Here, \({ B }_{ 0 }=300\mu T=3\times { 10 }^{ -4 }T\)
(i) Maximum value of electric field, \({ E }_{ 0 }=c{ B }_{ 0 }\)
\(\therefore { E }_{ 0 }=\left( 3\times { 10 }^{ 8 } \right) \times \left( 3\times { 10 }^{ -4 } \right) =9\times { 10 }^{ 4 }V{ m }^{ -1 }\)
(ii) Average energy density corresponding to electric field is
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 9\times { 10 }^{ 4 } \right) ^{ 2 }\)
\(=19.91\times { 10 }^{ -4 }=1.99\times { 10 }^{ -3 }J{ m }^{ -3 }\)

Total energy falling on the surface,
U = 18 x 20 x 30 x 60 J = 6.48 x 10⁵ J
Total momentum delivered to the surface is
\(p=\frac { U }{ c } =\frac { 6.48\times { 10 }^{ 5 } }{ 3\times { 10 }^{ 8 } } =2.16\times { 10 }^{ -3 }kg{ ms }^{ -1 }\)
The average force exerted on the surface is
\(F=\frac { p }{ t } =\frac { 2.16\times { 10 }^{ -3 } }{ 30\times 60 } =1.2\times { 10 }^{ -6 }N\)
It the surface is a perfect reflector, the change of momentum will be = p - (- p)
= 2 p = 2 x 2.16 x 10^-3 kg ms^-1
Now average force,
\(F=\frac { 2\times 2.16\times { 10 }^{ -3 } }{ 30\times 60 } =2.4\times { 10 }^{ -6 }N\)

Given :\({ I }_{ D }=2A,\ C=3 \ \mu F=3\times { 10 }^{ -6 }F\) 
The displacement current is given by
\({ I }_{ D }=\varepsilon _{ 0 }\frac { d\phi _{ e } }{ dt } \) 
\(=\varepsilon _{ 0 }\frac { d }{ dt } (EA)\)
\( =\varepsilon _{ 0 }\frac { d }{ dt } \left( \frac { V }{ d } A \right) \)
\( =\frac { \varepsilon _{ 0 }A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(\therefore \frac { dV }{ dt } =\frac { { I }_{ D } }{ C } =\frac { 2 }{ 3\times 10^{ -6 } } \) 
\(=6.67\times 15^{ 5 }Vs^{ -1 }\)
Thus to set up an instantaneous displacement current of 2.0 A between the plates of the current, the potential difference across the plates should be changed at the rate \(6.67\times 10^{ 5 }Vs^{ -1 }\) 

Consider an electric charge at rest so that at a point P some distance away, we have electric field but no magnetic field. Let, at time \(t=0\) , an impulse be given to the charge such that it starts moving with some finite velocity. For a moving charge, we expect at P both electric and magnetic fields, but we cannot immediately decide whether the magnetic field at P will change from zero to finite value instantaneously at \(t=0\) or after some time.
Instantaneous change means infinite rate of change. If the change is instantaneous at all points then considering any loop, we will conclude from Faraday's law that an infinite e.m.f. and infinite electric field is set up. This in turn would imply an infinite magnetic field as seen from the result. Fields are always finite away from charges and clearly the situation just described is inconsistent with known laws of electricity and magnetism.
\(\oint { \overset { \rightarrow }{ B } } .\overset { \rightarrow }{ dl } ={ \mu }_{ 0 }{ \varepsilon }_{ 0 }\frac { d\phi _{ e } }{ dt } \)
The moving charge sets up a magnetic field in its neighbourhood which in turn creates an electric field in the neighbourhood. The process continues since both time-varying electric and magnetic fields act as sources of each other. Thus an electromagnetic wave is started when a charge is accelerated. It is only when the wave reaches the point P that the magnetic field at P changes.
This shows that an accelerated charge emits an electromagnetic wave. It can also be shown that the electromagnetic wave and the oscillator will have the same frequency.

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

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

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

(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 \({ U }_{ E }={ U }_{ B }\).

During charging or discharging of a capacitor. Increasing or decreasing current in a circuit with time flows due to conduction current in wire and displacement current between the plates of a capacitor. when capacitor get fully charged both conduction and displacement current becomes zero. That is why galvanometer shows a momentary. deflection at the time of charging or discharging.
The expression to explain this observation is \(\oint { \overset { \rightarrow }{ B } .\overset { \rightarrow }{ dt } } ={ \mu }_{ 0 }(I+{ I }_{ D })\)

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

Maxwell, Hertz, Bose and Marconi

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

\(v=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}=\frac{c}{\sqrt{\mu_r \varepsilon_r}}\)
\(\left[\therefore c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\right]\)

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