Physics M.C.Q (1 Marks)

2. 2.2A

Explanation:

Displacement current $=\text{I}_\text{D}=\text{C}\frac{\text{dV}}{\text{dT}}=\text{C}\omega\text{V}_\text{o}=\frac{\text{V}_\text{o}}{\text{X}_\text{c}}=\frac{220\text{V}}{100\Omega}=2.2\text{A}$ 

As we are asked amplitude of displacement current. So, we don't have to worry about charge on capacitor.

1. $2\times10^{-8}\text{T}$

Explanation:

Velocity of EM wave $\text{v}=\frac{3\times10^8\text{m}}{\text{s}}$

Electric field vector $\text{E}=\frac{6\text{V}}{\text{m}}$

Thus magnetic field vector $\text{B}=\frac{\text{E}}{\text{v}}$

$\therefore\text{B}=\frac{6}{3\times10^8}=2\times10^{-8}\text{T}$

4. Double the frequency of the wave.

Explanation:

The energy per unit volume of an electromagnetic wave,

$\text{u}=\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}$

The energy of the given volume can be calculated by multiplying the volume with the above expression.

$\text{U}=\text{u}\times\text{V}=\Big(\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}\Big)\times\text{V}\ ....(\text{i})$

Let the direction of propagation of the electromagnetic wave be along the z-axls. Then, the electric and magnetic fields at a particular point are given by,

$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$

$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$

Substituting the values of electric and magnetic fields in (1) we get,

$\text{U}=\Big(\frac{1}{2}\in_0\big(\text{E}_0^2\sin^2(\text{kz}-\omega\text{t})+\frac{\text{B}^2_0\sin^2(\text{kz}-\omega\text{t})}{2\mu_0}\Big)\times\text{V}$

$\text{U}=\Big(\in_0\text{E}^2_0\frac{(1-\cos(2\text{kz}-2\omega\text{t}))}{4}+\frac{\text{B}_0^2(1-\cos(2\text{kz}-2\omega\text{t}))}{4\mu_0}\Big)\times\text{V}$

From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency $2\omega$ Thus. the frequency of the energy of the electromagnetic wave will also be double.

2. 36 × 10^-4 kg m/s.

Solution:

Givne energy flux $\phi=20\frac{\text{W}}{\text{cm}^2}$

Area, A = 30cm²

Time, t = 30min = 30 × 60s

Now, total energy falling on the surface in time t is,

$\text{U}=\phi\text{At}=20\times30\times(30\times60)\text{J}$

Momentum of the incident light $=\frac{\text{U}}{\text{c}}$

$=\frac{20\times30\times(30\times60)}{3\times10^{8}}=36\times10^{-4}\text{kg-ms}^{-1}$

Momentum of the reflected light = 0

$\therefore$ Momentum delivered to the surface

$=36\times10^{-4}-0=36\times10^{-4}\text{kg-ms}^{-1}$

Important points

Mass of photon:

Actually rest mass of the photon is zero. But its effective mass is given as $\text{E}=\text{mc}^2=\text{hv}\Rightarrow\ \text{m}=\frac{\text{E}}{\text{C}^2}=\frac{\text{hv}}{\text{C}^2}=\frac{\text{h}}{\text{c}\lambda}$. Thsi mass is also known as kintic mass of the photon.

Momentum of the photon:

Momentum $\text{p}=\text{m}\times\text{c}=\frac{\text{E}}{\text{c}}=\frac{\text{hv}}{\text{c}}=\frac{\text{h}}{\lambda}$

Number of emitted photons:

The number of photons emitted per second from a source of monochromatic radiation of wavelengh $\lambda$ and power P is given as $\text{(n)}=\frac{\text{P}}{\text{E}}=\frac{\text{P}}{\text{hv}}=\frac{\text{P}\lambda}{\text{hc}}$; where E = energy of each photon

Inrensity of light (I):

Energy crossing per unit area normally per second is called intensity or energy flux

i.e. $\text{I}=\frac{\text{E}}{\text{At}}=\frac{\text{P}}{\text{A}}\Big(\frac{\text{E}}{\text{t}}=\text{P}=\text{radiation Power}\Big)$

At a distance r from a point source of power P intensity is given by $\text{I}=\frac{\text{P}}{4\pi\text{r}^2}\Rightarrow\ \text{I}\propto\frac{1}{\text{r}^2}$.

3. $\frac{\text{L}^2}{\text{T}^2}$

Explanation:

The speed of light, $\text{C}=\frac{1}{\sqrt{\mu_0\in_0}}$

The dimensions of $\frac{1}{\sqrt{\mu_0\in_0}}$ are of velocity, i.e., $\frac{\text{L}}{\text{T}}$

Therefore, $\frac{1}{\in_0\mu_0}$ will have dimensions $\frac{\text{L}^2}{\text{T}^2}$

4. 7×10⁻⁶N

Explanation:

For a perfectly absorbing surface,

$\text{F}=\frac{\text{IA​}}{\text{C}}$

$=\frac{(1400\text{W/m}^2×1.5\text{m}^2)​}{(3×10^8\text{m/s})}$

$=7\times10^{−6}\text{N}.$

3. 1 : 1.

Solution:

The intensity of electromagnetic wave is given by,

I = U_avc, where U_av = Average energy and c = speed of light

Intensity in relation with electric field $\text{U}_\text{av}=\frac{1}{2}\in_0\text{E}_0^2$

Intensity relation with magnetic field $\text{U}_\text{av}=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}$

Now taking the intensity in terms of electric field,

$(\text{U}_\text{av})_\text{electric field}=\frac{1}{2}\in_0\text{E}_0^2=\frac{1}{2}\in_0(\text{cB}_0)^2\ \ (\because\ \text{E}_0=\text{cB}_0)$

But, $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}$

$\therefore\ (\text{U}_\text{av})_\text{Electric field}=\frac{1}{2}\in_0\times\frac{1}{\mu_0\in_0}\text{B}_0^2=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}$

$=(\text{U}_\text{av})_\text{magnetic field}$

Hence the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.

It means the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is 1:1.

Impotant point:

Propertioe of EM waves

Speed: In free, its speed $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{\text{E}_0}{\text{B}_0}=3\times10^8\text{m/s}$.

In medium $\text{v}=\frac{1}{\sqrt{\mu\in}}$; where $\mu_0$ = Absolute permeability, $\in_0$ = Absolute permittivity, E₀ and B₀ = Amplitude of electric of field and magnetic field vectors.

Energy: The energy in an EM waves is divided equally between the electric and magnetic fields.

Energy density of electric field $\text{u}_\text{e}=\frac{1}{2}\in_0\text{E}^2$, Energy density of magnetic field $\text{u}_\text{B}=\frac{1}{2}\frac{\text{B}^2}{\mu_0}$

It is found that u_e = u_B. Also $\text{u}_\text{av}=\text{u}_\text{e}+\text{u}_\text{B}=2\text{u}_\text{e}=2\text{u}_\text{B}=\in_0\text{E}^2=\frac{\text{B}^2}{\mu_0}$

Intensity (I): The energy crossing per unit area unit time, perpendicular to the direction of propagation of EM wave is called intensity.

$\text{I}=\text{u}_\text{av}\times\text{c}=\frac{1}{2}\in_0\text{E}^2\text{c}=\frac{1}{2}\frac{\text{B}^2}{\mu_0}.\text{c}$

Momentum: EM waves also carries momentum, if a portion of Em wave of energy u propagating with speed c, then linear momentum $=\frac{\text{Energy (u)}}{\text{Speed (c)}}$

When the incident EM wave is completely absorbed by a surface, it delivers energy u and momentum u/c to the surface.

When a wave of energy us is totally reflected from the surface, the momentum delivered to surface is 2u/c.

Poynting vector $(\vec{\text{S}})$: In EM waves, the rate of flow of energy crossing a unit area is described by the poynting vector. Its unit is watt/m² and $\vec{\text{S}}=\frac{1}{\mu_0}(\vec{\text{E}}\times\vec{\text{B}})=\text{c}^2\in_0(\vec{\text{E}}\times\vec{\text{B}})$. Because in EM waves, $\vec{\text{E}}$ and $\vec{\text{B}}$ are perpendicular to each other, the magnitude of $\vec{\text{S}}$ is $|\vec{\text{S}}|=\frac{1}{\mu_0}\text{E B}\sin90^\circ=\frac{\text{EB}}{\mu_0}=\frac{\text{E}^2}{\mu_\text{C}}$.

The direction of the Poynting vector $\vec{\text{S}}$ at any point gives the wave's direction of travel and direction of energy transport the point.

Radiation Pressure: Is the momentum imparted per second per unit area on which the light falls.

For a perfectly reflecting surface $\text{P}_\text{r}=\frac{2\text{S}}{\text{c}}$; S = Poynting vector; c = speed of light

For a perfectly absorbing surface $\text{P}_\text{a}=\frac{\text{S}}{\text{c}}$.

The radiation pressure is real that's why tails of comet point away from the sun.

2. $\pi$

Explanation: 

The phase difference between electric and magnetic fields in an electromagnetic wave is$\pi$.

3. increases or decreases

Explanation:

Displacement current inside a capacitor is given by:-

$\text{i}_\text{d}=\in_0\frac{\phi\text{E}}{\text{dt}}$

where $\phi_\text{E}$​ is the electric flux inside the capacitor

The displacement current is developed inside a capacitor when there is a change in the electric flux linked with the capacitor.

The change in electric flux can occur in both the cases either the charge increases or decreases on the capacitor. This will lead to a change in flux linked with the coil.

4. Both of them.

Explanation:

According to Ampere-Maxwell's Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by,

$\text{i}_\text{d}=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}}$ $\big(\because\phi_\text{E}$ is the electric flux$\big)$

Thus, the magnetic field is produced by the moving charge as well as the electric field.

3. 7.5m

Explanation

Using $\text{v}=\text{v}\lambda$ where v is the speed of EM wave.

As $\text{V}=\text{c}=3\times\frac{10\text{m}}{\text{s}}$

$\Rightarrow3\times10^8=40\times10^6\lambda$

$\Rightarrow\lambda=7.5\text{m}$

2. $\text{E}_\text{r}=\text{E}_0\hat{\text{i}}\cos(\text{kz}+\omega\text{t})$

Solution:

Key concept: When a wave is reflected from a denser medium or perfectly reflecting wall made with optically inactive material, then the type of wave doesn't change but only its phase changes by 180º or $\hat{\text{I}}€$ radian.

2. ​​​​$\frac{\pi}{2}$

Explanation:

Electromagnetic waves are formed when an electric field couples with a magnetic field. The magnetic and electric fields of an electromagnetic wave are perpendicular to each other and to the direction of the wave, as shown in figure.

2. perpendicular to the electric field $\overrightarrow{\text{S}}$

Explanation:

The electric field is always perpendicular to the magnetic field, and both fields are directed at right-angles to the direction of propagation of the wave. In fact, the wave propagates in the direction $\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$ Electromagnetic waves are clearly a type of transverse wave.

2. E_micro < E_infrared < E_visible < E_ultraviolet < E_gamma

Explanation:

The energy of electromagnetic waves is directly proportional to the frequency of the electromagnetic waves. So the order of frequency is given as:

v_micro < v_infrared < v_visible < v_ultraviolet < v_gamma

Since E = hv

$➔ \text{E} \propto\text{v}$

The order of energy is as follows:

E_micro < E_infrared < E_visible < E_ultraviolet < E_gamma

1. The associated magnetic field is given as $\text{B}=\frac{1}{\text{c}}\big(\text{E}_1\hat{\text{i}}+\text{E}_2\hat{\text{j}}\big)\cos(\text{kz}-\omega\text{t})$.

4. ​​​​​​​The given electromagnetic wave is plane polarised.

Solution:

We are given that the electric field vector of an electromagnetic wave travels in a vacuum along z-direction as,

$\vec{\text{E}}=\big(\text{E}_2\hat{\text{i}}+\text{E}_2\hat{\text{j}}\big)\cos(\text{kz}-\omega\text{t})$

The magnitude of the electric and the magnetic fields in an electronagnetic wave are related as

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

$\vec{\text{B}}=\frac{\vec{\text{E}}}{\text{c}}=\frac{\text{E}_1\text{i}+\text{E}_2\text{i}}{\text{c}}\cos(\text{kz}-\omega\text{t})$

Also, $\vec{\text{E}}$ and $\vec{\text{B}}$ are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to $\vec{\text{E}}$ as well as $\vec{\text{B}}$, so the given electromagnetic wave is plane polarized.

4. 60°

Explanation:

Given,

$\text{I}=\frac{\text{I}_\text{0}}{2}....(\text{i})$

$\text{I}=\text{I}\cos^2\theta$ $\Big(\because\text{I}=\frac{\text{I}_0}{8}\Big)$

$\therefore\frac{\text{I}_0}{8}=\frac{\text{I}_0}{2}\cos^2\theta$

From the equation (i), we have

$\frac{1}{4}=\cos^2\theta$

$\Rightarrow\cos\theta=\frac{1}{2}$

$\Rightarrow\cos\theta=\cos60^\circ$

$\Rightarrow\theta=60^\circ$

1. $\frac{0.166\times10^{-8}\text{N}}{\text{m}^2}$

Explanation:

Intensity or power per unit area of the radiations,

P = pv

$\Rightarrow\text{P}=\frac{\text{P}}{\text{v}}=\frac{0.5}{3\times10^8}=\frac{0.166\times10^{-8}\text{N}}{\text{m}^2}$

1. Will have frequency of 10⁹Hz.

3. Wil​​​l have a wavelength of 0.3 m.
4. Fall in the region of radio waves.

Solution:

Here we are given the frequency by which the charged particles oscillates about its mean equilibrium position, it is equal to 10⁹Hz. The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.

So, frequency of electromagnetic waves produced by the charged particle is v = 10⁹Hz.

Wavelength $\lambda=\frac{\text{c}}{\text{v}}=\frac{3\times10^8}{10^9}=0.3\text{m}$

The frequency of 10⁹Hz falls in the region of radiowaves.

4. $\sqrt{2}\text{E}.$

Solution:

We know the electric field intensity on a surface due to incident rediation is,

$\text{I}_\text{av}\propto\text{E}_0^2$

$\frac{\text{P}_\text{av}}{\text{A}}\propto\text{E}_0^2$

Here $\text{P}_\text{av}\propto\text{E}_0^2$ [$\because$ A is same in both cases]

We know that, $\text{E}_0\propto\sqrt{\text{P}_\text{av}}$

$\therefore\ \frac{(\text{E}_0)_1}{(\text{E}_0)_1}=\sqrt{\frac{(\text{P}_\text{av})_1}{(\text{P}_\text{av})_2}}\ .....(\text{i})$

$\Rightarrow\ \frac{\text{E}}{(\text{E}_0)_2}=\sqrt{\frac{1000}{5}}$

$(\text{E}_0)_2=\frac{\text{E}}{\sqrt{2}}$

Nowa according to question, P' = 50W, P = 100W

$\therefore$ Putting these value in Eq. (i), we get

$\frac{\text{E}'}{\text{E}}=\frac{50}{100}\Rightarrow\ \frac{\text{E}'}{\text{E}}=\frac{1}{2}\Rightarrow\ \text{E}'=\frac{\text{E}}{2}$

4. 3660cm

Explanation:

Given, frequency of EM waves

v = 8.196 × 10⁶Hz

velocity of EM waves (v) = 3 × 10⁸m/s

Wavelength of EM waves $\lambda=\frac{\text{v}}{\text{v}}$

$=\frac{3 × 10^8​}{8.196 × 10^6}$

= 36.60m

= 3660cm.