Physics M.C.Q (1 Marks)

2. 105cm^-1

Explantion:

The propagation constant can be written as

$\text{K}=\frac{2pi}{\lambda}=\frac{60284}{6284\times10^{-8}}=10^5\text{cm}^{-1}$

3. Mutually perpendicular directions and are in phase

Explanation:

$\overrightarrow{E}$ 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.

1. n₁ > n₂ > n₃

Explanation:

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

1. $2\mathring{\text{A}}$

Explanation:

Penetrating power is greater for lower wavelength.

2. Halved

Explanation:

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.

3. $\text{p}\neq0,\text{ E}\neq0$

Explanation:

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, $\text{p}\neq0,\text{ E}\neq0$

4. Displacement current

Explanation:

By Maxwell

$\text{I}_\text{d}=\frac{\in\text{dE}}{\text{dt}}$

dE is electric field

I_d​ is displacement current per unit area.

Hence changing electric field gives displacement current.

4. Both (b) and (©)

Explanation:

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. $\hat{\text{k}}.\text{E}=0,\hat{\text{k}}.\text{B}=0.$

Solution:

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.

$\vec{\text{B}}=\text{B}\hat{\text{j}}=\text{B}(\hat{\text{k}}\times\hat{\text{i}})=\frac{\text{E}}{\text{C}}(\hat{\text{k}}\times\hat{\text{i}})$

$=\frac{1}{\text{c}}[\text{k}\times\text{E}\hat{\text{i}}]=\frac{1}{\text{c}}[\hat{\text{k}}\times\vec{\text{E}}]\bigg(\text{as}\frac{\text{E}}{\text{B}}=\text{c}\bigg)$

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 $\hat{\text{k}}\times\vec{\text{B}}=\hat{\text{k}}\times(\text{B}\hat{\text{j}})=\text{B}(\hat{\text{k}}\times\hat{\text{j}})=-\text{B}\hat{\text{i}}$.

1. $\text{E}_0\text{k}=\text{B}_0\omega$

Explanation:

The relation between E₀ and B₀ 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

$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$

$\text{E}_0\text{k}=\text{B}_0\omega$

3. $\frac{1}{\text{r}}$

Explanation:

An antenna that produces the Electromagnetic wave are radiated outwards. The amplitude of electric field vector (E₀​). This electric field vector transports the energy from the source through the medium.

The electric field intensity of the wave from the source at a distance is inversely proportional to the distance between the source and the point.

$\text{E}_0=\frac{1}{\text{r}}$

2. $\lambda_1$ and $\lambda_2$ are variable

Explanation:

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.

3. 4 × 10¹⁴Hz

Explanation:

Velocity of wave = wavelength × frequency

$\text{v}=3\times10^8\frac{\text{m}}{\text{s}}$

$\lambda=7500\text{A}=7500\times10^{-10}\text{m}$

$\text{f}=\frac{3\times10^8}{7500\times10^{-10}}=4\times10^{14}\text{Hz}$

1. $\text{E}_0\text{k}=\text{B}_0\omega$

Explanation:

$\frac{\text{E}_0}{\text{B}_0}=\text{c},$ also $\frac{2\pi}{\lambda}$ and $\omega=2\pi\text{v}.$

$\Rightarrow\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}$

$\text{E}_0\text{k}=\text{B}_0\omega$

3. $\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$

Explanation:

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 $\overrightarrow{\text{B}}$

2. Reciprocal of speed of light in vacuum

Explanation:

Both electric field and magnetic field are vector quantities.

They both are components of electromagnetic waves.

We know that,

$\text{c}=\frac{\text{E}_0}{\text{B}_0}$

$\frac{\text{B}_0}{\text{E}_0}=\frac{1}{\text{c}}$
 

3. 2 × 106m^-1

Explanation:

Wave number $=\frac{1}{\text{wavelength}}$

$=\frac{1}{5000\times10^{-10}}$

$=2\times10^6\text{m}^{-1}$

4. 2×10⁻⁷T

Explanation:

Given: Electric field = 60v/m

C = 3×10⁸m/s²

To find: Magnetic field

Solution:

We know, $\text{C}=\frac{\text{E}_0}{\text{B}_0}$

Therefore, $\text{B}_0=\frac{\text{E}_0}{\text{C}}$

$\frac{60}{3\times10^8}=2\times10^7\text{T}$

1. $\frac{26.5\text{W}}{\text{m}^2}$

Explanation:

Maximum rate of energy flow, S = E₀ ​× H₀

Given, $\text{E}_0=\frac{100\text{V}}{\text{m}},\text{H}_0=\frac{0.265\text{A}}{\text{m}}$

$\therefore\text{S}=100\times0.265=\frac{26.5\text{W}}{\text{m}^2}$​

1. Towards +X direction

Explanation:

The direction of EM wave is given by the direction of $\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$

3. Ultraviolet region.

Solution:

Here it is given, the energy required to dissociate a carbon monoxide molecule into catbon and oxygon atoms is E = 11 eV

We know that, E = hf = 6.62 × 10^-34J-s

f = frequency

⇒ 11eV = hf

$\text{f}=\frac{11\times1.6\times10^{-19}}{6.62\times10^{-34}}=2.65\times10^{15}\text{Hz}$

This frequency radiation belongs to ultraviolet region.

Important point:

4. $\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$

Explanation:

Electromagnet waves have electric field as well as magnetic field which are perpendicular to each other and the electromagnetic waves propagate in a direction

which is perpendicular to both the fields.

Thus the propagation vector of EM waves $\overrightarrow{\text{k}}=\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$