Physics M.C.Q (1 Marks)

Light consists of electric and magnetic field that are perpendicular $90^\circ $ to each other.
$\text{APPOACH}$ by example
Electric field inside plates. The magnetic field this given rise to via the displacement current is along the perimeter of the circle parallel to capauatates plates.
So $B$ and $E$ are perpendicular in this case.

electromagnetic radiation consists of electromagnetic waves, which are synchronized oscillations of electric and magnetic fields that propagate at the speed of light through a vaccum. The oscillations of the two fields are perpendicular to each other and perpendicular to the direction of energy and wave propagation, forming a transverse wave. so if propogation is along $Y-$direction ,Electric field will be along $X$ or $Z$, if it is along $Z -$direction than Magnetic field has to be in $X -$direction.

The communication and broadcasting following the base on generation, propagation, and detection of electromagnetic waves.
The electromagnetic spectrum describes a different range of electromagnetic waves. These $\text{EM}$ waves are a special type of wave that can travel without a medium.
Electromagnetic waves are named like this due to the fact that they have both an electric and a magnetic component. In a vacuum, $\text{EM}$ waves always travel at the same speed i.e. the speed of light. So, other $\text{EM}$ waves besides light are infrared, ultraviolet, radio waves, and microwaves.
Therefore radio and television both are based on $\text{EM}$ wave properties. Other options like lasers, reactors, and computers are not guided by $\text{EM}$ waves.

Given, frequency of $EM$ waves
$v=8.196 \times 10^6 Hz$
velocity of EM waves $(v)=3 \times 10^8 m / s$
Wavelength of $EM$ waves $\lambda=\frac{ v }{ v }$
$=\frac{3 \times 10^8}{8.196 \times 10^6}$
$=36.60 m$
$=3660 \ cm$

Electric field and magnetic field vectors for an electromagnetic wave are cross $-$ field vectors.
So, the direction of an electromagnetic wave is given by the product of electric field vector and magnetic field vector.
According to the question, electric field vector is directed upwards and $\text{EM}$ wave is directed towards North. So, according to the right $-$ hand thumb rule, the magnetic field vector points towards the East.

The propagation constant can be written as
$\text{K}=\frac{2pi}{\lambda}=\frac{60284}{6284\times10^{-8}}=10^5\text{cm}^{-1}$

$\overrightarrow{E}$ and $\overrightarrow{B}$ are mutually perpendicular to each other and are in phase i.e., they become zero and minimum at the same place and at the same time.

They vary with time following a wave function $($sinuosoidal$)$ and average value of these function is zero and also we can see in figure they are mutually perpendicular and also perpendicular to direction of propagation.

When an electromagnetic wave strikes a material surface, it transports the momentum, as well as the energy, to the surface. The striking electromagnetic wave exerts pressure on the surface. The total energy transferred to the surface by the electromagnetic wave is given by $\text{E}=\text{pc}$ Therefore, $\text{p}\neq0,\text{ E}\neq0$

From electromagnetic spectrum, frequencies of $\gamma-$rays is greater than frequency of $X-$rays. Frequency of Xrays is greater than frequency of ultraviolet rays.

Energy of a photon,
$\text{E} = \text{hv} = \frac {\text{hc}}{\lambda}.$
$\text{E}\propto\frac{1}{\lambda}.$
When the wavelength of electromagnetic radiation is doubled, the energy of the photons is halved.

For a linearly polarised, plane electromagnetic wave,
$E = E _0 \sin \omega\left( t -\frac{ x }{ c }\right)$
$B = B _0 \sin \omega\left( t -\frac{ x }{ c }\right)$
The average value of either $E$ or Bover a cycle is zero $($average of $\sin (\theta)$ over a cycle is zero$).$
Also the electric energy density $\left(U_E\right)$ and magnetic energy density $\left(U_B\right)$ are equal.
$u _{ E }=\frac{1}{2} \in_0 E ^2=\frac{ B ^2}{2 \mu_0}= u _{ B }$
Energy can be found out by integrating energy density over the entire volume of full space. As the energy of the electromagnetic wave is equally shared between electric and magnetic field so their average values will also be equal.

By Maxwell
$I _{ d }=\frac{ EdE }{ dt }$
$d E$ is electric field
$I_d$ is displacement current per unit area.
Hence changing electric field gives displacement current.

The average of $\sin\theta$ and $\cos\theta$ for whole cycle is is zero.
Step $1:$ Analyzing the average value of Kinetic energy.
Kinetic Energy is always a positive quantity, therefore its average will also be a positive quantity.
Step $2:$ Finding the average of electric and the magnetic field.
The equations for the electric field and the magnetic field are given as

1. The associated magnetic field is given as $\text{B}=\frac{1}{\text{c}}\text{k}\times\text{E}=\frac{1}{\omega}(\hat{\text{k}}\times\text{E})$.
2. The electromagnetic field can be written in terms of the associated magnetic field as $\text{E}=\text{c}(\text{B}\times\hat{\text{k}})$.
3. $\hat{\text{k}}.\text{E}=0,\hat{\text{k}}.\text{B}=0.$

1. The direction of propagation of an eletromagnetic wave is always along the direction of vector product $\vec{\text{E}}\times\vec{\text{B}}$. Refer to Figure.

2. $\vec{\text{E}}=\text{E}\hat{\text{i}}=\text{cB}(\hat{\text{j}}\times\hat{\text{k}})=\text{c}(\text{B}\hat{\text{j}}\times\text{k})=\text{c}(\vec{\text{B}}\times\hat{\text{k}})$
3. $\hat{\text{k}}.\vec{\text{E}}=\hat{\text{k}}.(\text{E}\hat{\text{i}})=0,\vec{\text{k}}.\vec{\text{B}}=\vec{\text{k}}.(\text{B}\hat{\text{j}})=0$
4. $\hat{\text{k}}\times\vec{\text{E}}=\hat{\text{k}}\times(\text{E}\hat{\text{i}})=\text{E}(\hat{\text{k}}\times\hat{\text{i}})=\text{E}\hat{\text{j}}$ and $\hat{\text{k}}\times\vec{\text{B}}=\hat{\text{k}}\times(\text{B}\hat{\text{j}})=\text{B}(\hat{\text{k}}\times\hat{\text{j}})=-\text{B}\hat{\text{i}}$.

The relation between $E _0$ and $B _0$ id given by $\frac{\text{E}_0}{\text{B}_0}=\text{c}\ ....(\text{i})$
Here, $c =$ Speed of the electromagnetic wave,
The relation between $\omega ($the angular frequency$)$ and $k\ ($wave number$),$
$\frac{\omega}{\text{k}}=\text{c}\ ...(\text{ii})$
Therefore, from $(i)$ and $(ii),$ we get
$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$
$\text{E}_0\text{k}=\text{B}_0\omega$

According to Maxwell's hypothesis, changing of electric field gives rise to Magnetic field.
We know that $F = qE$,, where $F$ is force and $E$ is electric field.
We can relate magnetic field and force by $F = qvB$, where $v$ is velocity and $B$ is the magnetic field.
Therefore we can obtain magnetic field by changing electric field.

Velocity of a wave is given by:
$\text{v}=\frac{\text{E}}{\text{B}}$
Hence wave velocity change in both the cases.
Frequency of the wave remains the same.
Using $\text{v}=\text{f}\lambda,$ it can be concluded that both $\lambda_1$ and $\lambda_2$ are variable.

Velocity of wave $=$ wavelength $\times $ frequency
$\text{v}=3\times10^8\frac{\text{m}}{\text{s}}$
$\lambda=7500\text{A}=7500\times10^{-10}\text{m}$
$\text{f}=\frac{3\times10^8}{7500\times10^{-10}}=4\times10^{14}\text{Hz}$

The emitted $X-$rays transfer energy to the material on which it is falling.

The direction of propagation of electromagnetic wave is perpendicular to the variation of electric field $\overrightarrow{\text{E}}$ as well as to the magnetic field $\overrightarrow{\text{B}}$

The ionosphere can reflect electromagnetic waves of frequency less than $40MHz$ but not of frequency more than $40MHz.$

The frequency range of Gamma rays is $3 \times 10^{18}$ to $5 \times 10^{22} Hz$. These rays have a wavelength of $6 \times 10^{-13}$ to $10^{-10} m$. Gamma rays are produced in nuclear reactions and are also emitted by radioactive nuclei.

Light wave is an electromagnetic wave in which $E$ and $B$ are at right angles to each other as well as at right angles to the direction of wave propagation. So from the given information in the question, the direction of $B$ vector is in positive $z$ direction.

According to Maxwell's hypothesis, a displacement current will flow through a capacitor when the potential difference across its plates is varying. Thus a varying electric field will exist between the plates and this displacement current is same in magnitude to the current flowing in outer circuit. When a $D.C$ voltage applied across its plates, constant voltage appears across its plates and so there will be no displacement current flowing through the capacitor. Thus the displacement current will flow when the potential is increasing with time.

Wave number $=\frac{1}{\text{wavelength}}$
$=\frac{1}{5000\times10^{-10}}$
$=2\times10^6\text{m}^{-1}$