Physics 5 Marks Questions

1. Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
2. Radio waves; it belongs to the short wavelength end.
3. Temperature, T = 2.7 °K

$\lambda_\text{m}$ is given by Planck's law as:

$\lambda_\text{m}=\frac{0.29}{2.7}=0.11 \ \text{cm}$

This wavelength corresponds to microwaves.

4. This is the yellow light of the visible spectrum.
5. Transition energy is given by the relation,

E = hv

Where,

h = Planck's constant = 6.6 × 10^-34 Js

v = Frequency of radiation

Energy, E = 14.4 K eV 

$\therefore \ \text{v}=\frac{\text{E}}{\text{h}}$

$=\frac{14.4\times10^3\times1.6\times10^{-19}}{6.6\times10{-34}}$

= 3.4 × 10¹⁸ Hz

This corresponds to X-rays.

1. Wavelength of a wave is given as:

$\lambda=\frac{\text{c}}{\text{v}}$

$=\frac{3\times10^8}{2\times10^{10}}=0.015 \ \text{m}$

2. Magnetic field strength is given as:

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

$=\frac{48}{3\times10^{8}}=1.6\times10^{-7}\ \text{T}$

3. Energy density of the electric field is given as:

$\text{U}_E=\frac{1}{2}\in_0\text{E}^2$

And, energy density of the magnetic field is given as:

$\text{U}_\text{B}=\frac{1}{2\mu_0}\text{B}^2$

Where, $\in_0$ = Permittivity of free space

$\mu_0$ = Permeability of free space

We have the relation connecting E and B as:

E = cB … (1)

Where,

$\text{c}=\frac{1}{\sqrt{\in_0 \ \mu_0}}\dots(2)$

Putting equation (2) in equation (1), we get

$\text{E}=\frac{1}{\sqrt{\in_0 \ \mu_0}}\text{B}$

Squaring both sides, we get

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

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

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

$\Rightarrow \ \text{U}_\text{E}=\text{U}_\text{B}$

1. Magnitude of magnetic field strength is given as:

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

$=\frac{120}{3\times10^8}$

$=4\times10^{-7}\text{T}=400 \ \text{nT}$

Angular frequency of source is given as:

$\omega=2\text{nv}=2\text{n}\times50\times10^6$

$=3.14\times10^8 \ \text{rad}/ \text{m}$

Propagation constant is given as:

$\text{k}=\frac{\omega}{\text{c}}$

$=\frac{3.14\times10^8}{3\times10^8}=1.05 \ \text{red}/\text{m}$

Wavelength of wave is given as:

$\lambda=\frac{\text{c}}{\text{v}}$

$=\frac{3\times10^8}{50\times10^6}=6.0 \ \text{m}$

2. Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

$\vec{\text{E}}=\text{E}_0\sin(\text{kx}-\omega\text{t})\hat{\text{j}}$

$=120\sin[1.05\text{x}-3.14\times10^8\text{t}]\hat{\text{j}}$

And, magnetic field vector is given as:

$\vec{\text{B}}=\text{B}_0\sin(\text{kx}-\omega\text{t})\hat{\text{k}}$

$\vec{\text{B}}=\Big(4\times10^{-7}\Big)\sin\Big[1.05\text{x}-3.14\times10^8\text{t}\Big]\hat{\text{k}}$

1. In satellite communications, microwave is widely used. Hence λ₁, is the wavelength of microwave.
2. In water purifier, ultraviolet rays are used to kill germs. So, λ₂ is the wavelength of UV rays.
3. X-rays are used to detect leakage of oil in underground pipelines. So, λ₃  is the wavelength of X-rays.
4. Infrared rays are used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.

2. Wavelength of X-rays < wavelength of UV < wavelength of infrared < wavelength of microwave.

$\Rightarrow\ \lambda_3<\lambda_2<\lambda_4<\lambda_1$

3. Radiation	Uses
   $\gamma-\text{rays}$	Givens informations on nuclear structure, medical treatment etc.
   X-rays	Medical diagnosis and treatment study of crystal structure, inductrial radiograph.
   UV-rays	Preserce food, sterlizing the surgical instruments, detecting the invisible writings, finger prints etc.
   Visible light	To see objects.
   Infrated rays	To treat, muscular strain for taking photogtaphy during the fog, haze etc.
   Micro wave and radio wave	In radar and telecommunication.

1. Let electromagnetic wave is propagating along z-axis, in this case electric field vector $(\vec{\text{E}})$ be along x-axis magnetic field vector $(\vec{\text{B}})$ along y-axis,

i.e., $\vec{\text{E}}=\text{E}_0\hat{\text{i}}\text{ and }\vec{\text{B}}=\text{B}_0\hat{\text{j}}.$

The line integral of $\vec{\text{E}}$ over the closed rectangular path 1234 in x-z plane of the figure is

$\oint\vec{\text{E}}.\text{d}\vec{\text{l}}=\int_1^2\vec{\text{E}}.\text{d}\vec{\text{l}}+\int_2^3\vec{\text{E}}.\text{d}\vec{\text{l}}+\int_3^4\vec{\text{E}}.\text{d}\vec{\text{l}}+\int_4^1\vec{\text{E}}.\text{d}\vec{\text{l}}$

$=\int_1^2{\text{E}}.\text{d}{\text{l}}\cos90^\circ+\int_2^3{\text{E}}.\text{d}{\text{l}}\cos0^\circ+\int_3^4{\text{E}}.\text{d}{\text{l}}\cos90^\circ+\int_4^1{\text{E}}.\text{d}{\text{l}}\cos180^\circ$

$\oint\vec{\text{E}}.\text{d}\vec{\text{l}}=\text{E}_0\text{h}[\sin(\text{kz}_2-\omega\text{t})-\sin(\text{kz}_1-\omega\text{t})\ .....(\text{i})$

2. Now let us evaluate $\int\vec{\text{B}}.\text{d}\vec{\text{s}}$, let us consider the rectangle 1234 to be made strips of are ds = hdz each.

$\int\vec{\text{B}}.\text{d}\vec{\text{s}}=\int\text{B.ds}\cos0=\int\text{b.ds}$

$=\int_{\text{Z}_1}^{\text{Z}_2}\text{B}_0\sin(\text{kz}-\omega\text{t})\text{hdz}$

$\int\vec{\text{B}}.\text{d}\vec{\text{s}}=\frac{-\text{B}_0\text{h}}{\text{k}}[\cos(\text{kz}_2-\omega\text{t})-\cos(\text{kz}_1-\omega\text{t})\ .....(\text{ii})$

3. We are given $\oint\text{E.dl}=\frac{-\text{d}\phi_\text{B}}{\text{dt}}=-\frac{\text{d}}{\text{dt}}\oint\text{B.ds}$

Substituting the values from Eqs. (i) and (ii), we get

$\text{E}_0\text{h}[\sin(\text{kz}_2-\omega\text{t})-\sin(\text{kz}_1-\omega\text{t})]$

$=\frac{-\text{d}}{\text{dt}}\bigg[\frac{\text{B}_0\text{h}}{\text{k}}\{\cos(\text{kz}_2-\omega\text{t})-\cos(\text{kz}_1-\omega\text{t})\}\bigg]$

$=\frac{\text{B}_0\text{h}}{\text{k}}\omega[\sin(\text{kz}_2-\omega\text{t})-\sin(\text{kz}_1-\omega\text{t})]$

$\Rightarrow\ \text{E}_0=\frac{\text{B}_0\omega}{\text{k}}=\text{B}_0\text{c} \ \ \Big(\because\ \frac{\omega}{\text{k}}=\text{c}\Big)$

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

4. For evaluating $\oint\vec{\text{B}}.\text{d}\vec{\text{l}}$, let us consider a loop 1234 in y-z plane as shown in figure given below.

$\oint\vec{\text{B}}.\text{d}\vec{\text{l}}=\int_1^2\vec{\text{B}}.\text{d}\vec{\text{l}}+\int_2^3\vec{\text{B}}.\text{d}\vec{\text{l}}+\int_3^4\vec{\text{B}}.\text{d}\vec{\text{l}}+\int_4^1\vec{\text{B}}.\text{d}\vec{\text{l}}$

$=\int_1^2{\text{E}}.\text{d}{\text{l}}\cos0^\circ+\int_2^3{\text{E}}.\text{d}{\text{l}}\cos90^\circ+\int_3^4{\text{E}}.\text{d}{\text{l}}\cos180^\circ+\int_4^1{\text{E}}.\text{d}{\text{l}}\cos90^\circ$

$\oint\vec{\text{B}}.\text{d}\vec{\text{l}}=\text{B}_0\text{h}[\sin(\text{kz}_2-\omega\text{t})-\sin(\text{kz}_1-\omega\text{t})\ .....(\text{iii})$

Now to evaluate $\phi_\text{E}=\int\vec{\text{B}}.\text{d}\vec{\text{s}}$, let us consider the rectangle 1234 to be made of strips of area hds reach.

$\phi_\text{E}=\int\vec{\text{E}}.\text{d}\vec{\text{s}}=\int\text{Eds}\cos0=\int\text{Eds}$

$=\int_{\text{z}_1}^\text{z}\text{E}_0\sin(\text{kz}_1-\omega\text{t})\text{hdz}$

$\oint\vec{\text{E}}.\text{d}\vec{\text{s}}=-\frac{\text{E}_0\text{h}}{\text{k}}[\cos(\text{kz}_2-\omega\text{t})-\cos(\text{kz}_2-\omega\text{t})]$

$\therefore\ \frac{\text{d}\phi_\text{E}}{\text{dt}}=-\frac{\text{E}_0\text{h}\omega}{\text{k}}[\sin(\text{kz}_2-\omega\text{t})-\sin(\text{kz}_2-\omega\text{t})] \ .....(\text{iv})$

Let $\oint\text{B.dl}=\mu_0\Big(\text{I}+\frac{\in_0\text{d}\phi_\text{E}}{\text{dt}}\Big)$ where I = conduction current

= 0 in vacuum

$\therefore\ \oint\text{B.dl}=\mu_0\in\frac{\text{d}\phi_\text{E}}{\text{dt}}$

Using relations obtainned in Eqs. (iii) and (iv) and simplifying, we get

$\text{B}_0=\text{E}_0\frac{\omega\mu_0\in_0}{\text{k}}$

$\Rightarrow\ \frac{\text{E}_0}{\text{B}_0}\frac{\omega}{\text{k}}=\frac{1}{\mu_0\in_0}$

But $\frac{\text{E}_0}{\text{B}_0}=\text{c}\text{ and }\omega=\text{ck}\Rightarrow\ \text{c.c}=\frac{1}{\mu_0\in_0}$

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

1. We are given, the induced electric field at a distance r from the wire inside the cable.

$\vec{\text{E}}(\text{s},\text{t})=\mu_0\text{I}_0\text{v}\cos(2\pi\text{vt})\text{In}\Big(\frac{\text{s}}{\text{a}}\Big)\hat{\text{k}}$

The displacement current density is given by $\vec{\text{J}}_\text{d}=\in_0\frac{\text{d}\vec{\text{E}}}{\text{dt}}$

$=\in_0\frac{\text{d}}{\text{dt}}\bigg[\mu_0\text{I}_0\text{v}\cos(2\pi\text{vt})\text{In}\Big(\frac{\text{s}}{\text{a}}\Big)\hat{\text{k}}\bigg]$

$=\in_0\mu_0\text{I}_0\text{v}\frac{\text{d}}{\text{dt}}[\cos2\text{vt}]\text{In}\Big(\frac{\text{s}}{\text{a}}\Big)\hat{\text{k}}$

$=\frac{1}{\text{c}}^2\text{I}_0\text{v}^22\pi[-\sin2\pi\text{vt}]\text{In}\Big(\frac{\text{s}}{\text{a}}\Big)\hat{\text{k}}$

$=\frac{\text{v}^2}{\text{c}^2}2\pi\text{I}_0\sin2\pi\text{vt}\text{In}\Big(\frac{\text{a}}{\text{s}}\Big)\hat{\text{k}}$

$=\frac{\text{I}}{\lambda^2}2\pi\text{I}_0\text{In}\Big(\frac{\text{a}}{\text{s}}\Big)\sin2\pi\text{vt}\hat{\text{k}}$

$\Rightarrow\ \vec{\text{J}}_\text{d}=\frac{2\pi\text{I}_0}{\lambda^2}\text{In}\frac{\text{a}}{\text{s}}\sin2\pi\text{vt}\hat{\text{k}}$

2. Total displacement current, $\text{I}^\text{d}=\int\text{J}_\text{d}2\pi\text{sds}$

$\text{I}^\text{d}=\int\limits_0^\text{a}\bigg(\frac{2\pi\text{I}_0}{\lambda^2}\text{In}\frac{\text{a}}{\text{s}}\sin2\pi\text{vt}\bigg)2\pi\text{sds}$

$=\int\limits_0^\text{a}\bigg[\frac{2\pi}{\lambda^2}\text{I}_0\int^\text{a}_{\text{s}=0}\text{In}\Big(\frac{\text{a}}{\text{s}}\Big)\text{sds}\sin2\pi\text{vt}\bigg]=2\pi$

$=\Big(\frac{2\pi}{\lambda}\Big)^2\text{I}_0\int_0^\text{a}\Big(\frac{\text{a}}{\text{s}}\Big)\frac{1}{2}\text{d}(\text{s}^2)\sin2\pi\text{vt}$

$=\Big(\frac{\text{a}}{2}\Big)^2\Big(\frac{2\pi}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}\int_0^\text{a}\text{In}\Big(\frac{\text{a}}{\text{s}}\Big).\text{d}\Big(\frac{\text{s}}{\text{a}}\Big)^2$

$=\frac{\text{a}^2}{4}\Big(\frac{2\pi}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}\int_0^\text{a}\text{In}\Big(\frac{\text{a}}{\text{s}}\Big)^2.\text{d}\Big(\frac{\text{s}}{\text{a}}\Big)^2$

$=\frac{\text{a}^2}{4}\Big(\frac{2\pi}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}\times(1) \ \ \bigg[\because\ \int_{\text{s}=0}^\text{a}\text{In}\Big(\frac{\text{s}}{\text{a}}\Big)^2\text{d}\Big(\frac{\text{s}}{\text{a}}\Big)^2=1\bigg]$

$\therefore\ \text{I}^\text{d}=\frac{\text{a}^2}{4}\Big(\frac{2\pi}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}$

$\Rightarrow\ \text{I}^\text{d}=\Big(\frac{\pi\text{a}}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}$

3. The displacement current,

$\text{I}_\text{d}=\Big(\frac{\pi\text{a}}{\lambda}\Big)^2\text{I}_0\sin2\pi\text{vt}=\text{I}_0^\text{d}\sin2\pi\text{vt}$

Here, $\text{I}_0^\text{d}=\Big(\frac{\text{a}\pi}{\lambda}\Big)^2\text{I}_0$

$\Rightarrow\ \frac{\text{I}_0^\text{d}}{\text{I}_0}=\Big(\frac{\text{a}\pi}{\lambda}\Big)^2$

1. Total energy carried by electromagnetic wave is due to electric field vector and magnetic field vector. In electromagnetic wave, E and B vary from point to point and from moment to moment.

The energy density due to electric field E is

$\text{u}_\text{E}=\frac{1}{2}\in_0\text{E}^2$

The enerrgy density due to magnetic field B is

$\text{u}_\text{B}=\frac{1}{2}\frac{\text{B}^2}{\mu_0}$

Total energy density of electromagnetic wave

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

Let the EM wave be propagatine along z-direction. The electric field moment. Hence, the effective values of E² and B² are their time averages over complete cycle.

We know, $\langle\sin^2\theta\rangle=\frac{\int\limits_0^{2\pi}\sin^2\theta\text{d}\theta}{2\pi}=\frac{1}{2}$

and $\langle\cos^2\theta\rangle=\frac{\int\limits_0^{2\pi}\cos^2\theta\text{d}\theta}{2\pi}=\frac{1}{2}$

Hence, the time average value of E² over complete cycle,

$\langle\text{E}^2\rangle=\frac{\int\limits_0^\text{T}[\text{E}_0\sin(\text{kz}-\omega\text{t})]^2\text{dt}}{\text{T}}=\frac{\text{E}_0^2}{2}$

And, the time average value of B² over complere cycle,

$\langle\text{B}^2\rangle=\frac{\int\limits_0^\text{T}[\text{B}_0\sin(\text{kz}-\omega\text{t})]^2\text{dt}}{\text{T}}=\frac{\text{B}_0^2}{2}$

The time average of energy density over complere cycle

$\text{u}_\text{av}=\frac{1}{2}\frac{\in_0\text{E}_0^2}{2}+\frac{1}{2}\mu_0\Big(\frac{\text{B}_0^2}{2}\Big)$

$\Rightarrow\ \text{u}_\text{av}=\frac{1}{4}\in_0\text{E}_0^2+\frac{1}{4}\frac{\text{B}_0^2}{\mu_0}$

2. We know $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{\text{E}_0}{\text{B}_0}$

where $\mu_0$ = Absolute permeability, $\in_0$ = Absolute permittivity, E₀ and B₀ = Amplitudes of electric field and magnetic field vectors.

The time average of energy density due to magnetic field B is

$\text{u}_\text{B}=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}=\frac{1}{2}\frac{\Big(\frac{\text{E}_0^2}{\text{c}^2}\Big)}{\mu_0}$

$=\frac{\text{E}_0^2}{4\mu_0}\times\mu_0\in_0=\frac{1}{4}\in_0\text{E}_0^2$

Hence u_B = u_E; the time average of energy density due to magnetic field is equal to the time average of energy density due to electric field.

$\Rightarrow\ \text{u}_\text{av}=\frac{1}{4}\in_0\text{E}_0^2+\frac{1}{4}\frac{\text{B}_0^2}{\mu_0}$

$=\frac{1}{4}\in_0\text{E}_0^2+=\frac{1}{4}\in_0\text{E}_0^2$

$=\frac{1}{4}\in_0\text{E}_0^2=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}$

Time average intensity of the wave

$\text{I}_\text{av}=\text{u}_\text{av}\text{c}=\Big(\frac{1}{2}\in_0\text{E}_0^2\Big)\text{c}=\frac{1}{2}\text{c}\in_0\text{E}_0^2$

$\Rightarrow\ \text{S}_\text{av}=\frac{\text{c}^2}{2}\in_0\text{E}_0\Big(\frac{\text{E}_0}{\text{c}}\Big) \ \ \bigg[\text{As c}=\frac{\text{E}_0}{\text{B}_0}\bigg]$