Answer

Radius of each circular plate, $r=12 \mathrm{~cm}=12 \times 10^{-2} \mathrm{~m}$

Distance between the plates, $d=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$

Charging current, $I=0.15 \mathrm{~A}$

Permittivity of free space, $\varepsilon_{0}=8.85 \times 10^{-12} C^{2} N^{-1} m^{-2}$

(a) Capacitance between the two plates is given by the relation,

$A=$ Area of each plate $=\pi r^{2}$

$C=\frac{\varepsilon_{0} \pi r^{2}}{d}$

$=\frac{8.85 \times 10^{-12} \times 3.14 \times\left(12 \times 10^{-2}\right)^{2}}{5 \times 10^{-2}}$

$C=8.0032 \times 10^{-12} F=8 p F$

Charge on each plate, $q=C V$

Where,

$\mathrm{V}=$ Potential difference across the plates

Differentiation on both sides with respect to time $(t)$ gives:

$\frac{d q}{d t}=C \frac{d V}{d t}$

But, $\frac{d q}{d t}=$ current $(I)$

$\therefore \frac{d V}{d t}=\frac{I}{C}$

$\Rightarrow \frac{0.15}{80.032 \times 10^{-12}}=1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$

Therefore, the change in potential difference between the plates is $1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is $0.15 \mathrm{~A}$.

(c) Yes

Kirchhoff’s first rule is valid at each plate of the capacitor since displacement current is equal to conduction current.

Answer Radius of each circular plate, $R=6.0 \mathrm{~cm}=0.06 \mathrm{~m}$

Capacitance of a parallel plate capacitor, $C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}$

Supply voltage, $V=230 \mathrm{~V}$

Angular frequency, $\omega=300 \mathrm{rad}_{s^{-1}}$

(a) Rms value of conduction current, $I_{r m s}=\frac{V_{r m s}}{X_{c}}$

Where,

$X_{C}=$ Capacitive reactance

$=\frac{1}{\omega C}$

$\therefore I=V_{r m s} \times \omega C$

$=230 \times 300 \times 100 \times 10^{-12}$

$=6.9 \times 10^{-6} \mathrm{~A}$

$=6.9 \mu \mathrm{A}$

Hence, the rms value of conduction current is $6.9 \mu \mathrm{A}$.

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

(c) Magnetic field is given as:

$B=\frac{\mu_{o} r}{2 \pi R^{2}} I_{o}$

Where,

$\mu_{0}=$ Free space permeability $=4 \pi \times 10^{-7} \mathrm{NA}^{-2}$

$I 0=$ Maximum value of current $=\sqrt{2} l$ $r=$ Distance between the plates from the axis $=3.0 \mathrm{~cm}=0.03 \mathrm{~m}$

$B=\frac{\mu_{0} I_{0} r}{2 \pi R^{2}}=\frac{\mu_{0} I_{r m s} \sqrt{2} r}{2 \pi R^{2}}$

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

$=1.63 \times 10^{-11} \mathrm{~T}$ Hence, the magnetic field at that point is $1.63 \times 10^{-11} \mathrm{~T}$.

Answer

The electromagnetic wave travels in a vacuum along the z-direction. The electric field $(E)$ and the magnetic field $(H)$ are in the $x-y$ plane. They are mutually perpendicular.

Frequency of the wave, $v=30 \mathrm{MHz}=30 \times 10^{6} \mathrm{~s}^{-1}$

Speed of light in a vacuum, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Wavelength of a wave is given as:

$$ \begin{aligned} & \lambda=\frac{c}{v} \\ & =\frac{3 \times 10^{8}}{30 \times 10^{6}}=10 \mathrm{~m} \end{aligned} $$

Answer

The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., $10^{9} \mathrm{~Hz}$.

Answer

Amplitude of magnetic field of an electromagnetic wave in a vacuum,

$B 0=510 \mathrm{nT}=510 \times 10^{-9} \mathrm{~T}$

Speed of light in a vacuum, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Amplitude of electric field of the electromagnetic wave is given by the relation,

$c=\frac{E_{0}}{B_{0}}$

$E_{0}=c B_{0}$

$=3 \times 10^{8} \times 510 \times 10^{-9}=153 \mathrm{~N} / \mathrm{C}$

Therefore, the electric field part of the wave is $153 \mathrm{~N} / \mathrm{C}$.

Answer

Electric field amplitude, $E_{0}=120 \mathrm{~N} / \mathrm{C}$

Frequency of source, $v=50.0 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}$

Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

(a) Magnitude of magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

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

$=4 \times 10^{-7} \mathrm{~T}=400 \mathrm{nT}$

Angular frequency of source is given as:

$\omega=2 n v=2 \pi \times 50 \times 10^{6}$

$=3.14 \times 10^{8} \mathrm{rad} / \mathrm{s}$

Propagation constant is given as: $k=\frac{2 \pi \nu}{c}=\frac{\omega}{c}$

$=\frac{3.14 \times 10^{8}}{3 \times 10^{8}}=1.05 \mathrm{rad} / \mathrm{m}$

Wavelength of wave is given as:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{50 \times 10^{6}}=6.0 \mathrm{~m}$

(b) Suppose the wave is propagating in the positive $x$ direction. Then, the electric field vector will be in the positive $y$ direction and the magnetic field vector will be in the positive $z$ direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

$E=E_{0} \operatorname{Sin}(k x-\omega t) \hat{j}$

$\vec{E}=120 \operatorname{Sin}\left(1.05 x-3.14 \times 10^{8} t\right) \hat{j} N / C$

And, magnetic field vector is given as:

$B=B_{0} \operatorname{Sin}(k x-\omega t) \hat{k}$

$\vec{B}=\left(4 \times 10^{-7}\right) \operatorname{Sin}\left(1.05 x-3.14 \times 10^{8} t\right) \hat{k}$ Tesla

Answer

Energy of a photon is given as: $E=h v=\frac{h c}{\lambda}$

Where,

$h=$ Planck’s constant $=6.6 \times 10^{-34} \mathrm{Js}$

$c=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

$\lambda=$ Wavelength of radiation

$\therefore E=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{\lambda}=\frac{19.8 \times 10^{-26}}{\lambda}$

$=\frac{19.8 \times 10^{-26}}{\lambda \times 1.6 \times 10^{-19}}=\frac{12.375 \times 10^{-7}}{\lambda} \mathrm{eV}$

The given table lists the photon energies for different parts of an electromagnetic spectrum for different $\lambda$.

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

Answer

Frequency of the electromagnetic wave, $v=$ $$ 2 \times 10^{10} \mathrm{~Hz} $$

Electric field amplitude, $$ E_0=48 \mathrm{Vm}^{-1} $$

Speed of light, $c=$ $$ 3 \times 10^8 \mathrm{~m} / \mathrm{s} $$

(a) Wavelength of a wave is given as: $$ \begin{aligned} & \lambda=\frac{c}{v} \ & = \ & \frac{3 \times 10^8}{2 \times 10^{10}}=0.015 \mathrm{~m} \end{aligned} $$ (b) Magnetic field strength is given as: $$ \begin{aligned} & B_0=\frac{E_0}{c} \ & = \ & \frac{48}{3 \times 10^8}=1.6 \times 10^{-7} \mathrm{~T} \end{aligned} $$ (c) Energy density of the electric field is given as: $$ U_E=\frac{1}{2} \epsilon_0 E^2 $$

And the energy density of the magnetic field is given as: $$ U_B=\frac{1}{2 \mu_0} B^2 $$

Where, $\epsilon_0$ $=$ Permittivity of free space

$\mu_0$ $=$ Permeability of free space $$ \mathrm{E}=\mathrm{CB} $$

Where, $$ c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \quad \quad (….2)$$

Putting equation (2) in equation (1), we get $$ E=\frac{1}{\sqrt{\epsilon_0 \mu_0}} B $$

Squaring on both sides, we get $$ \begin{aligned} & E^2=\frac{1}{\epsilon_0 \mu_0} B^2 \ & \epsilon_0 E^2=\frac{B^2}{\mu_0} \ & \frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \frac{B^2}{\mu_0} \ & => \ & U_E=U_B \end{aligned} $$