Numerical problems-1 - Chapter 5 - Electromagnetic Waves - 12th Physics Guide Samacheer Kalvi Solutions

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Numerical problems
Question 1.
Consider a parallel plate capacitor whose plates are closely spaced. Let $\mathrm{R}$ be the radius of the plates and the current in the wire connected to the plates is $5 \mathrm{~A}$, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.
Solution:
The conduction current $\mathrm{I}_{\mathrm{c}}=5 \mathrm{~A}$
According to Gauss's Law,
Electric flux, $\theta_{\mathrm{E}}=\frac{q}{\varepsilon 0}$
$
\begin{aligned}
& \mathrm{I}_d=\varepsilon_0 \cdot \frac{d \phi_{\mathrm{E}}}{d t}=\varepsilon_0 \frac{d}{d t}\left(\frac{q}{\varepsilon_0}\right)=\frac{\varepsilon_0}{\varepsilon_0}\left(\frac{d q}{d t}\right) \\
& \mathrm{I}_c=\frac{q}{t} \Rightarrow q=\mathrm{I}_c t \\
& \mathrm{I}_d=\frac{d}{d t}\left(\mathrm{I}_c t\right)=\mathrm{I}_c \frac{d(t)}{d t}=5 \mathrm{~A} \\
& \mathrm{I}_d=\mathrm{I}_c=5 \mathrm{~A}
\end{aligned}
$
Question 2.
A transmitter consists of LC circuit with an inductance of $1 \mu \mathrm{H}$ and a capacitance of $1 \mu \mathrm{F}$. What is the wavelength of the electromagnetic waves it emits?
Solution:
Inductance of $\mathrm{LC}$ circuit, $\mathrm{L}=1 \mu \mathrm{H}=1 \times 10^{-6} \mathrm{H}$
Capacitance of LC circuit, $\mathrm{C}=1 \mu \mathrm{F}=1 \times 10^{-6} \mathrm{~F}$

Wave length of the electromagnetic wave $\mathrm{X}=\lambda=\frac{C}{f}$
Velocity of light $\mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$
Frequency of electromagnetic wave, $\mathrm{f}=\frac{1}{2 \pi \sqrt{L C}}$
$
=\frac{1}{2 \times 3.14 \sqrt{1 \times 10^{-6} \times 10^{-6}}}=\frac{1}{6.28 \times 10^{-6}} \Rightarrow f=15.92 \times 10^4 \mathrm{~Hz}
$
Wave Length $\lambda=\frac{\mathrm{C}}{f}=\frac{3 \times 10^8}{15.92 \times 10^4}$
$
\begin{aligned}
& =0.1884 \times 10^4 \\
& \lambda=18.84 \times 10^2 \mathrm{~m}
\end{aligned}
$
Question 3.
A pulse of light of duration $10^{-6} \mathrm{~s}$ is absorbed completely by a small object initially at rest. If the power of the pulse is $60 \times 10^{-3} \mathrm{~W}$, calculate the final momentum of the object.
Solution:
Duration of the absorption of light pulse, $\mathrm{t}=10^{-6} \mathrm{~s}$
Power of the pulse $\mathrm{P}=60 \times 10^{-3} \mathrm{~W}$
Final momentum of the object, $\mathrm{P}=\frac{U}{c}$
Velocity of light, $\mathrm{C}=3 \times 10^8$
Energy $\mathrm{U}=$ power $\mathrm{x}$ time
Momentum, $\mathrm{P}=\frac{60 \times 10^{-3} \times 10^{-6}}{3 \times 10^8}$
$\mathrm{P}=20 \times 10^{-17} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

Question 4.
Let an electromagnetic wave propagate along the $\mathrm{x}$ direction, the magnetic field oscillates at a frequency of $10^{10} \mathrm{~Hz}$ and has an amplitude of $10^{-5} \mathrm{~T}$, acting along the $\mathrm{y}$-direction. Then, compute the wavelength of the wave. Also write down the expression for electric field in this case.
Solution:
Frequency of electromagnetic wave, $\mathrm{v}=10^{10} \mathrm{~Hz}$
Amplitude of Oscillating magnetic field, $\mathrm{B}_0=10^{-5} \mathrm{~T}$
Wave length of the wave, $\lambda=\frac{C}{f}=\frac{3 \times 10^8}{10^{10}}=3 \times 10^{-2} \mathrm{~m}$
Amplitude of oscillating electric field, $\mathrm{E}_0=\mathrm{B}_0 \mathrm{C}$
$
\mathrm{C}=\frac{E_0}{B_0}
$
$
\begin{aligned}
& \mathrm{E}_0=10^{-5} \times 3 \times 10^8 \\
& \mathrm{E}_0=3 \times 10^3=\mathrm{NC}^{-1}
\end{aligned}
$
Experession for electric field in Oscillataing wave
$
\begin{aligned}
& \mathrm{E}=\mathrm{E}_0 \sin (\mathrm{kx}-\mathrm{wt}) \\
& \mathrm{K}=\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{3 \times 10^{-2}}=209 \times 10^2 \\
& \mathrm{~W}=2 \pi f=2 \times 3.14 \times 10^{10}=6.28 \times 10^{10} \\
& \vec{E}=3 \times 10^3 \sin \left(2.09 \times 10^2 \mathrm{x}-6.28 \times 10^{10} \mathrm{t}\right) \hat{i} \mathrm{NC}^{-1} .
\end{aligned}
$

Question 5 .
If the relative permeability and relative permittivity of the medium is 1.0 and 2.25 , respectively. Find the speed of the electromagnetic wave in this medium.
Solution:
Relative permeability of the medium, $\mu_{\mathrm{r}}=1$
Relative permitivity of the medium, $\varepsilon_{\mathrm{r}}=2.25$
$
\left(\varepsilon_r=\frac{\varepsilon}{\varepsilon_0} \& \mu_r=\frac{\mu}{\mu_0}\right)
$
Speed of electromagnetic wave, $\mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}}$
$
\begin{aligned}
& =\frac{1}{\sqrt{\mu_r \mu_{\circ} \varepsilon_r \varepsilon_{\circ}}}=\frac{C}{\sqrt{\mu_r \varepsilon_r}} \quad\left[\text { Where, } C=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right] \\
& =\frac{3 \times 10^8}{\sqrt{1 \times 2.25}}=\frac{3 \times 10^8}{1.5} \\
v & =2 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$