Electromagnetic Waves

$|B|\, = {{|E|} \over C} = {6 \over {3 \times {{10}^8}}}$

= 2 $\times$ 10^$-$8 T

$\therefore$ x = 2

E₀ = 200

$I = {1 \over 2}{\varepsilon _0}E_0^2.C$

Radiation pressure

$P = {{2I} \over C}$

$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$

$ = {\varepsilon _0}E_0^2$

$ = 8.85 \times {10^{ - 12}} \times {200^2}$

$ = 8.85 \times {10^{ - 8}} \times 4$

$ = {{354} \over {{{10}^9}}}$

$I = {1 \over 2}C{ \in _0}{E^2}$

${E^2} \propto I$

$I = {{Power} \over {Area}}$

${E^2} \propto {P \over A}$

$E \propto \sqrt P $

${{E'} \over E} = \sqrt {{{60} \over {100}}} $

$E' = \sqrt {{3 \over 5}} E$

So the value of x = 3

Given f = 9 $\times$ 10² Hz

$ \in $ = $ \in $₀$ \in $_r

$ \in $ = 80 $ \in $₀

So $ \in $_r = 80

$\rho $ = 0.25 $\Omega$m

V(t) = V₀ sin (2$\pi$ft)

${I_d} = {{dq} \over {dt}} = {{cdv} \over {dt}}$

${I_d} = {{{ \in _0}{ \in _r}A} \over d}{d \over {dt}}({v_0}\sin (2\pi ft))$

${I_d} = {{{ \in _0}{ \in _r}A} \over d}{V_0}(2\pi f)\cos (2\pi ft)$ .......... (1)

& ${I_c} = {V \over R}$

${I_c} = {{{V_0}\sin (2\pi ft)} \over {\rho {d \over A}}} = {{A{v_0}\sin (2\pi ft)} \over {\rho d}}$ ....... (2)

divide equation (1) and (2)

${{{I_d}} \over {{I_c}}} = { \in _0}{ \in _r}2\pi f(\rho )\cot (2\pi ft)$

${{{I_d}} \over {{I_c}}} = {1 \over {4\pi \times 9 \times {{10}^9}}} \times 80 \times 2\pi \times 9 \times {10^2} \times (0.25) \times \cot (2\pi \times 9 \times {10^2} \times {1 \over {800}})$

$ = {{{{10}^3}} \over {{{10}^9}}}\left( {\cot \left( {{{9\pi } \over 4}} \right)} \right)$

$ = {{{{10}^3}} \over {{{10}^9}}}$

${{{I_d}} \over {{I_c}}} = {1 \over {{{10}^6}}}$

${I_c} = {10^6}{I_d}$

So x = 6

Pressure = ${{Intensity} \over C}$ (for absorbing surface)

I = P $\times$ C

I = ${{2.5 \times {{10}^{ - 6}}} \over {30c{m^2}}}$ N $\times$ 3 $\times$ 10⁸ m/s

I = 25 W/cm²

Intensity of electro magnetic wave is,

$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$

${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$

${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$

$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$

$E_0^2 = {{375} \over 2} = 187.5$

${E_0} = 13.69$

${E_0} \approx 137 \times {10^{ - 1}}$ v/m

Firstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\pi r^2$ where r is the distance from the source. So,

$I = \frac{P}{4\pi r^2} \tag{1}$ ........(1)

This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :

$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$ .........(2)

where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space.

From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :

$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$

Since $ \varepsilon_0=\frac{1}{\mu_0 c^2} $

Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.

So, substituting $P = 0.8\, \text{W}$, $r = 10\, \text{m}$, $c = 3 \times 10^8\, \text{m/s}$, and $\mu_0 = 4\pi \times 10^{-7}\, \text{T m/A}$, we have :

$E_{\text{peak}} = \sqrt{\frac{2 \times 0.8 \times 4\pi \times 10^{-7} \times 3 \times 10^8}{4\pi \times 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$

Comparing with the expression in the question, we find that $x = 2$.

Therefore, the answer is $x = 2$.

Given wavelength of an x-ray beam = 10$\mathop A\limits^o $

$ \because $ E = ${{hc} \over \lambda } = m{c^2}$

$ \Rightarrow $ m = ${h \over {c\lambda }}$

The mass of a fictitious particle having the same energy as that of the x-ray photons = ${x \over 3}$h kg

$ \therefore $ ${x \over 3}h = {h \over {c\lambda }}$

$x = {3 \over {c\lambda }}$

$ = {3 \over {3 \times {{10}^8} \times 10 \times {{10}^{ - 10}}}}$

x = 10

Given, frequency of wave, f = 3 GHz = 3 $\times$ 10⁹ Hz

Relative permittivity, $\varepsilon $_r = 2.25

Since, f = C/$\lambda$

$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$ m

$\because$ $\lambda$_m (wavelength of wave in a medium) = $\lambda$/$\mu$ and we know that, $\mu = \sqrt {{\mu _r}{\varepsilon _r}} $

As, dielectric is non-magnetic, $\mu$_r = 1

$ \Rightarrow \mu = \sqrt {2.25} = 1.5$

$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$ m

= 6.67 cm = 667 $\times$ 10^$-$2 cm

Given, $\mu$_r = $\varepsilon $_r = 2

where, $\mu$_r is relative permeability, $\varepsilon $_r is relative permittivity.

Speed of electromagnetic wave v is given by

$v = {c \over n}$

where, n = refractive index = $\sqrt {{\mu _r}{\varepsilon _r}} = \sqrt 4 = 2$

$ \Rightarrow v = {{3 \times {{10}^8}} \over 2}$ = 15 $\times$ 10⁷ m/s

$\because$ x $\times$ 10⁷ = 15 $\times$ 10⁷

$\Rightarrow$ x = 15

I = ${1 \over 2}$$\varepsilon $₀$E_0^2$c

$ \Rightarrow $ E₀ = $\sqrt {{{2I} \over {{\varepsilon _0}c}}} $

$ \therefore $ E_rms = ${{{E_0}} \over {\sqrt 2 }}$ = $\sqrt {{I \over {{\varepsilon _0}c}}} $

= $\sqrt {{{{{315} \over \pi }} \over {8.86 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} $

= 194

As the cube has unit volume so energy density, $u=\frac{E}{1}=E$

where, $E=$ total energy

Energy of a photon, $E_1=h \nu$
So for n photon, $E=n h \nu$

Here, the question mentions "average energy contained in the electromagnetic waves within the same volume" which means we have to consider standing wave in a cavity (since the photons are in a cube).

Average energy density for a standing wave : $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 E_0{ }^2+\frac{1}{2} \frac{B_0{ }^2}{\mu_0}\right)$

The factor $\frac{1}{2}$ comes from averaging $\sin ^2(k x)$

over the volume. $\left\langle\sin ^2(k x)\right\rangle=\frac{1}{2}$

For an $E M$ wave $E=C B$

$ \text { so } E_0=C B_0 $

Hence, $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0\left(c B_0\right)^2+\frac{B_0^2}{2 \mu_0}\right)$

$ =\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right) $

$ \begin{aligned} \Rightarrow\langle u\rangle=\frac{1}{2} & {\left[\frac{1}{2}\left(\frac{1}{\mu_0 c^2}\right) c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right]\left(\begin{array}{l} As, c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \\ \text { so } \varepsilon_0=\frac{1}{\mu_0 c^2} \end{array}\right.} \\ \Rightarrow\langle u\rangle & =\frac{1}{2}\left[\frac{B_0^2}{2 \mu_0}+\frac{B_0^2}{2 \mu_0}\right] \\ & \Rightarrow\langle u\rangle=\frac{B_0^2}{2 \mu_0} \end{aligned} $

Given that this is equal to the photon energy density. so $\frac{B_0^2}{2 \mu_0}=n h \nu$

Total energy of photons.

$ \begin{aligned} E & =n h \nu \\ & =35 \times 10^7 \times 6 \times 10^{-34} \times 1.15 \\ \Rightarrow E & =210 \times 10^{-12} \mathrm{~J} \end{aligned} $

For $E M$ waves : $E=\frac{B_0{ }^2}{2 \mu_0}$

So $\frac{B_0^2}{2 \mu_0}=210 \times 10^{-12}$

$ \begin{aligned} & \Rightarrow B_0^2=2 \times 4 \times \frac{22}{7} \times 10^{-7} \times 210 \times 10^{-12} \\ & \Rightarrow B_0^2=5280 \times 10^{-19} \\ & \Rightarrow B_0^2=528 \times 10^{-18} \\ & \Rightarrow B_0=\sqrt{528} \times 10^{-9} \\ & \Rightarrow B_0=22.978 \times 10^{-9} \mathrm{~T} \\ & \\ & \text { = } \alpha \times 10^{-9} \mathrm{~T} \text { (given) } \\ &So, \alpha=22.98 \text { Ans. } \end{aligned} $