Electromagnetic Waves

8.8 Electric field amplitude, $E_0$ = 120 N/C

Frequency of source, $\nu$ = 50 MHz = 50 $*{10}^6$ Hz

Speed of light, c = 3 $*{10}^8$ m/s

Magnitude of magnetic field strength is given as

$B_0=$ $\frac{E_0}{c}$ = $\frac{120}{3\mathrm{}*{10}^8}$ = 4 $*{10}^{-7}$ T = 400 $*{10}^{-9}$ T = 400 nT

$\mathrm{A}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{a}\mathrm{s}\mathrm{}$ $\omega$ = 2 $\pi \nu$ = 2 $*\pi *$ 50 $*{10}^6$

= 3.14 $*{10}^8$ rad/s

Propagation constant is given as

k = $\frac{\omega }{c}$ = $\frac{3.14\mathrm{}*{10}^8}{3\mathrm{}*{10}^8}$ = 1.05 rad/m

Wavelength of wave is given as

$\lambda =\frac{c}{\nu }$ = $\frac{3\mathrm{}*{10}^8}{50\mathrm{}*{10}^6}$ = 6.0 m

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:

$\stackrel{?}{E}$ = $E_0\sin ?(kx-\omega t)\stackrel{?}{j}$

$=$ 120 $*\mathrm{s}\mathrm{i}\mathrm{n}?(1.05*x-$ 3.14 $*{10}^{8}t)\stackrel{?}{j}$

Equation of magnetic field vector is given as:

$\stackrel{?}{B}$ = $B_0\sin ?(kx-\omega t)\stackrel{?}{k}$

= (400 $*{10}^{-9})\sin ?(1.05*x-$ 3.14 $*{10}^{8}t$ ) $\stackrel{?}{k}$

8.2 Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of parallel capacitor, C = 100 pF = 100 $*{10}^{-12}$ F

Supply voltage, V = 230 V

Angular frequency, $\omega$ = 300 rad/s

rms value of the conduction current, I = $\frac{V}{X_c}$ , where $X_c=\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\mathrm{}\frac{1}{\mathrm{\omega }\mathrm{C}}$

Hence, I = V $*\omega *C$ = 230 $*300*$ 100 $*{10}^{-12}$ = 6.9 $*{10}^{-6}$ A = 6.9 $\mu A$

Yes, the conduction current is equal to the displacement current.

Magnetic field is given as, B = $\frac{{\mu }_{0}r}{2\pi R^2}{I}_0$ , where

${\mu }_0$ = Free space permeability = 4 $\pi *{10}^{-7}$ N $A^{-2}$

$I_0$ = Maximum value of current = $\surd 2I$

r = distance between two plates on the axis = 3.0 cm = 0.03 m

then, B = $\frac{4\pi *{10}^{-7}*0.03}{2\pi *{0.06}^2}$ $*\surd 2*$ 6.9 $*{10}^{-6}$ = 1.626 $*{10}^{-11}$ T

8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, ${?}_0$ = 8.85 $*{10}^{-12}$ $C^2{N}^{-1}{m}^{-2}$

Capacitance between two plates is given by the relation,

C = $\frac{{?}_{0}A}{d}$  , where A = Area of each plate = $?r^2$ = $?*{0.12}^2$

C = $\frac{8.85\mathrm{}*{10}^{-12}*\mathrm{}?*{0.12}^2}{0.05}$  = 8.007 $*{10}^{-12}$

F

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{dq}{dt}$ = C  $\frac{dV}{dt}$

But $\frac{dq}{dt}$  = I, therefore

$\frac{dV}{dt}=\frac{I}{C}$  = $\frac{0.15}{8.007\mathrm{}*{10}^{-12}}$  = 1.87 $*{10}^{10}$  V/s$$

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 1.87 $*{10}^{10}$ V/s

The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff's first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.