what is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field ? (c) Show that the total average energy density of the electric field equals the average energy density of magnetic field. (c = 3 × 10^{8} m/s)

Sol: (a) $\large \lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{2 \times 10^{10}} $

= 1.5 x 10^{-2} m

(b) $\large B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8}$

= 1.6 x 10^{-7} T

(c) The average energy density of electric field,

$\large U_E = \frac{1}{4} \epsilon_0 E_0^2 $

As , $\large E_0 = c B_0$

$\large U_E = \frac{1}{4} \epsilon_0 ( c^2 B_0^2 ) $

Speed of electromagnetic waves , $\large c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

Hence , $\large U_E = \frac{1}{4} \epsilon_0 B_0^2 \times \frac{1}{\mu_0 \epsilon_0} $

$\large U_E = \frac{1}{4} \frac{B_0^2}{\mu_0} = U_B $

Where , U_{B} is average energy density of magnetic field

Thus, the average energy density of the Electric field equals the average energy density of magnetic field.