Properties of Electromagnetic Waves Questions in English

(B) The total power of the bulb is $P_{total} = 100 \ W$.
Since $20 \%$ of the power is converted to visible radiation,the power of visible radiation $P_{vis}$ is:
$P_{vis} = 100 \ W \times \frac{20}{100} = 20 \ W$.
The intensity $I$ of radiation emitted isotropically at a distance $r$ from a point source is given by the formula:
$I = \frac{P_{vis}}{4 \pi r^2}$.
Given $r = 5 \ m$,we substitute the values:
$I = \frac{20}{4 \pi \times (5)^2} = \frac{20}{4 \pi \times 25} = \frac{5}{25 \pi} \ W/m^2$.
Comparing this with the given expression $\frac{\alpha}{25 \pi} \ W/m^2$,we find that $\alpha = 5$.

(D) The correct statement is given in option $(D)$.
Analysis of options:
Option $(A)$ is incorrect because photons have zero rest mass,yet they carry momentum due to their energy $(p = E/c)$.
Option $(B)$ is incorrect because the electromagnetic force is much stronger than the weak nuclear force.
Option $(C)$ is incorrect because the strong nuclear force is responsible for the stability of nuclei,not the weak nuclear force.
Option $(D)$ is correct because electromagnetic forces are long-range forces and do not require a material medium for propagation.

(A) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \varepsilon_0 E_0^2 c = \frac{1}{2} \frac{B_0^2 c}{\mu_0}$.
Radiation pressure $P$ on a totally absorbing surface is defined as the momentum transferred per unit area per unit time,which is given by $P = \frac{I}{c}$.
Substituting the expression for intensity,we get $P = \frac{\frac{1}{2} \varepsilon_0 E_0^2 c}{c} = \frac{1}{2} \varepsilon_0 E_0^2$.
Thus,the magnitude of the average momentum transferred per unit area and per unit time is $\frac{1}{2} \varepsilon_0 E_0^2$.

(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}}$.
Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3 \times 10^8 \text{ m/s}$,we can write $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Given $\epsilon_r = 8$ and $\mu_r = 200$,the refractive index $n = \sqrt{\mu_r \epsilon_r} = \sqrt{200 \times 8} = \sqrt{1600} = 40$.
Thus,the speed $v = \frac{3 \times 10^8}{40} = 0.075 \times 10^8 = 7.5 \times 10^6 \text{ m/s}$.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f}$.
Given $f = 100 \text{ MHz} = 10^8 \text{ Hz}$.
Therefore,$\lambda = \frac{7.5 \times 10^6}{10^8} = 7.5 \times 10^{-2} \text{ m} = 7.5 \text{ cm}$.

(C) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula: $I = \frac{1}{2} c \epsilon_0 E_0^2$.
Given: $I = \frac{15}{\pi} \text{ W m}^{-2}$, $c = 3 \times 10^8 \text{ m s}^{-1}$, and $\epsilon_0 = \frac{1}{36\pi} \times 10^{-9} \text{ F m}^{-1}$.
Substituting the values: $\frac{15}{\pi} = \frac{1}{2} \times (3 \times 10^8) \times (\frac{1}{36\pi} \times 10^{-9}) \times E_0^2$.
$\frac{15}{\pi} = \frac{3 \times 10^{-1}}{72\pi} \times E_0^2$.
$\frac{15}{\pi} = \frac{1}{240\pi} \times E_0^2$.
$E_0^2 = 15 \times 240 = 3600$.
$E_0 = \sqrt{3600} = 60 \text{ N C}^{-1}$.

(A) The given electric field is $E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \ Vm^{-1}$.
Comparing this with the standard equation $E_z = E_0 \sin(kx + \omega t)$,we get $E_0 = 60 \ Vm^{-1}$.
The amplitude of the magnetic field $B_0$ is given by $B_0 = \frac{E_0}{c}$,where $c = 3 \times 10^8 \ ms^{-1}$ is the speed of light.
$B_0 = \frac{60}{3 \times 10^8} = 20 \times 10^{-8} = 2 \times 10^{-7} \ T$.
Since the wave propagates in the negative $x$-direction (indicated by the $+kx$ term) and the electric field is in the $z$-direction,the magnetic field must be in the $y$-direction to satisfy the direction of propagation $\vec{E} \times \vec{B} \propto \vec{v}$.
Specifically,$\hat{k} \times \hat{i} = \hat{j}$ (for propagation in $-x$,$\hat{E} \times \hat{B} = -\hat{i}$,so $\hat{k} \times \hat{j} = -\hat{i}$).
Thus,the magnetic field is $B_y = B_0 \sin(kx + \omega t) = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \ T$.

(B) The relationship between the amplitude of the electric field $(E_0)$ and the amplitude of the magnetic field $(B_0)$ for an electromagnetic wave in vacuum is given by the equation: $E_0 = c B_0$,where $c$ is the speed of light in vacuum.
Given:
$B_0 = 270 \ nT = 270 \times 10^{-9} \ T$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$E_0 = (3 \times 10^8 \ m/s) \times (270 \times 10^{-9} \ T)$
$E_0 = 810 \times 10^{-1} \ NC^{-1} = 81 \ NC^{-1}$
Therefore,the amplitude of the electric field is $81 \ NC^{-1}$.

(B) For a plane electromagnetic wave incident on a completely absorbing surface,the momentum $p$ delivered is given by the relation $p = \frac{U}{c}$,where $U$ is the energy transferred and $c$ is the speed of light in vacuum.
We know that for an electromagnetic wave,the relationship between the electric field amplitude $E_0$ and magnetic field amplitude $B_0$ is given by $c = \frac{E_0}{B_0}$.
Substituting this value of $c$ into the momentum equation:
$p = \frac{U}{E_0 / B_0} = \frac{U B_0}{E_0}$.
Thus,the magnitude of the total momentum delivered is $\frac{U B_0}{E_0}$.

(D) For an electromagnetic wave,the relationship between the peak electric field $E_0$ and the peak magnetic field $B_0$ is given by the equation $E_0 = c B_0$,where $c$ is the speed of light in a vacuum.
Given:
$B_0 = 30 \times 10^{-9} \ T$
$c = 3 \times 10^8 \ m/s$
Substituting these values into the formula:
$E_0 = (3 \times 10^8 \ m/s) \times (30 \times 10^{-9} \ T)$
$E_0 = 90 \times 10^{-1} \ Vm^{-1}$
$E_0 = 9 \ Vm^{-1}$
Therefore,the peak value of the electric field is $9 \ Vm^{-1}$.

(A) Given frequency, $f = 3 \text{ GHz} = 3 \times 10^9 \text{ Hz}$.
Speed of light, $c = 3 \times 10^8 \text{ m/s}$.
The relationship between wavelength $\lambda$, frequency $f$, and speed of light $c$ is given by $\lambda = \frac{c}{f}$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{3 \times 10^9} = 10^{-1} \text{ m} = 0.1 \text{ m}$.

(D) is incorrect because electromagnetic $(EM)$ waves do not require any medium to travel; thus,they can travel in a vacuum.
$B$ is incorrect because $EM$ waves are transverse waves,not longitudinal.
$C$ is incorrect because $EM$ waves are produced by an accelerating charge,not by charges moving with uniform velocity.
$D$ is correct because $EM$ waves carry both energy and momentum as they propagate through space.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$,where $c$ is the speed of light in vacuum,$\mu_r$ is the relative permeability,and $\epsilon_r$ is the relative permittivity.
Given: $v = 1.5 \times 10^8 \ m/s$,$c = 3 \times 10^8 \ m/s$,and $\epsilon_r = 2$.
Substituting the values: $1.5 \times 10^8 = \frac{3 \times 10^8}{\sqrt{\mu_r \times 2}}$.
Squaring both sides: $(1.5)^2 = \frac{3^2}{2\mu_r} \Rightarrow 2.25 = \frac{9}{2\mu_r}$.
Solving for $\mu_r$: $2\mu_r = \frac{9}{2.25} = 4 \Rightarrow \mu_r = 2$.
The magnetic susceptibility $\chi_m$ is related to relative permeability by $\mu_r = 1 + \chi_m$.
Therefore,$\chi_m = \mu_r - 1 = 2 - 1 = 1$.

(B) We know that for an electromagnetic wave,the relation between the amplitudes of electric and magnetic fields is $E_0 = B_0 c$.
Given $B_0 = 2 \times 10^{-8} \text{ T}$ and $c = 3 \times 10^8 \text{ m/s}$.
$E_0 = (2 \times 10^{-8}) \times (3 \times 10^8) = 6 \text{ V m}^{-1}$.
The direction of propagation of the wave is given by the direction of the vector $\vec{E} \times \vec{B}$.
Here,the wave propagates along $-\hat{j}$ and the magnetic field is along $\hat{k}$.
Let the direction of $\vec{E}$ be $\hat{n}$. Then $\hat{n} \times \hat{k} = -\hat{j}$.
Since $\hat{i} \times \hat{k} = -\hat{j}$,the electric field must be along the $\hat{i}$ direction.
The phase of the wave remains the same,so the argument of the cosine function is $\pi \times 10^{15}(t + \frac{y}{c})$.
Thus,$\vec{E} = (6) \cos [\pi \times 10^{15}(t + \frac{y}{c})] \hat{i} \text{ V m}^{-1}$.

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is the direction of $\vec{E} \times \vec{B}$.
First,we find the unit vectors for the electric and magnetic fields:
$\hat{E} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$
$\hat{B} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$
The direction of propagation $\hat{n}$ is given by $\hat{E} \times \hat{B}$:
$\hat{n} = \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) \times \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$
$\hat{n} = \frac{1}{2} [(\hat{i} \times \hat{i}) - (\hat{i} \times \hat{j}) + (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{j})]$
Using the cross product rules $\hat{i} \times \hat{i} = 0$,$\hat{j} \times \hat{j} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{j} \times \hat{i} = -\hat{k}$:
$\hat{n} = \frac{1}{2} [0 - \hat{k} - \hat{k} - 0] = \frac{1}{2} [-2\hat{k}] = -\hat{k}$.

(B) The intensity $I$ of an electromagnetic wave is related to the amplitude of the magnetic field $B_0$ by the formula: $I = \frac{B_0^2}{2 \mu_0} c$.
Given: $I = 2.1 \times 10^{15} \ W/m^2$,$c = 3 \times 10^8 \ m/s$,and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values: $2.1 \times 10^{15} = \frac{B_0^2 \times 3 \times 10^8}{2 \times 4 \pi \times 10^{-7}}$.
$2.1 \times 10^{15} = \frac{B_0^2 \times 3 \times 10^8}{8 \pi \times 10^{-7}}$.
$B_0^2 = \frac{2.1 \times 10^{15} \times 8 \pi \times 10^{-7}}{3 \times 10^8} = \frac{16.8 \pi \times 10^8}{3 \times 10^8} = 5.6 \pi \approx 5.6 \times 3.14 = 17.584 \approx 17.64$.
$B_0 = \sqrt{17.64} = 4.2 \ T$.

(B) The average intensity $I$ of an electromagnetic wave is related to the amplitude of the magnetic field $B_0$ by the formula: $I = \frac{B_0^2 C}{2 \mu_0}$.
Given,$I = \frac{6}{\pi} \times 10^3 \text{ W/m}^2$,$C = 3 \times 10^8 \text{ m/s}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Rearranging the formula for $B_0^2$: $B_0^2 = \frac{2 \mu_0 I}{C}$.
Substituting the values: $B_0^2 = \frac{2 \times (4\pi \times 10^{-7}) \times (\frac{6}{\pi} \times 10^3)}{3 \times 10^8}$.
Simplifying the expression: $B_0^2 = \frac{8\pi \times 10^{-7} \times 6 \times 10^3}{\pi \times 3 \times 10^8} = \frac{48 \times 10^{-4}}{3 \times 10^8} = 16 \times 10^{-12}$.
Taking the square root: $B_0 = \sqrt{16 \times 10^{-12}} = 4 \times 10^{-6} \text{ T}$.

(B) Given: $E_0 = 150 \text{ V/m}$,$f = 30 \text{ MHz} = 30 \times 10^6 \text{ Hz}$.
$1$. Calculate the magnetic field amplitude: $B_0 = \frac{E_0}{c} = \frac{150}{3 \times 10^8} = 5 \times 10^{-7} \text{ T}$.
$2$. Calculate angular frequency: $\omega = 2\pi f = 2\pi \times 30 \times 10^6 = 60\pi \times 10^6 \text{ rad/s} = 6\pi \times 10^7 \text{ rad/s}$.
$3$. Calculate wave number: $k = \frac{\omega}{c} = \frac{6\pi \times 10^7}{3 \times 10^8} = 0.2\pi = \frac{\pi}{5} \text{ rad/m}$.
$4$. Direction: The wave propagates along $\hat{i}$ ($x$-axis) and $\vec{E}$ is along $\hat{j}$ ($y$-axis). Since $\vec{B}$ must be perpendicular to both $\vec{E}$ and the direction of propagation,$\vec{B}$ is along $\hat{j} \times \hat{i} = -\hat{k}$ or $\hat{k} \times \hat{i} = \hat{j}$. Specifically,$\vec{E} \times \vec{B}$ must be in the direction of propagation $(\hat{i})$. Thus,$\hat{j} \times \hat{k} = \hat{i}$,so $\vec{B}$ is along the $z$-axis.
$5$. The wave equation is $\vec{B} = B_0 \sin(kx - \omega t) \hat{k} = 5 \times 10^{-7} \sin \left[\pi \left(\frac{x}{5} - 6 \times 10^7 t\right)\right] \hat{z} \text{ T}$.

(A) The total power of the bulb is $P = 100 \pi \ W$.
The power converted to visible radiation is $10 \%$ of the total power,so $P' = 0.10 \times 100 \pi \ W = 10 \pi \ W$.
The intensity $I$ at a distance $d = 10 \ m$ from a point source is given by the formula $I = \frac{P'}{4 \pi d^2}$.
Substituting the values,we get $I = \frac{10 \pi}{4 \pi (10)^2} = \frac{10 \pi}{4 \pi \times 100} = \frac{10}{400} = 0.025 \ W \ m^{-2}$.

(A) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula:
$I = \frac{1}{2} \epsilon_0 E_0^2 c$
Given:
$I = 17.7 \times 10^{14} \ W/m^2$
$\epsilon_0 = 8.85 \times 10^{-12} \ C^2 / (N \cdot m^2)$
$c = 3 \times 10^8 \ m/s$
Substituting the values into the formula:
$17.7 \times 10^{14} = \frac{1}{2} \times (8.85 \times 10^{-12}) \times E_0^2 \times (3 \times 10^8)$
$17.7 \times 10^{14} = \frac{26.55 \times 10^{-4}}{2} \times E_0^2$
$17.7 \times 10^{14} = 13.275 \times 10^{-4} \times E_0^2$
$E_0^2 = \frac{17.7 \times 10^{14}}{13.275 \times 10^{-4}} = \frac{17.7}{13.275} \times 10^{18} \approx 1.333 \times 10^{18} = \frac{4}{3} \times 10^{18}$
Taking the square root:
$E_0 = \sqrt{\frac{4}{3} \times 10^{18}} = \frac{2}{\sqrt{3}} \times 10^9 \ N \ C^{-1}$

(A) The intensity $I$ of a parallel beam of light is given by the formula:
$I = \frac{1}{2} \varepsilon_0 E_0^2 c$ ... $(i)$
where $E_0$ is the amplitude of the electric field,$\varepsilon_0$ is the permittivity of free space,and $c$ is the speed of light.
Given:
$I = \frac{15}{\pi} \text{ W/m}^2$
$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \implies \varepsilon_0 = \frac{1}{36 \pi \times 10^9} \text{ F/m}$
$c = 3 \times 10^8 \text{ m/s}$
Rearranging equation $(i)$ for $E_0^2$:
$E_0^2 = \frac{2I}{\varepsilon_0 c} = \frac{2I \times 4 \pi}{(4 \pi \varepsilon_0) c}$
Substituting the values:
$E_0^2 = \frac{2 \times (15/\pi) \times 4 \pi}{(1 / (9 \times 10^9)) \times 3 \times 10^8}$
$E_0^2 = \frac{120}{3 \times 10^8 / 9 \times 10^9} = \frac{120}{1/30} = 120 \times 30 = 3600$
$E_0 = \sqrt{3600} = 60 \text{ N/C}$

(D) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is parallel to $\vec{E} \times \vec{B}$.
Given the direction of the magnetic field is $\hat{B} = \hat{i} - \hat{j}$.
Given the direction of the electric field is $\hat{E} = \hat{i} + \hat{j}$.
The direction of propagation $\hat{n}$ is given by the cross product of the unit vectors of the electric and magnetic fields:
$\hat{n} = \hat{E} \times \hat{B} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j})$.
Expanding the cross product:
$\hat{n} = (\hat{i} \times \hat{i}) - (\hat{i} \times \hat{j}) + (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{j})$.
Using the properties of unit vectors $\hat{i} \times \hat{i} = 0$,$\hat{j} \times \hat{j} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{j} \times \hat{i} = -\hat{k}$:
$\hat{n} = 0 - \hat{k} - \hat{k} - 0 = -2\hat{k}$.
Since the direction is $-2\hat{k}$,the wave is travelling along the $-z$-direction.

(B) The average power emitted by the source is $P_{\text{avg}} = 4 \% \text{ of } 100 \,W = \frac{4}{100} \times 100 \,W = 4 \,W$.
At a distance $r = 2 \,m$, the radiation spreads over a spherical surface area $A = 4 \pi r^2 = 4 \pi (2)^2 = 16 \pi \,m^2$.
The average intensity $I_{\text{avg}}$ is given by $I_{\text{avg}} = \frac{P_{\text{avg}}}{A} = \frac{4}{16 \pi} = \frac{1}{4 \pi} \,W/m^2$.
Also, the average intensity of an electromagnetic wave is related to the rms electric field $E_{\text{rms}}$ by $I_{\text{avg}} = \varepsilon_0 c E_{\text{rms}}^2$, where $c = 3 \times 10^8 \,m/s$ is the speed of light.
Equating the two expressions for intensity: $\frac{1}{4 \pi} = \varepsilon_0 c E_{\text{rms}}^2$.
Rearranging for $E_{\text{rms}}^2$: $E_{\text{rms}}^2 = \frac{1}{4 \pi \varepsilon_0 c} = \left( \frac{1}{4 \pi \varepsilon_0} \right) \times \frac{1}{c}$.
Substituting the given values: $E_{\text{rms}}^2 = (9 \times 10^9) \times \frac{1}{3 \times 10^8} = 3 \times 10 = 30$.
Therefore, $E_{\text{rms}} = \sqrt{30} \,V/m$.

(A) Given,frequency of the electromagnetic $(EM)$ wave,$f = 3 \ MHz = 3 \times 10^6 \ Hz$.
Permittivity of the non-magnetic medium,$\epsilon = 16 \epsilon_0$.
The wavelength of the $EM$ wave in vacuum is $\lambda = \frac{c}{f} = \frac{3 \times 10^8 \ m/s}{3 \times 10^6 \ Hz} = 100 \ m$.
The refractive index of the medium is $n = \sqrt{\epsilon_r \mu_r}$. Since the medium is non-magnetic,$\mu_r = 1$.
Thus,$n = \sqrt{\frac{\epsilon}{\epsilon_0}} = \sqrt{16} = 4$.
The wavelength in the medium is $\lambda' = \frac{\lambda}{n} = \frac{100 \ m}{4} = 25 \ m$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = 25 \ m - 100 \ m = -75 \ m$.

(B) Given,maximum magnetic field,$B_0 = 1.26 \times 10^{-4} \ T$ and permeability of free space,$\mu_0 = 1.26 \times 10^{-6} \ H/m$. The speed of light is $c = 3 \times 10^8 \ m/s$.
The intensity $I$ of an electromagnetic wave is given by the formula:
$I = \frac{1}{2} \frac{B_0^2 c}{\mu_0}$
Substituting the values:
$I = \frac{1}{2} \times \frac{(1.26 \times 10^{-4})^2 \times 3 \times 10^8}{1.26 \times 10^{-6}}$
$I = \frac{1}{2} \times \frac{1.26 \times 1.26 \times 10^{-8} \times 3 \times 10^8}{1.26 \times 10^{-6}}$
$I = \frac{1}{2} \times 1.26 \times 3 \times 10^2 \times 10^6$
$I = 0.63 \times 3 \times 10^6 = 1.89 \times 10^6 \ W/m^2$.

(B) The general equation for a plane electromagnetic wave is given by $B = B_0 \sin(kx + \omega t)$.
Comparing this with the given equation $B = 5 \times 10^{-6} \sin(0.6 \times 10^2 x + 0.5 \times 10^{10} t)$,we get:
Wave number $k = 0.6 \times 10^2 \text{ m}^{-1}$
Angular frequency $\omega = 0.5 \times 10^{10} \text{ rad/s}$
The speed of the wave $v$ is given by the relation $v = \frac{\omega}{k}$.
Substituting the values:
$v = \frac{0.5 \times 10^{10}}{0.6 \times 10^2}$
$v = \frac{5}{6} \times 10^8 \text{ m/s}$
$v \approx 0.833 \times 10^8 \text{ m/s}$.

(C) The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2\mu_0} B^2$.
Since the electric field $E$ and magnetic field $B$ oscillate as $E = E_0 \sin(\omega t)$ and $B = B_0 \sin(\omega t)$, the energy density involves terms like $\sin^2(\omega t)$.
Using the trigonometric identity $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$, we see that the energy density oscillates with an angular frequency of $2\omega$.
Therefore, the frequency of energy oscillation is $f_{energy} = 2f = 2 \times (4 \times 10^{14} \,Hz) = 8 \times 10^{14} \,Hz$.

(C) The given equation for the electric field is $E = i 30 \cos (k z - 5 \times 10^8 t)$.
Comparing this with the standard wave equation $E = E_0 \cos (k z - \omega t)$,we identify the angular frequency $\omega$ as $5 \times 10^8 \ rad/s$.
The relationship between the speed of light $c$,angular frequency $\omega$,and wave vector $k$ is given by $c = \frac{\omega}{k}$.
Rearranging to solve for $k$,we get $k = \frac{\omega}{c}$.
Substituting the given values: $k = \frac{5 \times 10^8 \ rad/s}{3 \times 10^8 \ m/s}$.
Thus,$k = \frac{5}{3} \approx 1.66 \ rad/m$.

(A) The velocity of an electromagnetic wave in free space is given by the relation:
$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
Squaring both sides,we get:
$c^2 = \frac{1}{\mu_0 \varepsilon_0}$
Since $c$ represents the speed of light,its dimensions are $[LT^{-1}]$.
Therefore,the dimensions of $\frac{1}{\mu_0 \varepsilon_0}$ are the same as the dimensions of $c^2$:
$[\frac{1}{\mu_0 \varepsilon_0}] = [c^2] = [LT^{-1}]^2 = [L^2 T^{-2}]$.

(D) Given: Power of the bulb $P = 60 \ W$,Efficiency $\eta = 16 \% = 0.16$,Distance $r = 2 \ m$.
First,calculate the intensity of the electromagnetic radiation emitted by the bulb at distance $r$:
The power radiated as electromagnetic waves is $P_{rad} = \eta \times P = 0.16 \times 60 = 9.6 \ W$.
The intensity $I$ at distance $r$ is given by $I = \frac{P_{rad}}{4 \pi r^2} = \frac{9.6}{4 \pi (2)^2} = \frac{9.6}{16 \pi} = \frac{0.6}{\pi} \ W/m^2$.
The intensity of an electromagnetic wave is also related to the peak electric field $E_0$ by $I = \frac{1}{2} c \epsilon_0 E_0^2$.
Equating the two expressions for intensity: $\frac{0.6}{\pi} = \frac{1}{2} c \epsilon_0 E_0^2$.
We know that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$,so $\epsilon_0 = \frac{1}{36 \pi \times 10^9}$.
Using $c = 3 \times 10^8 \ m/s$,we have $E_0^2 = \frac{2 \times I}{c \epsilon_0} = \frac{2 \times 0.6 / \pi}{3 \times 10^8 \times (1 / (36 \pi \times 10^9))}$.
$E_0^2 = \frac{1.2}{\pi} \times \frac{36 \pi \times 10^9}{3 \times 10^8} = 1.2 \times 12 \times 10 = 144$.
Therefore,$E_0 = \sqrt{144} = 12 \ Vm^{-1}$.

(B) The given electric field is $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.
Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec{k} \cdot \vec{r} - \omega t)}$,we identify the wave vector $\vec{k} = k_0(\hat{i} + \hat{j} + \hat{k})$.
The magnitude of the wave vector is $k = |k_0(\hat{i} + \hat{j} + \hat{k})| = k_0 \sqrt{1^2 + 1^2 + 1^2} = k_0 \sqrt{3}$.
The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{k_0 c}{k_0 \sqrt{3}} = \frac{c}{\sqrt{3}}$.
The refractive index $n = \frac{c}{v} = \sqrt{3}$.
Since the wave is transverse,$\vec{E} \cdot \vec{k} = 0$,which implies $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
Given that $\vec{E}$ is polarized in the $x-z$ plane,$\hat{n}$ must be in the $x-z$ plane,so $\hat{n} = a\hat{i} + b\hat{k}$.
From $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$,we get $a + b = 0$,so $a = -b$.
Normalizing $\hat{n}$,we get $\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}$.
Thus,both options $B$ and $C$ are correct.

(B) For an electromagnetic wave,the direction of propagation is given by the vector $\vec{k}_{dir} = \hat{k}$ (since the phase is $kz - \omega t$).
The electric field vector is $\vec{E} = E_0(\hat{i} + \hat{j})$.
The magnetic field $\vec{B}$ is perpendicular to both the direction of propagation and the electric field $\vec{E}$.
The direction of $\vec{B}$ is given by the cross product $\vec{k}_{dir} \times \vec{E}$.
$\vec{B}_{dir} = \hat{k} \times (\hat{i} + \hat{j}) = (\hat{k} \times \hat{i}) + (\hat{k} \times \hat{j}) = \hat{j} - \hat{i} = -\hat{i} + \hat{j}$.
Thus,the direction of the magnetic field is $-\hat{i} + \hat{j}$.

(B) Given,$\vec{E} = 90 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{k} \text{ V/m}$.
Comparing this with the standard wave equation $\vec{E} = E_{0} \sin (kx + \omega t) \hat{n}$,we get $E_{0} = 90 \text{ V/m}$.
The direction of propagation is along $-\hat{i}$ (since the argument is $kx + \omega t$).
The amplitude of the magnetic field is $B_{0} = \frac{E_{0}}{c} = \frac{90}{3 \times 10^{8}} = 3 \times 10^{-7} \text{ T}$.
The direction of propagation is given by the direction of $\vec{E} \times \vec{B}$.
Here,$\vec{E}$ is along $\hat{k}$ and the direction of propagation is $-\hat{i}$.
So,$\hat{k} \times \hat{B} = -\hat{i}$.
Since $\hat{k} \times \hat{j} = \hat{i}$,it follows that $\hat{k} \times (-\hat{j}) = -\hat{i}$.
Therefore,the magnetic field $\vec{B}$ must be along $\hat{j}$ (Wait,checking cross product: $\hat{k} \times \hat{j} = -\hat{i}$,so $\vec{B}$ is along $\hat{j}$).
Thus,$\vec{B} = 3 \times 10^{-7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{j} \text{ T}$.

(A) The direction of propagation of the wave is given by the wave vector $\vec{k}$,which is along the $z$-axis ($+\hat{k}$ direction).
The relationship between the electric field $\vec{E}$,magnetic field $\vec{B}$,and the direction of propagation $\hat{c}$ is given by $\vec{B} = \frac{1}{c} (\hat{c} \times \vec{E})$.
Here,$\hat{c} = \hat{k}$ and $\vec{E} = E_0 \sin(kz - \omega t) \hat{j}$.
The direction of $\vec{B}$ is $\hat{k} \times \hat{j} = -\hat{i}$.
The magnitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{54}{3 \times 10^8} = 18 \times 10^{-8} = 1.8 \times 10^{-7} \ T$.
Therefore,$\vec{B} = -1.8 \times 10^{-7} \sin(kz - \omega t) \hat{i} \ T$.

(C) The general equation for a wave is given by $E = E_0 \sin(\omega t - kx)$.
Comparing this with the given equation $\overline{E}(x,t) = 25 \sin(2.0 \times 10^{15}t - 10^{7}x)\hat{n}$,we get:
$\omega = 2.0 \times 10^{15} \text{ rad/s}$
$k = 10^{7} \text{ m}^{-1}$
The velocity of the wave in the medium is $v = \frac{\omega}{k}$.
$v = \frac{2.0 \times 10^{15}}{10^7} = 2.0 \times 10^8 \text{ m/s}$.
The refractive index $\mu$ is given by $\mu = \frac{c}{v}$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light in vacuum.
$\mu = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.

(D) The speed of electromagnetic waves in a vacuum is $C = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$.
The speed of electromagnetic waves in a medium is $V = \frac{1}{\sqrt{\mu\varepsilon}}$.
The ratio of speeds is $\frac{C}{V} = \sqrt{\frac{\mu\varepsilon}{\mu_{0}\varepsilon_{0}}} = \sqrt{\mu_{r}\varepsilon_{r}}$.
Given,dielectric constant $K = \varepsilon_{r} = 3$ and relative permeability $\mu_{r} = \frac{\mu}{\mu_{0}} = 2$.
Substituting these values,we get $\frac{C}{V} = \sqrt{3 \times 2} = \sqrt{6}$.
Therefore,the ratio is $\sqrt{6} : 1$.

(D) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 E_0^2 c$.
From this,the amplitude of the electric field $E_0$ is $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$.
We know that the relation between the amplitudes of the electric and magnetic fields is $E_0 = B_0 c$,which implies $B_0 = \frac{E_0}{c}$.
Substituting $E_0$,we get $B_0 = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2I}{\epsilon_0 c^3}}$.
Given $I = 4.0 \times 10^{14} \ W/m^2$,$\epsilon_0 = 8.85 \times 10^{-12} \ C^2/Nm^2$,and $c = 3 \times 10^8 \ m/s$:
$B_0 = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}} = \sqrt{\frac{8.0 \times 10^{14}}{8.85 \times 10^{-12} \times 27 \times 10^{24}}} = \sqrt{\frac{8.0 \times 10^{14}}{238.95 \times 10^{12}}} = \sqrt{\frac{800}{238.95}} \approx \sqrt{3.348} \approx 1.83 \ T$.

(D) The general equation for a plane wave is $E = E_0 \sin(kx - \omega t)$.
Comparing this with the given equation,we have the angular frequency $\omega = 4.5 \times 10^{14} \text{ rad/s}$ and the wave number $k = 3 \times 10^6 \text{ rad/m}$.
The phase velocity $v$ of the wave in the medium is given by $v = \frac{\omega}{k} = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8 \text{ m/s}$.
The refractive index $n$ of the medium is $n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$.
For a non-magnetic medium,the relative permeability $\mu_r = 1$. The refractive index is related to the dielectric constant (relative permittivity) $\varepsilon_r$ by $n = \sqrt{\mu_r \varepsilon_r} = \sqrt{\varepsilon_r}$.
Thus,$\varepsilon_r = n^2 = 2^2 = 4$.

(A) The given electric field is $E_y = E_0 \sin(kx - \omega t)$, where $E_0 = 69 \text{ V/m}$, $k = 0.6 \times 10^3 \text{ rad/m}$, and $\omega = 1.8 \times 10^{11} \text{ rad/s}$.
Since the wave propagates in the $+x$ direction $(\hat{i})$ and the electric field is in the $y$ direction $(\hat{j})$, the magnetic field must be in the $z$ direction $(\hat{k})$ because $\vec{B} = \frac{1}{c} (\hat{c} \times \vec{E}) = \frac{1}{c} (\hat{i} \times E_y \hat{j}) = \frac{E_y}{c} \hat{k}$.
The amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} = 2.3 \times 10^{-7} \text{ T}$.
The phase of the magnetic field is the same as the electric field, so $B_z = B_0 \sin(kx - \omega t) = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] \text{ T}$.

(B) The intensity $I$ of a point source at a distance $r$ is given by the formula $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source.
Since the detector is shifted to another location on the same spherical surface of radius $L$,the distance $r$ from the source remains constant $(r = L)$.
Because the intensity $I$ depends only on the distance $r$ for an isotropic point source,the intensity at the new location remains the same as the initial intensity $I_o$.
Therefore,the measured intensity at the new location is $I_o$.

(A) The relationship between the electric field $\vec{E}$ and the magnetic field $\vec{B}$ in an electromagnetic wave is given by $\vec{E} = c(\vec{B} \times \hat{n})$,where $\hat{n}$ is the direction of wave propagation.
Here,the wave propagates in the $+x$ direction,so $\hat{n} = \hat{i}$.
The magnetic field is given as $\vec{B} = B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{j}$.
Using the relation $\vec{E} = c(\vec{B} \times \hat{i})$,we substitute $\vec{B}$:
$\vec{E} = c B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) (\hat{j} \times \hat{i})$.
Since $\hat{j} \times \hat{i} = -\hat{k}$ and the speed of light $c = v\lambda$ (where $v$ is frequency and $\lambda$ is wavelength),we get:
$\vec{E} = -v\lambda B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{k}$.

(A) The electric force on the electron is given by $F_e = eE$,where $e$ is the charge of the electron and $E$ is the electric field amplitude.
The magnetic force on the electron is given by $F_m = evB$,where $v$ is the speed of the electron and $B$ is the magnetic field amplitude.
For an electromagnetic wave,the relationship between the amplitudes of electric and magnetic fields is $E = cB$,where $c$ is the speed of light $(c = 3 \times 10^8 \text{ m/s})$.
Therefore,the ratio of maximum forces is $\frac{F_e}{F_m} = \frac{eE}{evB} = \frac{E}{v(E/c)} = \frac{c}{v}$.
Given $c = 3 \times 10^8 \text{ m/s}$ and $v = 1.5 \times 10^6 \text{ m/s}$,the ratio is $\frac{3 \times 10^8}{1.5 \times 10^6} = \frac{300 \times 10^6}{1.5 \times 10^6} = 200$.

(B) The relationship between the electric field $\vec{E}$ and the magnetic field $\vec{B}$ in an electromagnetic wave is given by $\vec{E} = c(\vec{B} \times \hat{n})$,where $\hat{n}$ is the unit vector in the direction of wave propagation.
Here,the wave propagates along the $x$-direction,so $\hat{n} = \hat{i}$.
The speed of light in free space is $c = 3 \times 10^8 \text{ m/s}$.
Given $\vec{B} = 2 \times 10^{-7} \hat{j} \text{ T}$.
Substituting these values: $\vec{E} = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-7} \hat{j} \text{ T} \times \hat{i})$.
Since $\hat{j} \times \hat{i} = -\hat{k}$,we have $\vec{E} = (3 \times 10^8) \times (2 \times 10^{-7}) \times (-\hat{k}) = 60 \times (-\hat{k}) = -60\hat{k} \text{ V/m}$.

(B) For an electromagnetic wave,the relationship between the electric field $\vec{E}$,the magnetic field $\vec{B}$,and the propagation vector $\vec{k}$ is given by the Maxwell-Faraday equation.
According to Faraday's law in differential form,$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$.
For a plane electromagnetic wave,$\vec{E} = \vec{E}_0 \cos(\vec{k} \cdot \vec{r} - \omega t)$ and $\vec{B} = \vec{B}_0 \cos(\vec{k} \cdot \vec{r} - \omega t)$.
Substituting these into the differential equation,we get $\vec{k} \times \vec{E} = \omega \vec{B}$.
Therefore,the magnetic field is given by $\vec{B} = \frac{\vec{k} \times \vec{E}}{\omega}$.