Here, R = 10 cm = 0.1 cm;
C = 200 pF = 200 x 10^-12 F = 2 x 10^-10 F;
E_rms = 200 V; \(\omega\) = 200 rad s^-1 ;
r = 2.0 x 10^-2 m.
(a) \({ I }_{ rms }=\frac { { E }_{ rms } }{ 1/\omega C } =\omega C{ E }_{ rms }\)
= 200 x (2 x 10^-10) x 200
(b) Yes, because I_D = 1
(c) \({ I }_{ 0 }=\sqrt { 2 } { I }_{ rms }=\sqrt { 2 } \times 8\times { 10 }^{ -6 }\)
= 11.312 x 10^-6 A
(d) Consider a loop of radius r between two circular plates of parallel plate capacitor placed coaxially with them. The area of this loop \({ A }^{ \prime }=\pi { r }^{ 2 }\)
By symmetry, the magnetic field \(\overrightarrow { B } \) is equal in magnitude and is tangentially to the circle at every point. In this case, only a part of displacement current I_D will cross the loop of area \({ A }^{ \prime }\) . Therefore, the current passing through the area \({ A }^{ \prime }\)
\({ I }^{ \prime }=\frac { { I }_{ D } }{ \pi { R }^{ 2 } } \times \pi { r }^{ 2 }=\frac { { I }_{ D } }{ { R }^{ 2 } } { r }^{ 2 }\)
Using Ampere's Maxwell law we have, \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\times \)(total current through the area \({ A }^{ \prime }\)) or \(B=\frac { { \mu }_{ 0 }{ I }_{ 0 }r }{ 2\pi { R }^{ 2 } } =\frac { 4\pi \times { 10 }^{ -7 }\times 11.312\times { 10 }^{ -6 }\times 2\times { 10 }^{ -2 } }{ 2\pi \times { \left( 0.1 \right) }^{ 2 } } \)
or \(2\pi rB={ \mu }_{ 0 }\frac { { I }_{ 0 } }{ { R }^{ 2 } } { r }^{ 2 }\)
= 4.525 x 10^-12 T

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

Here, Maximum magnetic field,
B₀ = 5 x 10^-9T;
charge on alpha particle, q = + 2e
= 2 x 1.6 x 10^-19 = 3.2 x 10^-19C, 
v = 3 x 10⁷ ms^-1 
Maximum electric field,
E₀ = cB₀ = (3 x 10⁸) x (5 x 10^-19) = 150 Vm^-1
Maximum force on alpha particle due to electric field = q x E₀ = (3.2 x 10^-19) x 150
= 4.80 x 10^-17 N
Force on alpha particle due to magnetic field = qvB₀ = (3.2 x 10^-19) x ( 3 x 10⁷) x (5 x 10^-9) 
= 4.80 x 10^-19 N

Here, v = 6 x 10¹⁴ Hz, E₀ = 2 V m^-1
(i) Average energy density of the electric field
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times { 2 }^{ 2 }\)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)
(ii) Average energy density of magnetic field
\({ u }_{ B }=\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { \left( { E }_{ 0 }/c \right) }^{ 2 } }{ { \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { E }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 }{ c }^{ 2 } } \)
\(=\frac { 1 }{ 4 } \times \frac { { 2 }^{ 2 } }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times { \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } } } \)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)

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
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 9\times { 10 }^{ 4 } \right) ^{ 2 }\)
\(=19.91\times { 10 }^{ -4 }=1.99\times { 10 }^{ -3 }J{ m }^{ -3 }\)

Useful Intensity,
\(I=\frac { power }{ area } =\frac { 100\times \left( 2.5/100 \right) }{ 4\pi { \left( 3 \right) }^{ 2 } } =\frac { 2.5 }{ 36\pi } W{ m }^{ -2 }\)
Half of this intensity (I) belongs to electric field and half of that to magnetic field. Therefore,
\(\frac { I }{ 2 } =\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c \ or \ { E }_{ 0 }=\sqrt { \frac { 2I }{ { \varepsilon }_{ 0 }c } } \)
\(=\sqrt { \frac { 2\times \left( 2.5/36\pi \right) }{ \left( \frac { 1 }{ 4\pi \times 9\times { 10 }^{ 9 } } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } =4.08V{ m }^{ -1 }\)
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.08 }{ 3\times { 10 }^{ 8 } } =1.36\times { 10 }^{ -8 }T\)

Here, \({ I }_{ D }=2.0A, \ C=1\mu F={ 10 }^{ -6 }F.\)
We know, \({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { d }{ dt } \left( EA \right) \)
\(={ \epsilon }_{ 0 }A\frac { dE }{ dt } ={ \epsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) ={ \epsilon }_{ 0 }\frac { A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(\left( \because E=\frac { V }{ d } \right) and \ \left( C=\frac { { \epsilon }_{ 0 }A }{ d } \right) \)
\(or \ \frac { dV }{ dt } =\frac { { I }_{ D } }{ D } =\frac { 2.0 }{ { 10 }^{ -6 } } =2\times { 10 }^{ -6 }V{ s }^{ -1 }\)
Thus a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 x 10⁶ Vs^-1 .

\(C=2.0\mu F=2\times { 10 }^{ -6 }F,\)
\(\frac { dV }{ dt } =6V{ s }^{ -1 }\)
Displacement current,
\({ I }_{ D }={ \epsilon }_{ 0 }A\frac { dE }{ dt } ={ \epsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) \)
\(=\frac { { \epsilon }_{ 0 }A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(=\left( 2\times { 10 }^{ -6 } \right) \times 6=12\times { 10 }^{ -6 }A\)
\(=12\mu A\)

Here, \(r=6\times { 10 }^{ -2 }m;\)
\(A=\pi \times \left( 6\times { 10 }^{ -2 } \right) ^{ 2 }=36\pi \times { 10 }^{ -4 }{ m }^{ 2 }\)
\(\frac { dE }{ dt } ={ 10 }^{ 10 }V{ cm }^{ -1 }{ s }^{ -1 }={ 10 }^{ 12 }V{ m }^{ -1 }{ s }^{ -1 }\)
Displacement current,
\({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }A\frac { dE }{ dt } \)
\(=\left( 8.85\times { 10 }^{ -12 } \right) \times \left( 36\pi \times { 10 }^{ -4 } \right) \times { 10 }^{ 12 }\)
= 0.1 A

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, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)