Numerical Problems-2 - Chapter 5 Electromagnetic Waves 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A parallel plate capacitor has circular plates, each of radius $5.0 \mathrm{~cm}$. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^{12} \mathrm{~V} \mathrm{~m}^{-1} \mathrm{~s}^{-}$ ${ }^1$. What is the displacement current?
Solution:
Radius, $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$
The rate okf electric frield, $\frac{d E}{d t}=10^{12} \mathrm{~V} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$
Displacement current, $\mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{d \varphi_E}{d t}=\varepsilon_0 \frac{d}{d t}(\mathrm{EA})=\varepsilon_0\left(\pi \mathrm{r}^2\right) \frac{d E}{d t}$ $=8.85 \times 10^{-12} \times 3.14 \times\left(5 \times 10^{-2}\right)^2 \times 10^{12}$
$\mathrm{I}_{\mathrm{d}}=0.069$
$\mathrm{I}_{\mathrm{d}}=0.07$ (or) $70 \mathrm{~mA}$
Question 2.
The voltage between the plates of a parallel - plate capacitor of capacitance $1 \mu \mathrm{F}$ is changing at the rate of $5 \mathrm{Vs}^{-1}$. What is the displacement current in the capacitor?
Solution:
Capacitance of parallel plate capacitor, $\mathrm{C}=1 \mu \mathrm{F}$
$\mathrm{C}=1 \times 10^{-6} \mathrm{~F}$
The rate of voltage $\mathrm{b} / \mathrm{n}$ the plate, $\frac{d v}{d t}=5 \mathrm{Vs}^{-1}$
Displacement current,
$
\begin{aligned}
& \mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{d \varphi_E}{d t}=\varepsilon_0 \frac{d}{d t}(\mathrm{EA}) \\
& \varepsilon_0=\mathrm{A} \frac{d}{d t}\left(\frac{V}{d}\right) \mathrm{E}=\frac{V}{d} \\
& =\frac{\varepsilon_0 \mathrm{~A}}{d} \frac{d V}{d t}=\mathrm{C} \cdot \frac{d V}{d t} \\
& \mathrm{I}_{\mathrm{d}}=1 \times 10^{-6} \times 5 \\
& \mathrm{Id}=5 \mathrm{µA} .
\end{aligned}
$

Question 3.
Electromagnetic waves travels in a medium at a speed of $2 \times 10^8 \mathrm{~ms}^{-1}$. The relative permeability of the medium is 1 . Find the relative permitivity.
Solution:
Speed of an em wave in a medium is given by
$
\begin{aligned}
\mathrm{V} & =\sqrt{\frac{1}{\mu \varepsilon}}=\frac{1}{\sqrt{\mu_r \mu_0 \varepsilon_r \varepsilon_0}}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \frac{1}{\sqrt{\mu_r \varepsilon_r}} \\
V & =C \frac{1}{\sqrt{\mu_r \varepsilon_r}} \\
\mathrm{~V}^2 & =\frac{\mathrm{C}^2}{\mu_r \varepsilon_r}
\end{aligned}
$
Relative permitivity,
$
\begin{aligned}
& \varepsilon_r=\frac{\mathrm{C}^2}{\mathrm{~V}^2 \mu_r}=\frac{\left(3 \times 10^8\right)^2}{1 \times\left(2 \times 10^8\right)^2}=\frac{9 \times 10^{16}}{4 \times 10^{16}} \\
& \varepsilon_r=2.25 \text { (no unit) }
\end{aligned}
$

Question 4.
A radiation of energy $E$ falls normally on a perfectly reflecting surface. Find the momentum transferred to the surface.
Solution:
Momentum of radiation of energy $\mathrm{E}$ is $\mathrm{P}=\frac{E}{C}$
Since the radiation is completely reflected, its momentum changes by $\frac{2 E}{C}$
Therefore, by the law of conservation of momentum the momentum transferred to the surface is $\frac{2 E}{C}$.
Question 5.
The energy of the $\mathrm{EM}$ wave is of the order of $15 \mathrm{KeV}$. To which part of the spectrum does it belong?
Solution:
$
\begin{aligned}
& \mathrm{E}=\mathrm{hv}=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E} \\
& \mathrm{~h}=6.626 \times 10^{-34} \mathrm{Js} ; \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1} \\
& \mathrm{E}=15 \times 10^3 \mathrm{eV}=15 \times 10^3 \times 1.6 \times 10^{-19} \mathrm{~V} \\
& \lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{15 \times 10^3 \times 1.6 \times 10^{-19}}=\frac{19.878 \times 10^{-26}}{24 \times 10^{-16}}=0.8282 \times 10^{10} \mathrm{~m} \\
& \lambda=0.8282 \times \AA
\end{aligned}
$
Hence X-rays
Question 6.
Light With an energy flux of $25 \times 10^4 \mathrm{Wm}^{-2}$ falls on a perfectly reflecting surface at normal incidence. If surface area is $15 \mathrm{~cm}^2$, to calculate the average force exerted on the surface.
Solution:
Average force $=$ momentum transferred per second
$
\mathrm{F}_{\mathrm{av}}=\frac{P}{T}=\frac{2 u}{C}
$
Where $U$ is the energy falling on th surface per second.
$
\begin{aligned}
& \mathrm{F}_{\mathrm{av}}=\frac{2 \times 25 \times 10^4 \times 15 \times 10^{-4}}{3 \times 10^8}=250 \times 10^{-8} \\
& \mathrm{~F}_{\mathrm{av}}=2.5 \times 10^{-6} \mathrm{~N} .
\end{aligned}
$
Common Errors And Its Rectifications
Common Errors :
1. Students do the mistakes most of times in unit of frequency. They write the units in improper ways. Eg. Hertz (or) H. This is the wrong way.
2. They may confuse the frequency range of radiations and wavelength range of radiation.
Rectifications:
1. The correct way of unit is hertz (or) Hz. The unit of frequency is hertz (or) $\mathrm{Hz}$ (or) $\mathrm{s}^{-1}$.
2. The easy way to understand frequency and wavelength range of radiations are, Frequency increases the order of gamma, X-ray, UV, visible, IR, microwave, radiowave. Wavelength increases the order of the reverse of frequency order.