Electromagnetic Waves

$E_0=60 \mathrm{~V} / \mathrm{m}$

$ \therefore \quad B_0=\frac{E_0}{C}=\frac{60}{3 \times 10^8}=2 \times 10^{-7} \mathrm{~T} $

For a plane electromagnetic wave, the electric field, magnetic field and direction of propagation are mutually perpendicular. Since the electric field is along the $Z$-axis $\left(E_z\right)$ and the wave is propagating along the $X$-axis (indicated by the $k x$ term in the argument of the sine function), the magnetic field must be along the $Y$-axis $\left(B_y\right)$.

$ \begin{aligned} \therefore B_y=2 \times 10^{-7} \sin \left(0.5 \times 10^3 x\right. + & \left.1.5 \times 10^{11} f\right) \mathrm{T} \end{aligned} $

Step 1: Write the formula connecting intensity and electric field.

The formula for the intensity ($I$) of a light beam in terms of the amplitude of the electric field ($E_0$) is:

$ I = \frac{1}{2} \varepsilon_0 c E_0^2 $

where $\varepsilon_0$ is the permittivity of free space, and $c$ is the speed of light.

Step 2: Rearrange the formula to find $E_0$.

$ E_0 = \sqrt{ \frac{2I}{\varepsilon_0 c} } $

Step 3: Substitute the given values into the formula.

Given: $I = \frac{15}{\pi} \ \mathrm{Wm}^{-2}$, $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2$, $c = 3 \times 10^8 \ \mathrm{m/s}$

$ E_0 = \sqrt{ \frac{2 \times \frac{15}{\pi} }{ 8.85 \times 10^{-12} \times 3 \times 10^8 } } $

Step 4: Calculate the value.

When you solve the above, you get:

$ E_0 = 60 \ \mathrm{NC}^{-1} $

When light waves with a power of 600 W hit a surface that absorbs all the light (does not reflect any), the force on the surface can be found using a formula.

The formula is: $ F = \frac{P}{C} $

Here, $ F $ is the force, $ P $ is the power (600 W), and $ C $ is the speed of light ($ 3 \times 10^8 $ m/s).

Now substitute the values: $ F = \frac{600}{3 \times 10^8} $

Solve this to get: $ F = 2 \times 10^{-6} \text{ Newton} $

Velocity of electromagnetic wave in medium

$ \begin{aligned} v & =\frac{c}{\sqrt{\mu_r \varepsilon_r}}=\frac{3 \times 10^8}{\sqrt{200 \times 8}} \\ v & =7.5 \times 10^6 \mathrm{~m} / \mathrm{s} \\ \Rightarrow \quad v \lambda & =7.5 \times 10^6 \\ \Rightarrow \quad \lambda & =\frac{7.5 \times 10^6}{v}=\frac{7.5 \times 10^6}{100 \times 10^6} \\ & =0.075 \mathrm{~m}=7.5 \mathrm{~cm} \end{aligned} $

Given,

$ B_{0} $ (Peak value of magnetic field)

$ =30 \times 10^{-9} \mathrm{~T} $

Using the relation of $ B_{0} $ and $ E_{0} $ (Peak value of electric field)

$ E_{0}=c \times B_{0} $

where $ c $ is speed of light

$ \begin{array}{l} E_{0}=30 \times 10^{-9} \times 3 \times 10^{8} \\\\ E_{0}=9 \mathrm{~V} / \mathrm{m} \end{array} $

We know that the momentum transferred by EM wave $=\frac{U}{c}$ Also we know that, $\frac{E_0}{B_0}=c$

$ \therefore \text { Momentum transferred }=\frac{U}{E_0} \times B_0 $

Given that,

$ B_0=270 \mathrm{nT}=270 \times 10^{-9} \mathrm{~T} $

Speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$

Here, $E_0=B_0 \times c$

$ \begin{aligned} & E_0=270 \times 10^{-9} \times 3 \times 10^8=810 \times 10^{-1} \\ & E_0=81 \mathrm{NC}^{-1} \end{aligned} $

The amplitude of the electric fields part of the wave is $81 \mathrm{NC}^{-1}$.

A-R

X-rays are used to study the crystal structure by Bragg's law method.

B-Q

UV rays are used to study the finger print in forensic lab.

C-S

Radio waves are used in TV communication for sending the signals' to large distance.

D-P

IR rays are used in TV remotes to send signal for short distance.

Hence, A-R, B-Q, C-S, D-P is correct match.

Given, Intensity of EM radiation $=0.6 \mathrm{~W} / \mathrm{m}^2$ The radiation pressure is given by,

$ \begin{aligned} P & =\frac{\text { Intensity }}{\text { Speed of light }}=\frac{0.6}{3 \times 10^8} \\ & =2 \times 10^{-9} \mathrm{~N} / \mathrm{m}^2 \end{aligned} $

Given, speed of electromagnetic wave in medium

$ \begin{aligned} & v=1.5 \times 10^8 \mathrm{~ms}^{-1} \\ & \varepsilon_r=2 \end{aligned} $

We know that,

$ \begin{aligned} & & v & =\frac{c}{\sqrt{\mu_r \varepsilon_r}} \\ & \Rightarrow & \sqrt{\mu_r \varepsilon_r} & =\frac{c}{v}=\frac{3 \times 10^8}{1.5 \times 10^8}=2 \\ & \Rightarrow & \mu_r \varepsilon_r & =4 \\ & \Rightarrow & \mu_r & =\frac{4}{\varepsilon_r}=\frac{4}{2}=2 \end{aligned} $

We know that, magnetic susceptibility

$ \chi=\mu_r-1=2-1=1 $

Electromagnetic waves are unique in that they consist of oscillating electric and magnetic fields and do not require a medium to travel, which allows them to propagate through a vacuum.

These waves differ from mechanical waves as they are not longitudinal. Instead, electromagnetic waves are transverse, meaning that the oscillations of the fields are perpendicular to the direction of wave propagation.

Electromagnetic waves carry both energy and momentum. The energy carried by a wave is given by the equation:

$ E = h \nu $

where $ h $ is Planck's constant and $ \nu $ is the frequency of the wave.

The momentum carried by electromagnetic waves can be described using the equation:

$ p = \frac{h}{\lambda} $

where $ p $ is the momentum and $ \lambda $ is the wavelength of the wave. This relationship illustrates how electromagnetic waves convey momentum as they move through space.

To find the wavelength of a wave, you can use the formula:

$\lambda = \frac{c}{f}$

where:

$\lambda$ is the wavelength,

$c$ is the speed of light (in vacuum, which is $3 \times 10^8 \, \text{m/s}$),

$f$ is the frequency.

Given that:

$f = 3 \, \text{GHz} = 3 \times 10^9 \, \text{Hz}$

Plug the values into the formula:

$\lambda = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^9 \, \text{Hz}} = 0.1 \, \text{m}$

Thus, the wavelength of the wave is 0.1 m, which corresponds to Option A.

Given, frequency of electric field of EM wave,

$ f=30 \mathrm{MHz}=3 \times 10^7 \mathrm{~Hz} $

Amplitude, $E_0=150 \mathrm{~V} / \mathrm{m}$

∴ Amplitude of magnetic field vector,

$ B_0=\frac{E_0}{c}=\frac{150}{3 \times 10^8}=5 \times 10^7 \mathrm{~T} $

Angular frequency,

$ \begin{aligned} \omega & =2 \pi f=22 \pi \times 3 \times 10^7 \\ & =6 \pi \times 10^7 \mathrm{rad} / \mathrm{s} \end{aligned} $

∴ Propagation constant,

$ k=\frac{2 \pi}{\lambda}=\frac{2 \pi}{c / f}=\frac{2 \pi f}{c}=\frac{\omega}{c}=\frac{6 \pi \times 10^7}{3 \times 10^8}=\frac{\pi}{5} $

Since, $\mathbf{E}, \mathbf{B}$ and direction of propagation are mutually perpendicular to each other, hence, for the given situation $\mathbf{B}$ will be directed along $Z$-axis ∴ Expression of magnetic field,

$ \begin{aligned} B & =B_0 \sin [k x-\omega t] \hat{\mathbf{z}} \\ \Rightarrow \quad B & =5 \times 10^{-7} \sin \left[\frac{\pi}{5} x-6 \pi \times 10^7 t\right] \hat{\mathbf{z}} \\ & =5 \times 10^{-7} \sin \left[\pi\left(\frac{x}{5}-6 \times 10^7 t\right)\right] \hat{\mathbf{z}} \end{aligned} $

Average power delivered by sun to the top of earth's atmosphere,

$ \begin{gathered} P_{\text {avg }}=\frac{6}{\pi} \times 10^3 \mathrm{~W} / \mathrm{m}^2 \\ \frac{1}{2} \varepsilon_0 E_0^2 c=\frac{6}{\pi} \times 10^3 \\ E_0=\sqrt{\frac{6}{\pi} \times 10^3 \times \frac{2}{\varepsilon_0 \cdot c}} \\ =\sqrt{\frac{6}{\pi} \times 10^3 \times \frac{2}{8.85 \times 10^{-12} \times\left(3 \times 10^8\right)}} \\ \end{gathered} $

$ \begin{aligned} & =12 \times 10^2 \mathrm{~N} / \mathrm{C} \\ \therefore \quad B_0 & =\frac{E_0}{c} \\ & =\frac{12 \times 10^2}{3 \times 10^8}=4 \times 10^{-6} \mathrm{~T} \end{aligned} $

Intensity of laser beam,

$ I=2.1 \times 10^{15} \mathrm{~W} / \mathrm{m}^2 $

We know that, intensity of a EM wave

$ \begin{aligned} I & =\varepsilon_0 E_{\mathrm{rms}}^2 c \\ & =\varepsilon_0\left(\frac{E_0}{\sqrt{2}}\right)^2 c=\varepsilon_0 \frac{E_0^2}{2} c \\ & =\varepsilon_0 \frac{\left(B_0 c\right)^2}{2} c \quad\left[\because E_0=B_0 c\right] \end{aligned} $

$ \begin{aligned} & I=\varepsilon_0 \frac{B_0^2 c^3}{2} \\ & \Rightarrow B_0=\sqrt{\frac{2 I}{\varepsilon_0 c^3}}=\sqrt{\frac{2 I}{\varepsilon_0 \cdot \frac{1}{\mu_0 \varepsilon_0} \cdot c}} \\ & {\left[\because c^3=c^2 \cdot c=\frac{1}{\mu_0 \varepsilon_0} \cdot c\right] } \end{aligned} $

         $ \begin{aligned} & =\sqrt{\frac{2 \mu_0 I}{c}} \\ & =\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 2.1 \times 10^{15}}{3 \times 10^8}} \\ & =\sqrt{17.6}=4.195 \\ & =4.2 \mathrm{~T} \end{aligned} $

Given, power of bulb, $P=100 \mathrm{~W}$

$ r=5 \mathrm{~m} $

Power of visible radiation,

$ \begin{aligned} P^{\prime} & =20 \% \text { of } P \\ & =\frac{20}{100} \times 100=20 \mathrm{~W} \end{aligned} $

Average intensity of radiation,

$ \begin{aligned} I & =\frac{P^{\prime}}{4 \pi r^2} \\ & =\frac{20}{4 \pi \times 5^2}=\frac{5}{25 \pi} \mathrm{~W} / \mathrm{m}^2 \\ & =\frac{\alpha}{25 \pi} \mathrm{~W} / \mathrm{m}^2\,\,\,\text { (given) } \\ \therefore \alpha & =5 \end{aligned} $

Equations of EM waves are given as,

$ \begin{aligned} & \mathbf{E}(t)=\mathbf{E}_m \sin (k x-\omega t) \\ & \mathbf{B}(t)=\mathbf{B}_m \sin (k x-\omega t) \end{aligned} $

If $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ are unit vectors along $\mathbf{E}(t)$ and $\mathbf{B}(t)$ respectively, then

$ \hat{\mathbf{x}}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{1^2+1^2}}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}} $

and $\hat{\mathbf{y}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{1^2+1^2}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}$

∴ Unit vector in the direction of propagation of EM wave is $\hat{\mathbf{z}}$, then

$ \begin{aligned} \hat{\mathbf{z}} & =\hat{\mathbf{x}} \times \hat{\mathbf{y}} \\ & =\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \times\left(\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}\right) \\ & =\frac{1}{2}[\hat{\mathbf{i}} \times \hat{\mathbf{i}}-\hat{\mathbf{i}} \times \hat{\mathbf{j}}+\hat{\mathbf{j}} \times \hat{\mathbf{i}}-\hat{\mathbf{j}} \times \hat{\mathbf{j}}] \\ & =\frac{1}{2}[0-\hat{\mathbf{k}}-\hat{\mathbf{k}}-0]=\frac{1}{2} \times(-2 \hat{\mathbf{k}}) \\ & =-\hat{\mathbf{k}} \end{aligned} $

Hence, unit vector along the direction of propagation of EM wave will be $-\hat{\mathbf{k}}$.

Given, power of incident beam,

$ P=10 \mathrm{~W} $

Energy of EM wave, $E=\frac{h c}{\lambda}$

or Power or EM wave,

$ \begin{aligned} P & =\frac{E}{t}=\frac{h c}{\lambda t} \\ \Rightarrow \quad P & =\frac{h}{\lambda} \cdot\left(\frac{c}{t}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

But we know that, de-Broglie wavelngth

$ \lambda=\frac{h}{p} $

where, $p=$ momentum

$ \Rightarrow \quad p=\frac{h}{\lambda} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$ \begin{aligned} P & =p \frac{c}{t}=\frac{p}{t} \cdot c=F c \\ \Rightarrow \quad F & =\frac{P}{c} \end{aligned} $

$ \begin{aligned} & \therefore \text { Total effective force } \\ & \quad=F_{\text {absorbed }}+F_{\text {reflected }} \\ & \quad=70 \% \text { of } \frac{P}{c}+2 \times 30 \% \text { of } \frac{P}{c} \\ & \quad=0.7 \frac{P}{c}+0.6 \frac{P}{c} \\ & \quad=1.3 \frac{P}{c}=\frac{1.3 \times 10}{3 \times 10^8}=4.33 \times 10^{-8} \mathrm{~N} \end{aligned} $

Expression for the magnetic field of EM wave is given as

$ \mathbf{B}=\left(2 \times 10^{-8}\right) \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{k}} \mathrm{T} $

Magnitude of magnetic field vector,

$ B_0=2 \times 10^{-8} \mathrm{~T} $

Magnitude of electric field vector of EM wave,

$ E_0=B_0 c $

where, $c$ is speed of light

$ \begin{aligned} & =2 \times 10^{-8} \times 3 \times 10^8 \\ & =6 \mathrm{~V} / \mathrm{m} \quad\left(\because c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \end{aligned} $

If $\hat{\mathbf{n}}$ be the unit vector along the direction of electric field, then we know that unit vector along electric field vector $\mathbf{x}$

Unit vector along magnetic field vector $=$ Unit vector along the direction of propagation of EM wave.

$ \begin{aligned}i.e.\,\, \hat{\mathbf{n}} \times \hat{\mathbf{k}} & =-\hat{\mathbf{j}} \\clearly,\,\, \hat{\mathbf{n}} & =\hat{\mathbf{i}} \end{aligned} $

Hence, expression of electric field vector of EM wave,

$ \mathbf{E}=6 \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{i}} \mathrm{V} / \mathrm{m} $

Wavelength range of X-ray is $10^{-8} \mathrm{~m}$ to $10^{-12} \mathrm{~m}$. Hence, option (a) is correct.

Average intensity, $I_{\text {avg }}=\frac{P_{\text {avg }}}{A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $P_{\text {avg }}=$ average power generated by source

$ =4 \% \text { of } 100 \mathrm{~W}=\frac{4}{100} \times 100 \mathrm{~W}=4 \mathrm{~W} $

$A=$ area to which energy is transmitted

$ =4 \pi r^2=4 \pi \times(2)^2=4 \pi \times 4=16 \pi \mathrm{~m}^2 $

On putting the values into Eq. (i), we have

$ \begin{equation*} I_{\mathrm{avg}}=\frac{4}{16 \pi} \Rightarrow I_{\mathrm{avg}}=\frac{1}{4 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

Also average intensity,

$ \begin{equation*} I_{\mathrm{avg}}=\varepsilon_0 E_{\mathrm{rms}}^2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{equation*} $

where, $\varepsilon_0=$ permittivity of free space and $E_{\text {rms }}=$ root mean square (rms) value of electric field.

$ \begin{aligned} c & =\text { speed of light in vacuum } \\ & =3 \times 10^8 \mathrm{~m} / \mathrm{s} \end{aligned} $

On putting the values into Eq. (iii), we get

$ \begin{equation*} I_{\mathrm{avg}}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{equation*} $

On equating both the Eqs. (ii) and (iv), we get

$ \frac{1}{4 \pi}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 $

$ \begin{aligned} & \Rightarrow \frac{1}{4 \pi \varepsilon_0 \times 3 \times 10^8}=E_{\mathrm{rms}}^2 \\ & E_{\mathrm{rms}}^2= \frac{1}{4 \pi \varepsilon_0} \times \frac{1}{3 \times 10^8} \\ &= 9 \times 10^9 \times \frac{1}{3 \times 10^8} \\ &\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\ &= 3 \times 10=\sqrt{30} \mathrm{~V} / \mathrm{m} \end{aligned} $

Electric field between plates of a capacitor is

$ E=V / d $

But $V=\frac{Q}{C}$ and $C=\frac{\varepsilon_0 A}{d} \Rightarrow Q=A E \varepsilon_0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \therefore \quad E=\frac{Q d}{\varepsilon_0 A d}=\frac{Q}{A \varepsilon_0} $

Displacement currrent is

$ \begin{aligned} I_d & \left.=\frac{d Q}{d t}=\frac{d}{d t}\left(E \cdot A \varepsilon_0\right) \quad \text { [from Eq. (i)] }\right .\\ & =\varepsilon_0 A\left(\frac{\Delta E}{\Delta t}\right)=\frac{4 \pi \varepsilon_0 A}{4 \pi} \times\left(\frac{\Delta E}{\Delta t}\right) \\ & =\frac{4 \pi \varepsilon_0 \cdot \pi r^2}{4 \pi} \cdot\left(\frac{\Delta E}{\Delta t}\right) \\ \Rightarrow I_d & =4 \pi \varepsilon_0\left(\frac{r^2}{4}\right)\left(\frac{\Delta E}{\Delta t}\right) \end{aligned} $

Substituting values given, we have

$ \begin{aligned} I_d & =\frac{1}{9 \times 10^9}\left(\frac{\left(10 \times 10^{-2}\right)^2}{4}\right) \times 3.6 \times 10^{12} \\ & =1 \mathrm{~A} \quad\left[\text { Since, } \frac{\Delta E}{\Delta t}=3.6 \times 10^{12} \mathrm{~V} / \mathrm{ms}\right] \end{aligned} $

Intensity of parallel beam of light is given as

$ I=\frac{1}{2} \varepsilon_0 E_0^2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $E_0=$ amplitude of electric field.

Here, $\quad I=\frac{15}{\pi} \frac{\mathrm{~W}}{\mathrm{~m}^2}, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$

$ c=3 \times 10^8 \mathrm{~ms}^{-1} $

From Eq. (i), we get

$ \begin{aligned} & E_0^2=\frac{2 \times I}{\varepsilon_0 \times c}=\frac{2 \times 4 \pi \times I}{4 \pi \varepsilon_0 \times c} \\ & \quad=\left(\frac{2 \times 4 \pi \times \frac{15}{\pi} \times 9 \times 10^9}{3 \times 10^8}\right) \\ & \Rightarrow \quad E_0^2=3600 \Rightarrow E_0=60 \mathrm{~N} / \mathrm{C} \end{aligned} $