Here, R = 10 cm = 0.10 cm;
C = 200 pF = 200 x 10^-12 F, 
\(\omega\)=400 rad s^-1 , V_rms = 230 V.
(i) \({ I }_{ rms }=\frac { { V }_{ rms } }{ { X }_{ C } } =\frac { { V }_{ rms } }{ 1/\omega C } ={ V }_{ rms }\times \omega C\)
= 230 x 400 x (200 x 10^-12)
= 18.4 x 10^-6 A = 18.4 \(\mu A\)
(ii) Magnetic field at a distance r from the axis of plates is
\(B=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } I\)   [See Solved Example 5]
Amplitude of \(\overrightarrow { B } \) is given by
\({ B }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ rms }\sqrt { 2 } \)
\(\left[ \because { I }_{ rms }={ I }_{ 0 }/\sqrt { 2 } \right] \)
Here, r = 2.0 cm = 2.0 x 10^-2 m ; R = 0.10 m.
\(\therefore { B }_{ 0 }=\frac { \left( 4\pi \times { 10 }^{ -7 } \right) }{ 2\pi } \times \frac { 2.0\times { 10 }^{ -2 } }{ { \left( 0.10 \right) }^{ 2 } } \times \left( 18.4\times { 10 }^{ -6 } \right) \)
= 1.04 x 10^-11

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.

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

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

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 
\(B=\frac { \mu _{ 0 } }{ 4\pi } \frac { 2I_{ D } }{ r } =\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \frac { d\phi }{ dt }\)
\(=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \frac { d }{ dt } (E\pi r^{ 2 })\)
\( B=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \pi r^{ 2 }\left[ \frac { dE }{ dt } \right] =\frac { \mu _{ 0 }\varepsilon _{ 0 }r }{ 2 } \frac { dE }{ dt } \)
\(=\frac { r }{ { 2C }^{ 2 } } \frac { dE }{ dt } =\frac { 1 }{ 2\times 9\times 10^{ 16 } } \times { 10 }^{ 10 }=5.6\times 10^{ -8 }T\)

(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) Infrared rays are used in physical therapy i.e., to treat muscular strain
(ii) Gamma rays are used in the treatment of cancer and tumours
(iii) microwaves are used in a radar system for aircraft navigation
(iv) ultraviolet rays are used to destroy the bacteria and the sterling the surgical instruments.

The wavelength of electromagnetic waves ranges from.\(6\times { 10 }^{ -14 }m \ to \ 6\times { 10 }^{ 6 }m\) These waves travel at the same velocity in a\((=3\times { 10 }^{ 8 }{ ms }^{ -1 })\) vacuum but with different velocity in a medium. In fact, the velocity of an electromagnetic wave is less in a medium than vacuum and a medium provides different values of refractive index to the electromagnetic waves of different wavelengths.

While dealing with the charging of a parallel plate capacitor with varying current, it was found that Ampere's circuital law is not logically consistent, because \(\oint { \overset { \rightarrow }{ B } ,\overset { \rightarrow }{ dl } } \)has not same value on the two sides of a plate of charged capacitor. The inconsistency of Ampere's circuital law was removed by maxwell by predicting the presence of displacement current in the region between the plates of the capacitor. when the charge on the capacitor is changing with time. Maxwell also predicted that the sum of conduction current and displacement current has the property of continuity.
The generalised form of Ampere's circuital law, modified by Maxwell states that
\(\oint { \overset { \rightarrow }{ B } ,\overset { \rightarrow }{ dl } = } { \mu }_{ 0 }(I+{ I }_{ D })={ \mu }_{ 0 }\left( (I+{ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } ) \right) \)

Electric field of an electromagnetic wave is an oscillating field which causes force on the charged particle. This electric force averaged over an integral number of cycles is zero, because its direction changes with every half cycle. So, electric field is not responsible for radiation pressure.

(i) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(ii) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iii) \(4.42\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iv) \(3.3\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)