$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$ and
$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$
At t = 0, a particle of charge q is at origin with
a velocity $\overrightarrow v = 0.8c\widehat j$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$
At t = 0, a positively charged particle is at the point (x, y, z) = $\left( {0,0,{\pi \over k}} \right)$.
If its instantaneous velocity at (t = 0) is ${v_0}\widehat k$ , the force acting on it due to the wave is :
$\overrightarrow B $ = 3 $ \times $ 10-8 sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat j$ T, then what will be expression for electric field ?
$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$
The corresponding magnetic field $\overrightarrow B $ is then given by
$$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$
The associated electric field will be :-
[Given permittivity of space $ \in $0 = 9 $ \times $ 10–12 SI units, Speed of light c = 3 $ \times $ 108 m/s]
$\overrightarrow {{E_1}} $ = ${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$ in air and
$\overrightarrow {{E_2}} $ = ${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$ in medium,
where the wave number k and frequency $\nu $ refer to their values in air. The medium is non-magnetic. If ${\varepsilon _{{r_1}}}$ and ${\varepsilon _{{r_2}}}$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
It is polarized along the $\widehat k$ direction. The acceptable form for the magnetic field is :
$\overrightarrow E $ = 2 E0 $\widehat i$ cos kz cos $\omega $t
Its magnetic field $\overrightarrow B $ is then given by :
$\overrightarrow B $ = B0 sin (k x + $\omega $t) $\widehat j\,T$
Expression for corresponding electric field will be :
Where c is speed of light.
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$
and the magnetic field is $\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat y$
A : Blue light B : Yellow light
C : X-ray D : Radiowave.
The equation of the electric field of an electromagnetic wave propagating through free space is given by : $E=\sqrt{377} \sin \left(6.27 \times 10^3 t-2.09 \times 10^{-5} x\right) \mathrm{N} / \mathrm{C}$
The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^2$. The value of $\alpha$ is
$ \left(\text { Take } \sqrt{\frac{\mu_0}{\varepsilon_o}}=377 \text { in SI units }\right) $
Explanation:
The standard wave equation for the electric field component of an electromagnetic wave is given as $\mathrm{E}= \mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, where $\mathrm{E}_0$ is the amplitude of electric field, $\omega$ is the angular frequency and k is the wave number.
The give equation for the electric field of electromagnetic wave is $=\sqrt{377} \sin \left(6.27 \times 10^3 t-2.09 \times\right. \left.10^{-5} \mathrm{x}\right) \mathrm{N} / \mathrm{C}$
So, electric field amplitude is $\mathrm{E}_0=\sqrt{377} \mathrm{~N} / \mathrm{C}$.
The total energy density (energy per unit volume) of an electromagnetic wave is the sum of the electric field energy density ( $\mathrm{u}_{\mathrm{E}}$ ) and magnetic field energy density ( $\mathrm{u}_{\mathrm{B}}$ ).
$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \mu_0 \mathrm{~B}^2 $
In an electromagnetic wave traveling in a vacuum, the instantaneous electric and magnetic fields are related by the speed of light (c):
$ \mathrm{E}=\mathrm{cB} \Rightarrow \mathrm{~B}=\frac{\mathrm{E}}{\mathrm{c}} $
So, the energy density for magnetic field is,
$ u_B=\frac{1}{2} \mu_0\left(\frac{E}{c}\right)^2=\frac{E^2}{2 \mu_0 c^2} $
Since $c=\frac{1}{\sqrt{\mu_0} \epsilon_0} \Rightarrow c^2=\frac{1}{\mu_0 \epsilon_0} \Rightarrow \mu_0 c^2=\frac{1}{\epsilon_0}$.
$ \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2 $
Thus, the energy is equally shared: $\mathrm{u}_{\mathrm{E}}=\mathrm{u}_{\mathrm{B}}$.
The total instantaneous energy density is:
$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \epsilon_0 \mathrm{E}^2=\epsilon_0 \mathrm{E}^2 $
Since the electric field oscillates sinusoidally $\left(E=E_0 \sin \omega t\right)$, the average value of $E^2$ over one cycle is $\frac{E_0^2}{2}$.
$ \overline{\mathrm{u}}=\epsilon_0\left(\frac{\mathrm{E}_0^2}{2}\right)=\frac{1}{2} \epsilon_0 \mathrm{E}_0^2 $
Intensity is defined as the energy crossing a unit area per unit time. It is related to the average energy density by the speed of the wave $c$ :
$ \mathrm{I}=\overline{\mathrm{u}} \times \mathrm{c} $
$\Rightarrow $ $ I=\left(\frac{1}{2} \epsilon_0 E_0^2\right) c $
The characteristic impedance of free space, $\mathrm{Z}_0=\sqrt{\frac{\mu_0}{\epsilon_0}}$ (approx $377 \Omega$ ).
$I=\frac{1}{2} \epsilon_0 E_0^2\left(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\right)$
$\Rightarrow $ $I=\frac{E_0^2}{2}\left(\frac{\epsilon_0}{\sqrt{\mu_0 \epsilon_0}}\right)=\frac{E_0^2}{2} \sqrt{\frac{\epsilon_0}{\mu_0}}=\frac{(\sqrt{377})^2}{2} \times \frac{1}{377}=\frac{1}{2}$
$\Rightarrow $ $ \frac{1}{2}=\frac{1}{\alpha} \Rightarrow \alpha=2 $
Therefore, the value of $\alpha$ is 2.
The electric field of a plane electromagnetic wave, travelling in an unknown nonmagnetic medium is given by,
$ E_{\mathrm{y}}=20 \sin \left(3 \times 10^6 x-4.5 \times 10^{14} \mathrm{t}\right) \mathrm{V} / \mathrm{m} $
(where $x, \mathrm{t}$ and other values have S.I. units). The dielectric constant of the medium is $\_\_\_\_$
(speed of light in free space is $3 \times 10^8 \mathrm{~m} / \mathrm{s}$ )
Explanation:
The general standard mathematical form for a plane wave traveling in the positive x -direction is:
$ \mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) $
Where :
- A is the amplitude (maximum displacement).
k is the angular wave number, related to the wavelength $(\lambda)$ by $\mathrm{k}=\frac{2 \pi}{\lambda}$.
$\omega$ is the angular frequency, related to the time period (T) by $\omega=\frac{2 \pi}{T}$.
The speed of propagation of traveling wave is given as,
$ v=\frac{\omega}{k} $
From Maxwell's equations, the speed of an electromagnetic wave in a vacuum (c) depends on the fundamental constants of free space:
$ c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} $
Where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.
When light travels through a material medium, its speed (v) changes depending on the magnetic and electrical properties of that specific medium :
$ v=\frac{1}{\sqrt{\mu \epsilon}} $
Where $\mu$ is the absolute permeability and $\epsilon$ is the absolute permittivity of the medium.
Relative to a vacuum :
$\mu=\mu_0 \mu_{\mathrm{r}}$ (where $\mu_{\mathrm{r}}$ is relative permeability)
$\epsilon=\epsilon_0 \epsilon_{\mathrm{r}}$ (where $\epsilon_{\mathrm{r}}$ is relative permittivity, also called the dielectric constant, often denoted as K )
$\mathrm{v}=\frac{1}{\sqrt{\left(\mu_0 \mu_{\mathrm{r}}\right)\left(\epsilon_0 \epsilon_{\mathrm{r}}\right)}}$
$\Rightarrow $ $v=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \epsilon_r}}$
$\Rightarrow $ $ v=\frac{c}{\sqrt{\mu_r \epsilon_r}} $
As the medium is non-magnetic so its relative permeability is exactly 1 ( $\mu_{\mathrm{r}}=1$ ).
$ \mathrm{v}=\frac{\mathrm{c}}{\sqrt{1 \times \epsilon_{\mathrm{r}}}} $
$\Rightarrow $ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\epsilon_{\mathrm{r}}}}$
$\Rightarrow $ $\sqrt{\epsilon_{\mathrm{r}}}=\frac{\mathrm{c}}{\mathrm{v}}$
$\Rightarrow $ $ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}\right)^2 $
From the given electric field equation :
$ E_y=20 \sin \left(3 \times 10^6 x-4.5 \times 10^{14} t\right) $
By comparing this to the standard form $\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \sin (\mathrm{kx}-\omega \mathrm{t})$ :
$\mathrm{k}=3 \times 10^6 \mathrm{rad} / \mathrm{m}$
$\omega=4.5 \times 10^{14} \mathrm{rad} / \mathrm{s}$
The speed of the wave in the medium (v) :
$ \mathrm{v}=\frac{\omega}{\mathrm{k}} $
$\Rightarrow $ $\mathrm{v}=\frac{4.5 \times 10^{14}}{3 \times 10^6}$
$\Rightarrow $ $ \mathrm{v}=1.5 \times 10^8 \mathrm{~m} / \mathrm{s} $
We are given $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$.
$ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}\right)^2 $
$\Rightarrow $ $\epsilon_{\mathrm{r}}=\left(\frac{3 \times 10^8}{1.5 \times 10^8}\right)^2$
$\Rightarrow $ $ \epsilon_{\mathrm{r}}=(2)^2=4 $
Therefore, the dielectric constant of the medium is 4 .
Hence, the correct answer is 4 .
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma = 10 \,\mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$.
$ \left[\text { Take } \frac{1}{4 \pi \epsilon_0}=9 \times 10^9\, \mathrm{Nm}^2 / \mathrm{C}^2\right] $
Explanation:
According to Ohm's Law in point form, the conduction current density is proportional to the electric field (E) :
$ \mathrm{J}_{\mathrm{c}}=\sigma \mathrm{E} $
The maximum value occurs when the electric field is at its peak $\left(\mathrm{E}_0\right)$ :
$ \left(\mathrm{J}_{\mathrm{c}}\right)_{\max }=\sigma \mathrm{E}_0 $
Displacement current density arises from a time-varying electric field. For an electromagnetic wave, the electric field can be represented as $E=E_0 \sin (\omega t)$.
Maxwell defined displacement current density as :
$ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial \mathrm{E}}{\partial \mathrm{t}} $
$\Rightarrow $ $ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial}{\partial \mathrm{t}}\left(\mathrm{E}_0 \sin (\omega \mathrm{t})\right)=\epsilon \omega \mathrm{E}_0 \cos (\omega \mathrm{t}) $
The maximum value occurs when the cosine term is 1 :
$ \left(\mathrm{J}_{\mathrm{d}}\right)_{\max }=\epsilon \omega \mathrm{E}_0 $
The ratio of the maximum values is:
Ratio $=\frac{\left(\mathrm{J}_{\mathrm{c}}\right)_{\text {max }}}{\left(\mathrm{J}_{\mathrm{d}}\right)_{\text {max }}}=\frac{\sigma \mathrm{E}_0}{\epsilon \omega \mathrm{E}_0}=\frac{\sigma}{\epsilon \omega}$
The conductivity $(\sigma)$ of the medium is, $\sigma=10 \mathrm{mho} / \mathrm{m}$
The angular frequency $(\omega)$ is $2 \pi \mathrm{f}=2 \pi \times\left(100 \times 10^6\right)=2 \pi \times 10^8 \mathrm{rad} / \mathrm{s}$
Since no specific medium permittivity is given, we use the permittivity of free space ( $\epsilon_0$ ).
Given $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$, then $\epsilon_0=\frac{1}{36 \pi \times 10^9}$.
Substituting these values into the ratio:
Ratio $=\frac{10}{\frac{1}{36 \pi \times 10^9} \times 2 \pi \times 10^8}=\frac{10 \times 36 \pi \times 10^9}{2 \pi \times 10^8}$
Ratio $=180 \times 10^1=1800$
Therefore, the ratio of maximum conduction current density to maximum displacement current density is 1800 .
Hence, the correct answer is $\mathbf{1 8 0 0}$.
A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance $2.5 \mu \mathrm{~F}$. The dielectric constant of the medium between the capacitor plates is 1 . It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _________ $\mathrm{Vs}^{-1}$.
Explanation:
$ \frac{dV}{dt} = \frac{I}{C} $
Given that the displacement current is $ I = 0.25 \times 10^{-3} \, \text{A} $ and the capacitance is $ C = 2.5 \times 10^{-6} \, \text{F}, $ the rate of change of the potential difference is calculated as follows:
$ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}. $
Thus, the magnitude of the rate of change of the potential difference is $ 100 \, \text{V/s} $.
A parallel plate capacitor of area $A=16 \mathrm{~cm}^2$ and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area $\mathrm{A}_0=3.2 \mathrm{~cm}^2$ inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through $\mathrm{A}_0$ is __________ mA .
Explanation:
To determine the displacement current through a hypothetical plane surface within a parallel plate capacitor, follow these steps:
Current Density Calculation:
The current density ($ J_d $) is the current ($ I $) divided by the area ($ A $) of the capacitor plates.
Given: $ I = 6 \, \text{A} $ and $ A = 16 \, \text{cm}^2 $.
$ J_d = \frac{I}{A} = \frac{6 \, \text{A}}{16 \, \text{cm}^2} $
Displacement Current Through the Hypothetical Surface:
The hypothetical surface has an area $ A_0 = 3.2 \, \text{cm}^2 $.
The displacement current through this smaller area ($ I_{small} $) is the current density multiplied by the area of the hypothetical surface.
$ I_{small} = J_d \times A_0 = \frac{6 \, \text{A}}{16 \, \text{cm}^2} \times 3.2 \, \text{cm}^2 $
Calculation:
Simplify the expression to find the displacement current:
$ I_{small} = \left(\frac{6}{16}\right) \times 3.2 = 1.2 \, \text{A} = 1200 \, \text{mA} $
Thus, the displacement current through the hypothetical plane surface is 1200 mA.
In a medium the speed of light wave decreases to $0.2$ times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is $x: 1$. The value of $x$ is _________.
(Given speed of light in free space $=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$ and for the given medium $\mu_{\mathrm{r}}=1$)
Explanation:
Putting the values:
$0.2 c=\frac{c}{\sqrt{\varepsilon_{r}}}$
$\Rightarrow \sqrt{\varepsilon_{r}}=5$
$\Rightarrow$ Required ratio $=\frac{\varepsilon_{r}}{n}=\frac{\varepsilon_{r}}{\sqrt{\varepsilon_{r}}}=\sqrt{\varepsilon_{r}}=5$
$\Rightarrow x=5$
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $24 \mathrm{~W}$. The radius of curvature of hemisphere is $10 \mathrm{~cm}$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________ $\times~10^{-8} \mathrm{~N}$.
Explanation:
$ \begin{aligned} & \text { Force }=\int P d A \cos \theta \\\\ & =\frac{2 \mathrm{I}}{\mathrm{C}} \int \mathrm{dA} \cos \theta=\frac{2 \mathrm{I}}{\mathrm{C}} \pi \mathrm{R}^2=2 \frac{\mathrm{p}_0}{4 \pi \mathrm{R}^2} \cdot \frac{\pi \mathrm{R}^2}{\mathrm{C}} \\\\ & =\frac{\mathrm{p}_0}{2 \mathrm{C}}=\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} \mathrm{~N} \end{aligned} $
Nearly 10% of the power of a $110 \mathrm{~W}$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1 \mathrm{~m}$ from the bulb to a distance of $5 \mathrm{~m}$ is $a \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2}$. The value of 'a' will be _________.
Explanation:
$=\frac{10}{100} \times 110 \mathrm{~W}$
$=11 \mathrm{~W}$
$\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_1^2}-\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_2^2}$
$=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]$
$=\frac{11}{4 \pi} \times \frac{24}{25}$
$=\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$
The displacement current of 4.425 $\mu$A is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs$-$1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x $\times$ 10$-$3 m. The value of x is __________.
(Permittivity of free space, E0 = 8.85 $\times$ 10$-$12 C2 N$-$1 m$-$2).
Explanation:
$4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}$
$ \Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}$
$ \Rightarrow d = 8 \times {10^{ - 3}}$ m
$ \Rightarrow x = 8$
The intensity of the light from a bulb incident on a surface is 0.22 W/m2. The amplitude of the magnetic field in this light-wave is ______________ $\times$ 10$-$9 T.
(Given : Permittivity of vacuum $\in$0 = 8.85 $\times$ 10$-$12 C2 N$-$1-m$-$2, speed of light in vacuum c = 3 $\times$ 108 ms$-$1)
Explanation:
$I = {1 \over 2}{\varepsilon _0}E_0^2\,.\,c = {1 \over 2}{\varepsilon _0}{(c{B_0})^2}c$
$ \Rightarrow I = {1 \over 2}{\varepsilon _0}{c^3}B_0^2$
$ \Rightarrow 0.22 = {1 \over 2}\left( {8.85 \times {{10}^{ - 12}}} \right){\left( {3 \times {{10}^8}} \right)^3}B_0^2$
$ \Rightarrow {B_0} \simeq 43 \times {10^{ - 9}}$ T
The energy contained in a cylinder of volume V is 5.5 $\times$ 10$-$12 J. The value of V is _____________ cm3. (given $\in$0 = 8.8 $\times$ 10$-$12C2N$-$1m$-$2)
Explanation:
Energy density = ${1 \over 2}{ \in _0}E_0^2$
Energy of volume $V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$
${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$
$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$
= .0005 $\times$ 106 (c.m)3
= 500 (c.m)3