The electric and magnetic field vectors \(\overset { \rightarrow }{ E } \) and \(\overset { \rightarrow }{ B } \) must be perpendicular to each other a well as perpendicular to the direction of propagation of e.m.waves. Thus \(\overset { \rightarrow }{ E } \) and \(\overset { \rightarrow }{ B } \) must lie in x-y plane and are mutually perpendicular to each other
Wavelength, \(\lambda =c/v=3\times { 10 }^{ 8 }/30\times { 10 }^{ 6 }=10 \ m\)

\(Here,{ N }_{ 1 }=400, \ { N }_{ 2 }=700\)
\( I_{ 1 }=3.5A \ \phi =300 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\({ \phi }_{ 2 }=900 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\(M=\frac { { N }_{ 2 }{ \phi }_{ 2 } }{ I_{ 1 } } =\frac { 700 \times 90 \times { 10 }^{ -6 } }{ 3.5 } =1.8 \times { 10 }^{ -2 }H\)
\({ L }_{ 1 }=\frac { { N }_{ 1 }{ \phi }_{ 1 } }{ I_{ 1 } } =\frac { 400 \times 300 \times { 10 }^{ -6 } }{ 3.5 } =3.43 \times { 10 }^{ -2 }H\)
\({ e }_{ 2 }=M(\frac { dI_{ 1 } }{ dt } )=1.8 \times { 10 }^{ -2 } \times (0.5)\)
\(=9 \times { 10 }^{ -3 }V\)

\(Here,l=0.3m, \ A=1.2 \times { 10 }^{ -3 }{ m }^{ 2 },{ N }_{ 1 }=2500\)
\( { N }_{ 2 }=350,e=?,dI=-3-(3)=-6A,\)
\(dt=0.25s\)
\(e=-M\frac { dI }{ dt } =-\frac { { \mu }_{ 0 }{ N }_{ 1 }{ N }_{ 2 }A }{ l } (\frac { dI }{ dt } )\)
\(=\frac { -4\pi \times { 10 }^{ -7 } \times 2500 \times 350 \times 1.2 \times { 10 }^{ -3 }(-6) }{ 0.3 \times 0.25 } \\ dt=0.25s\)
\( e=-M\frac { dI }{ dt } =-\frac { { \mu }_{ 0 }{ N }_{ 1 }{ N }_{ 2 }A }{ l } (\frac { dI }{ dt } )\)
\( =\frac { -4\pi \times { 10 }^{ -7 } \times 2500 \times 350 \times 1.2 \times { 10 }^{ -3 }(-6) }{ 0.3 \times 0.25 } \)
\(e=0.106 \ V\)

\(7.896 \times { 10 }^{ -5 }H;19.74mV\)

\({ I }_{ D }={ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { EA }{ t } =\frac { { \epsilon }_{ 0 }(V/d)\times A }{ t }\)
\( =\frac { { \epsilon }_{ 0 }VA }{ td } =\frac { 8.85\times { 10 }^{ -12 }\times 400\times (60\times { 10 }^{ -4 } }{ } \)
\( =1.062\times { 10 }^{ -2 }\)

The total energy falling on the surface is
\(U=(18W/{ cm }^{ 2 })\times (20{ cm }^{ 2 })\times (30\times 60)\)
\( =6.48\times 10^{ 5 }J\)
Total momentum delivered (for complete absorption) is
\(P=\frac { U }{ c } =\frac { 6.48\times { 10 }^{ 5 }J }{ 3\times { 10 }^{ 8 }m/s } =2.16\times { 10 }^{ -3 } \ kgm/s\) 
The average force exerted on the surface is
\(F=\frac { P }{ t } =\frac { 2.16\times 10^{ -3 } }{ 0.18\times { 10 }^{ 4 } } =1.2\times 10^{ -6 } \ N\) 
If the surface is perfect reflector, the change in momentum 
\(=p-(-p)=2p=2+2.16\times { 10 }^{ -3 } \ kg \ ms^{ -1 }\) 
And average force
 \(F=\frac { 2p }{ t } =\frac { 2\times 2.16\times { 10 }^{ -3 } }{ 30\times 60 } \)
\(=2.4\times 10^{ -6 }N\)

Uses of ultraviolet rays:
(i) For checking mineral samples by making use of its property of causing fluorescence and also used for study of molecular structure.
(ii) For sterilizing the surgical instruments because U.V. rays destroy bacteria.
a. in food preservation.
b. in burglars alarm.
c. in the detection of forged documents, fingerprints in forensic laboratory.
(b) Values displayed by Rohan-Curiosity and sharing knowledge.

Given \(v=40\times { 10 }^{ 6 }Hz\) 
\(\therefore \)\(T=\frac { 1 }{ v } =\frac { 1 }{ 40\times 10^{ -6 } } =0.25\times 10^{ -6 }/s\) 
Magnetic field
\({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 750 }{ 3\times 10^{ 18 } } =2.5\times 10^{ -6 }T \ along \ (z \ direction)\) 
And angular frequency 
\(\omega =2\pi v=2\times \pi \times 40\times { 10 }^{ 6 }=8\pi \times { 10 }^{ 7 } \ Hz\)​​​​​

Magnetic field between the plates of a capacitor at distance \(r\) having varying electric field is given by 
\(B=\frac { \mu _{ 0 } }{ 4\pi } \frac { 2I_{ D } }{ r } =\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \frac { d\phi }{ dt }\)
\(=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \frac { d }{ dt } (E\pi r^{ 2 })\)
\( B=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \pi r^{ 2 }\left[ \frac { dE }{ dt } \right] =\frac { \mu _{ 0 }\varepsilon _{ 0 }r }{ 2 } \frac { dE }{ dt } \)
\(=\frac { r }{ { 2C }^{ 2 } } \frac { dE }{ dt } =\frac { 1 }{ 2\times 9\times 10^{ 16 } } \times { 10 }^{ 10 }=5.6\times 10^{ -8 }T\)

\({ I }_{ av }=\frac { c }{ 2 } \frac { { B }^{ 2 }_{ 0 } }{ { \mu ^{ 2 } }_{ 0 } } =\frac { 3\times 10^{ 8 }\times (12\times 10^{ 8 })^{ 2 } }{ 2\times 4\pi \times 10^{ -7 } } =1.7Wm^{ -2 }\)

There is no contradiction. Field lines inside the bar go away from S towards N. The next flux of B over any surface fully enclosing N or S must be identically zero. 

(a)  \(\lambda =\frac { c }{ v } =\frac { 3\times 10^{ 8 } }{ 2.0\times 10^{ 10 } } \)
\( =1.5\times 10^{ -2 }m\)
\(E=48Vm^{ -1 }\)
(b) \({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 48 }{ 3\times { 10 }^{ 8 } }\)
or  \({ B }_{ 0 }=1.6\times 10^{ -7 } \ T\)
(c) Energy density in E field, 
\({ U }_{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }\)
Energy density in B field, 
\({ U }_{ B }=\frac { 1 }{ 2\mu _{ 0 } } B^{ 2 }\)
Using \(E=cB\) and \(c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } ,\)
We find \({ U }_{ E }={ U }_{ B }\).

(i) x-rays have the wavelength range,\({ 10 }^{ 13 } \ m \ to \ 3\times { 10 }^{ -8 }m\) which is smaller than that of ultraviolet rays of a wavelength range \(6\times { 10 }^{ -9 }m \ to \ 4\times { 10 }^{ -7 }m, \ i.e., \ { \lambda }_{ x }<{ \lambda }_{ uv }\)
since energy. \(E=\frac { hc }{ \lambda } \ or \ E\propto \frac { 1 }{ \lambda } ; \ so \ \frac { { E }_{ x } }{ { E }_{ uv } } =\frac { { \lambda }_{ uv } }{ \lambda _{ x } } >1 \ or \ { E }_{ x }>{ E }_{ uv }\) Thus x-rays are more energetic than ultraviolet rays
(ii) The wavelength of radio waves is of range. \(0.3m \ to \ 6\times { 10 }^{ 2 }m\) which is very large as compared to the size of molecules of our blood, flesh and bones etc. As the energy of radio waves is quite small so these radio waves can not penetrate the blood of our body. That is why we can not use the radio waves to detect the fracture in bones
(iii) Just as different em waves have different application depending on their wavelength, in the same way, every person has different qualities, which make him suitable for different purpose/fields.
The aim of a teacher or manager is to is identify the quality in different children/persons and put them to their best use accordingly.

(i) The invisible em waves whose wavelength is lower than the smallest wavelength of visible light are ultraviolet radiation, x-rays and gamma rays
(ii) The white light electromagnetic wave. the speed of em wave in vacuum is given by 
\(c=\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \epsilon }_{ 0 } } } ,\ where \ { \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }T{ mA }^{ -1 } \ and \ { \epsilon }_{ 0 }=\frac { 1 }{ 4\pi \times 9\times { 10 }^{ 9 } } { C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }\)
\( \therefore \quad c=\frac { 1 }{ \sqrt { (4\pi \times { 10 }^{ -7 })\frac { 1 }{ (4\pi \times 9\times { 10 }^{ 9 }) } } } =3\times { 10 }^{ 8 }m{ s }^{ -1 }\)
(iii) From the above study, we find that nature/god has created all human beings alike. when they are exposed to different environments of the world and family. their thinking/views/speeds become different. God wants us to live in peace treating everyone as equal.

(i) Given induced electric field
\(\overset { \rightarrow }{ E } ={ \mu }_{ 0 }{ I }_{ 0 }v \ cos \ (2\pi vt) \ ln \ \frac { s }{ a } \overset { \wedge }{ k } \) 
So displacement current density
\(\overset { \rightarrow }{ J_{ d } } ={ \varepsilon }_{ 0 }\frac { d\overset { \rightarrow }{ E } }{ dt } ={ \mu }_{ 0 }{ \varepsilon }_{ 0 }{ I }_{ 0 } \ v \ ln \ \left( \frac { s }{ a } \right) \overset { \wedge }{ k } .\frac { d }{ dt } \ (cos \ 2\pi vt)\) 
\(=\frac { { I }_{ 0 }2\pi v^{ 2 } }{ { c }^{ 2 } } \ (-sin2\pi vt) \ ln \ \left( \frac { s }{ a } \right) \overset { \wedge }{ k }\)
\( = \ \left( \frac { v }{ c } \right) ^{ 2 }2\pi { I }_{ 0 } \ (sin \ 2\pi vt) \ l n \left( \frac { a }{ s } \right) \overset { \wedge }{ k }\)
\( \left[ \therefore \frac { s }{ a } =-ln\frac { a }{ s } \right] \)
or \(\overset { \rightarrow }{ J_{ d } } =\frac { 1 }{ { \lambda }^{ 2 } } 2\pi { I }_{ 0 } \ ln\frac { a }{ s } sin \ 2\pi vt \ \overset { \wedge }{ k }\) 
\( =\frac { 2\pi }{ { \lambda }^{ 2 } } { I }_{ 0 } \ ln \ \frac { a }{ s } \ sin \ (2\pi vt)\overset { \wedge }{ k } \)
(ii) \({ I }_{ d }=\int _{ s=0 }^{ a }{ { J }_{ D } } s \ ds\int _{ 0 }^{ 2\pi }{ d\theta =\int _{ 0 }^{ a }{ { J }_{ { D }^{ S } }ds.2\pi } } \)

\(=2\pi \int _{ 0 }^{ a }{ \frac { 2 }{ \lambda ^{ 2 } } } { I }_{ 0 }sin(2\pi vt) \ ln\frac { a }{ s } .sds\) 
\(=\left( \frac { 2\pi }{ \lambda } \right) ^{ 2 }{ I^{ 2 } }_{ 0 }sin(2\pi vt)\int _{ 0 }^{ a }{ { a }^{ 2 } } ln\left( \frac { a }{ s } \right) .\frac { 1 }{ 2 } \frac { d }{ ds } \left( \frac { s^{ 2 } }{ a^{ 2 } } \right) \) 
\(=-\left( \frac { 2\pi }{ \lambda } \right) ^{ 2 }{ I }_{ 0 }\frac { { a }^{ 2 } }{ 4 } sin(2\pi vt)\int _{ 0 }^{ a }{ ln } \left( \frac { s }{ a } \right) ^{ 2 }\frac { d }{ ds } \left( \frac { s }{ a } \right) ^{ 2 }\) 
\(=\frac { { a }^{ 2 } }{ 4 } \left( \frac { 2\pi }{ \lambda } \right) ^{ 2 }{ I }_{ 0 }sin2\pi vt\times (-1)\) 
\(\left[ \therefore \int _{ 0 }^{ a }{ ln } \left( \frac { s }{ a } \right) ^{ 2 }\frac { d }{ ds } \left( \frac { s }{ a } \right) ^{ 2 }=-1 \right] \)  
or \({ I }_{ d }=\left( \frac { 2\pi a }{ 2\lambda } \right) ^{ 2 }{ I }_{ 0 } \ sin \ 2\pi vt\) 
(iii) Displacement current
\({ I }_{ d }=\left( \frac { 2\pi a }{ 2\lambda } \right) ^{ 2 }{ I }_{ 0 } \ sin \ 2\pi vt={ I }_{ od } \ sin \ 2\pi vt\) 
where
\({ I }_{ 0d }=\left( \frac { 2\pi a }{ 2\lambda } \right) ^{ 2 }{ I }_{ 0 }=\left( \frac { \pi a }{ \lambda } \right) ^{ 2 }{ I }_{ 0 }\)
\( \frac { { I }_{ od } }{ I_{ 0 } } =\left( \frac { \pi a }{ \lambda } \right) ^{ 2 }\)

(i) The given equation signifies that the electromagnetic wave is moving along Y-axis and also in negative direction, so it moves in - \(\hat{j}\) direction.
(ii) The electric part of electromagnetic wave in vacuum.
E = [3.1 cos{1.8 y + (5.4 \(\times\)10⁶ t)}] \(\hat{i}\)
Comparing with standard equation,
E = E₀ cos (ky + \(\omega\)t), we get
Angular frequency, \(\omega\) = 5.4 \(\times\)10⁶ rad/s
Wave number, k = 1.8 rad/m
The amplitude of the electric field part of the wave,
E₀ = 3.1 N/C
\(\begin{array}{rlrl} \lambda & =\frac{2 \pi}{k}=\frac{2 \pi}{1.8}=3.491 \mathrm{~m} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \lambda =3.5 \mathrm{~m} \end{array}\)
(iii) Angular frequency, \(\omega\) = 2\(\pi\)v
\(v=\frac{\omega}{2 \pi}=\frac{5.4 \times 10^6 \times 7}{2 \times 22}\)
= 0.86 \(\times\)10⁶ Hz
(iv) As, \(c=\frac{E_0}{B_0}\)
Amplitude of magnetic field,
\(\begin{aligned} B_0 & =\frac{E_0}{c}=\frac{3.1}{3 \times 10^8} \end{aligned}\)
\(\begin{aligned} =1.03 \times 10^{-8} \mathrm{~T} \end{aligned}\)
(v) Expression for the magnetic field part of wave,
B = B₀ cos (ky + \(\omega\)t) \(\hat{k}\)
B = 1.03 \(\times\)10^-8 cos (1.8 y + 5.4 \(\times\)10⁶ t) \(\hat{k}\)

\(3.3\times { 10 }^{ -9 }N\)

Eye is most sensitive to the light of wavelength \(\lambda=5600 \dot{A}\)
\( T=\frac{1}{v}=\frac{\lambda}{c}=\frac{5600 \times 10^{-10}}{3 \times 10^8} \\ =1.87 \times 10^{-15} \mathrm{~s} \)

Radio waves,  microwaves, UV rays, X-rays,

Ratio = 1; because of the velocity of both the wave. length in a vacuum is same \(\left(=3 \times 10^8 \mathrm{~ms}^{-1}\right)\)

Pressure exerted by reflected wave on the surface is 
\(P=\frac{2 l}{c}=\frac{2 \times 1}{3 \times 10^8}=6.67 \times 10^{-9} \mathrm{~N} / \mathrm{m}^2\)

Time taken by laser beam to move through a distance 90 cm is
\(t=\frac{90}{c}=\frac{90 \mathrm{~cm}}{3 \times 10^{10} \mathrm{~cm} / \mathrm{s}}=3 \times 10^9 \mathrm{~s}\)
The energy contained in 90 cm length of laser beam is
\( U=p t p(10 \mathrm{~mW}) \times\left(3 \times 10^{-9} \mathrm{~s}\right) \\ =\left(10 \times 10^{-3} \mathrm{Js}^{-1}\right) \times\left(3 \times 10^{-9} \mathrm{~s}\right) \\ =30 \times 10^{-12} \mathrm{~J} \)

The basic source of an electromagnetic wave is the time varying electric field produces magnetic field and vice versa

Not necessarily, A displacement current such as that between the plates of a charging capacitor) can also produce loops of \(\vec B\)

\(\text { (i) } \oint_s \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\epsilon_0}\)
\(\text { (ii) } \oint \vec{E} \cdot \overrightarrow{d t}=0\)