Chapter 8: Electromagnetic Waves

Ans: Radius of each circular plates, \[r = 12{\text{ cm}} = 0.12{\text{ m}}\]

Distance between the given plates, \[d = 5{\text{ }}cm = 0.05{\text{ }}m\]

Charging current, $I = 0.15A$

The permittivity of free space, ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$

Capacitance between the two plates can be given as: 

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

Where, 

$A = \pi {r^2}$. 

Hence,

$C = \frac{{{\varepsilon _0}\pi {r^2}}}{d}$

$C = \frac{{8.85 \times {{10}^{ - 12}} \times {{(0.12)}^2}}}{{0.05}}$

$C = 8.0032 \times {10^{ - 12}}F$   

$C = 80.032pF$

Charge on each plate is given as, \[q = CV\], where,

V = Potential difference across the plates

Differentiating both sides with respect to time (t), we get:

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

As, $\frac{{dq}}{{dt}} = I$, so, the rate of change of potential difference between the plates can be given as:

$\frac{{dV}}{{dt}} = \frac{I}{C} = \frac{{0.15}}{{80.032 \times {{10}^{ - 12}}}}$

$\therefore \frac{{dV}}{{dt}} = 1.87 \times {10^9}{\kern 1pt} V/s$

1. Obtain the Displacement Current Across the Plates.

Ans: The displacement current across the plates would be the same as the conduction current. Hence, the displacement current, ${I_d}$ would be \[0.15{\text{ }}A\].

1. Is Kirchhoff’s First Rule (junction Rule) Valid at Each Plate of the Capacitor? Explain.

Ans: Yes, Kirchhoff’s first rule would be valid at each plate of the capacitor, provided that sum of conduction and displacement currents, i.e., $I = {I_c} + {I_d}$ (junction rule of Kirchhoff’s law).

Ans: Radius of each circular plate, \[R = 6.0{\text{ }}cm = 0.06{\text{ }}m\]

Capacitance of a parallel plate capacitor, \[C = 100{\text{ }}pF = 100 \times {10^{ - 12}}F\]

Supply voltage, \[V = 230{\text{ }}V\]

Angular frequency, $\omega  = 300{\kern 1pt} {\kern 1pt} rad{\kern 1pt} {\kern 1pt} {s^{ - 1}}$

Rms value of conduction current can be given as: $I = \frac{V}{{{X_C}}}$.

Where, ${X_C}$ = capacitive currant $ = \frac{1}{{\omega C}}$

$\therefore I = V \times \omega C$    

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

$I = 6.9 \times {10^{ - 6}}$    

$I = 6.9\mu A$

Therefore, the RMS value of conduction current will be $6.9\mu A$.

1. Is the Conduction Current Equal to the Displacement Current?

Ans: Yes, the conduction current will be the same as the displacement current. i.e., the conduction current is equal to the displacement current.

1. Determine the Amplitude of ${\overrightarrow B}$ at a point \[{\mathbf{3}}.{\mathbf{0}}{\text{ }}{\mathbf{cm}}\] from the Axis Between the Plates.

Ans: Magnetic field is given as: $\overrightarrow B = \frac{{{\mu _0}r}}{{2\pi {R^2}}}{I_0}$

Where,

${\mu _0}$= Free space permeability $ = 4\pi  \times {10^{ - 7}}N{A^{ - 2}}$

${I_0}$ = Maximum value of current = $\sqrt 2 I$

$r$= Distance between the plates from the axis = $3.0cm = 0.03cm$.

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

${\overrightarrow B} = 1.63 \times {10^{ - 11}}T$

Therefore, the magnetic field at that point will be $1.63 \times {10^{ - 11}}T$.

Ans: The speed of light (\[3 \times {10^8}m/s\]) is independent of the wavelength in the vacuum. Hence, it is the same for all wavelengths in a vacuum.

Ans: The electromagnetic wave is travelling along the z-direction, in a vacuum. The electric field (E) and the magnetic field (H) will lie in the x-y plane and they will be mutually perpendicular.

Frequency of the wave, \[\nu  = 30{\text{ }}MHz = 30 \times {10^6}{s^{ - 1}}\]

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

Wavelength, $\lambda $ of the wave can be given as: $\lambda  = \frac{c}{\nu }$

$\therefore \lambda  = \frac{{3 \times {{10}^8}}}{{30 \times {{10}^6}}} = 10m$

Ans: A radio can tune to minimum frequency, \[{\nu _1} = 7.5{\text{ }}MHz = 7.5 \times {10^6}{\kern 1pt} {\kern 1pt} Hz\]

Maximum frequency, \[{\nu _2} = 12{\text{ }}MHz = 12 \times {10^6}Hz\]

Speed of light, \[c = 3 \times {10^8}m/s\]

Wavelength for frequency, ${\nu _1}$ can be calculated as: 

${\lambda _1} = \frac{c}{{{\nu _1}}}$    

${\lambda _1}= \frac{{3 \times {{10}^8}}}{{7.5 \times {{10}^6}}}$    

${\lambda _1} = 40m$

Wavelength for frequency, ${\nu _2}$ can be calculated as: 

${\lambda _2} = \frac{c}{{{\nu _2}}}$

${\lambda _2} = \frac{{3 \times {{10}^8}}}{{12 \times {{10}^6}}}$      

$ {\lambda _2} = 25m$

Therefore, the corresponding wavelength band would be between $40m$ to $25m$.

Ans: The frequency of an electromagnetic wave produced by the oscillator will be equal to the frequency of charged particle oscillating about its mean position i.e., \[{10^9}Hz\].

Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum, is:

${B_0} = 510nT = 510 \times {10^{ - 9}}T$

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

Amplitude of electric field of the electromagnetic wave can be given as:

$E = cB = 3 \times {10^8} \times 510 \times {10^{ - 9}} = 153N/C$.

Ans: Electric field amplitude, ${E_0} = 120N/C$

Frequency of source, \[\nu  = 50.0MHz = 50 \times {10^6}Hz\]

Speed of light, \[c = 3 \times {10^8}m/s\]

Magnetic field strength can be given as:

${B_0} = \frac{{{E_0}}}{c}$     

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

${B_0} = 4 \times {10^{ - 7}}T$

${B_0} = 400nT$

Angular Frequency of Source Can be Given As:

$\omega  = 2\pi \nu$    

$\omega = 2\pi (50 \times {10^6})$    

$\omega = 3.14 \times {10^8}{\kern 1pt} {\kern 1pt} rad/s$

Propagation Constant Can be Given As:

$k = \frac{\omega }{c}$

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

$k = 1.05{\kern 1pt} {\kern 1pt} rad/m$

Wavelength of Wave Can be Given As:

$\lambda  = \frac{c}{\nu }$   

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

$\lambda = 6.0{\kern 1pt} {\kern 1pt} m$

1. Find Expressions for E and B.

Ans: If the wave propagates in the positive x-direction. Then, the electric field vector would be in the positive y-direction and the magnetic field vector will lie in the positive z-direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector can be given as:

\[\overrightarrow E  = \;{E_0}sin(kx - \omega t)\hat j\]

$\overrightarrow E  = 120\sin \left[ {1.05x - 3.14 \times {{10}^8}t} \right]\hat j$

And, magnetic field vector can be given as:

\[\overrightarrow B  = \;{B_0}sin(kx - \omega t)\hat k\]

$\overrightarrow B = (4 \times {10^{ - 7}})\sin \left[ {1.05x - 3.14 \times {{10}^8}t} \right]\hat k$

Ans: The energy of photon can be given as:

$E = h\nu  = \frac{{hc}}{\lambda }$

Where,

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

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

$\lambda $ = wavelength of radiation

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

$E = \frac{{19.8 \times {{10}^{ - 26}}}}{\lambda }J$

$E = \frac{{19.8 \times {{10}^{ - 26}}}}{{\lambda  \times 1.6 \times {{10}^{ - 19}}}}{\kern 1pt} {\kern 1pt} eV$

$E = \frac{{12.375 \times {{10}^{ - 7}}}}{\lambda }{\kern 1pt} {\kern 1pt} eV$

For Different Values of in an Electromagnetic Spectrum, Photon Energies 

are Listed in Below Table:

Ans: Frequency of the electromagnetic wave, \[\nu  = 2.0 \times {10^{10}}Hz\]

Electric field amplitude, \[{E_0} = 48{\text{ }}V{m^{ - 1}}\]

Speed of light, \[c = 3 \times {10^8}m/s\]

Wavelength of wave can be given as:

$\lambda  = \frac{c}{\nu }$

$\lambda = \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{10}}}}$

$\lambda = 0.015m$

1. What is the Amplitude of the Oscillating Magnetic Field?

Ans: Magnetic field strength can be given as:

${B_0} = \frac{{{E_0}}}{c}$

      $ = \frac{{48}}{{3 \times {{10}^8}}}$

      $ = 1.6 \times {10^{ - 7}}T$

1. Show That the Average Energy Density of the E Field Equals the Average Energy Density of the B field. C = 3 x 108m/s

Ans: Energy density of the electric field is given as:

${U_E} = \frac{1}{2}{ \in _0}{E^2}$

Energy density of the magnetic field can be given as:

${U_B} = \frac{1}{{2{\mu _0}}}{B^2}$

Where,

${ \in _0}$ = Permittivity of free space

${\mu _0}$ = Permeability of free space

The relation between E and B can be given as:

$E = cB$  ….(1)

Where,

$c = \frac{1}{{\sqrt {{ \in _0}{\mu _0}} }}$   ….(2)

Substituting equation (2) in equation (1), we get:

$E = \frac{1}{{\sqrt {{ \in _0}{\mu _0}} }}B$

Squaring both the sides, we get:

${E^2} = \frac{1}{{{ \in _0}{\mu _0}}}{B^2}$

${E^2}{ \in _0} = \frac{{{B^2}}}{{{\mu _0}}}$

$\frac{1}{2}{E^2}{ \in _0} = \frac{1}{2}\frac{{{B^2}}}{{{\mu _0}}}$

$ \Rightarrow {U_E} = {U_B}$.