Electromagnetic Waves

In an electromagnetic wave traveling through a vacuum, the amplitudes of the electric field ( $\mathrm{E}_0$ ) and the magnetic field ( $B_0$ ) are related by the speed of light ( $c$ ):

$ c=\frac{E_0}{B_0} \Rightarrow B_0=\frac{E_0}{c} $

Given that $\mathrm{E}_0=54 \mathrm{~V} / \mathrm{m}$ and $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$

$ B_0=\frac{54}{3 \times 10^8} $

$ \mathrm{B}_0=1.8 \times 10^{-7} \mathrm{~T} $

In a transverse electromagnetic wave, the electric field (E), magnetic field (B), and the direction of propagation (v) are all mutually perpendicular. The direction follows the right-hand rule:

$ \widehat{\mathrm{E}} \times \widehat{\mathrm{B}}=\widehat{\mathrm{v}} $

The variable $z$ in phase expression ( $k z-\omega t$ ) indicates the wave is moving along the $+z$ direction ( $\hat{k}$ ).

The unit vector along direction of electric field ( $\overrightarrow{\mathrm{E}}$ ) is given as $\hat{\mathrm{j}}$.

We need a unit vector such that $\hat{\mathrm{j}} \times(\widehat{\mathrm{B}})=\hat{\mathrm{k}}$.

We know that $\hat{\jmath} \times \hat{i}=-\hat{\mathrm{k}}$ and $-\hat{\jmath} \times \hat{i}=\hat{\mathrm{k}}$

Therefore, if $\widehat{\mathrm{E}}=\hat{\mathrm{j}}$ and $\widehat{\mathrm{v}}=\widehat{\mathrm{k}}$, then $\widehat{\mathrm{B}}$ must be $-\hat{\mathrm{i}}$ because $\hat{\mathrm{j}} \times(-\mathrm{i})=-(\hat{\mathrm{j}} \times \hat{\mathrm{i}})=-(-\widehat{\mathrm{k}})=\widehat{\mathrm{k}}$.

Combining the magnitude and the direction:

$ \overrightarrow{\mathrm{B}}=-1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{~T} $

Hence, the correct option is (A).

The standard equation for a wave travelling in the +x direction is :

$ \mathrm{E}(\mathrm{x}, \mathrm{t})=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx}) $

The given equation of electromagnetic wave is $\overrightarrow{\mathrm{E}}(\mathrm{x}, \mathrm{t})=25 \sin \left(2.0 \times 10^{15} \mathrm{t}-10^7 \mathrm{x}\right) \hat{\mathrm{n}}$ :

Comparing both the equations,

Angular frequency $=\omega=2.0 \times 10^{15} \mathrm{rad} / \mathrm{s}$

Wave number $=\mathrm{k}=10^7 \mathrm{~m}^{-1}$

The speed of the electromagnetic wave in the medium is given by the ratio of angular frequency to wave number:

$ v=\frac{\omega}{k} $

$ \mathrm{v}=\frac{2.0 \times 10^{15}}{10^7}=2.0 \times 10^8 \mathrm{~m} / \mathrm{s} $

The refractive index of a medium is the ratio of the speed of light in a vacuum (c) to the speed of light in that medium (v) :

$ \mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}} $

Taking the speed of light in vacuum as $\mathrm{c}=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$ :

$ \mathrm{n}=\frac{3.0 \times 10^8}{2.0 \times 10^8} $

$ \mathrm{n}=1.5 $

Hence, the correct option is (C).

Radio-waves

• 1. Radio waves have the longest wavelengths in the EM spectrum. They are artificially generated by transmitters. The core process involves an oscillating electric current in a special conductor called an antenna.

• 2. According to electromagnetic theory, an accelerating electric charge produces electromagnetic waves. In an antenna, electrons are rapidly accelerated back and forth.

• 3. So, the radio waves are produced due to rapid acceleration of electrons.

Micro-waves

• 1. Micro-waves are high-frequency radio waves. While they can be produced by acceleration, high-power microwaves (like those in radar or microwave ovens) are generated using special vacuum tubes.

• 2. The most common device for this is the Magnetron (or Klystron). A Magnetron uses a magnetic field to control the flow of electrons in a vacuum, causing them to oscillate at microwave frequencies.

• 3. Hence the Micro-wave is produced by Magnetron valve.

Infrared-waves

• 1. Infrared waves are often perceived as heat. All objects above absolute zero emit infrared radiation.

• 2. This emission arises from the thermal motion of atoms and molecules. When the atoms in a molecule vibrate or rotate, they change energy states. The transition between these vibrational and rotational energy levels results in the emission of infrared photons.

• 3. Hence, the infrared waves are produced due to change in the vibrational modes of atoms.

X-rays

• 1. X-rays have very high energy and short wavelengths. They are typically produced in X-ray tubes.

• 2. There are two main ways X-rays are produced:

1. (A) Braking Radiation: When high-speed electrons are suddenly stopped by a metal target.

2. (B) Characteristic X-rays: When a high-energy electron knocks an electron out of an inner shell (like the K-shell) of an atom, a vacancy is created. An electron from a higher energy outer shell falls into this vacancy. The large energy difference is released as an X-ray photon.

• 3. Hence, it is due to inner shell electrons moving from higher energy level to lower energy level.

So,

A (Radio-wave) → IV

B (Micro-wave) → I

C (Infrared-wave) → II

D (X-ray) → III

Hence, the correct option is (C).

The speed of light (or any electromagnetic wave) in a vacuum is given by the formula :

$ c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} $

Where :

• 1. $\mu_0$ is the permeability of free space.

• 2. $\epsilon_0$ is the permittivity of free space.

The speed of light in a specific medium is determined by that medium's permeability ( $\mu$ ) and permittivity ( $\epsilon$ ):

$ v=\frac{1}{\sqrt{\mu \epsilon}} $

For the given medium :

Permeability : $\mu=2 \mu_0$

Dielectric Constant $\mathrm{k}=3$. The dielectric constant (relative permittivity $\epsilon_{\mathrm{r}}$ ) is defined as $\epsilon_{\mathrm{r}}=\frac{\epsilon}{\epsilon_0}$. Therefore, $\epsilon=\mathrm{k} \epsilon_0=3 \epsilon_0$.

We need to find the ratio $c: v$, which is the same as the refractive index $n$ of the medium.

$ \frac{c}{v}=\frac{1 / \sqrt{\mu_0 \epsilon_0}}{1 / \sqrt{\mu \epsilon}} $

$\Rightarrow $ $ \frac{c}{v}=\sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}}=\sqrt{\frac{\left(2 \mu_0\right)\left(3 \epsilon_0\right)}{\mu_0 \epsilon_0}} $

For the medium $\mu=2 \mu_0$ and $\epsilon=3 \epsilon_0$ :

$ \frac{c}{v}=\sqrt{\frac{\left(2 \mu_0\right)\left(3 \epsilon_0\right)}{\mu_0 \epsilon_0}}=\frac{\sqrt{6}}{1} $

The ratio of the speed in vacuum to the speed in the medium is $\sqrt{6}: 1$.

Hence, the correct option is (C).

For relation A

The equation $\oint \vec{E} \cdot d \vec{l}=-\frac{d}{d t}[\oint \vec{B} \cdot d \vec{a}]$ mathematically states that the line integral of the electric field around a closed loop is equal to the negative rate of change of magnetic flux through the loop. This is the integral form of Faraday's Law of Electromagnetic Induction, i.e., $\mathrm{A} \rightarrow$ II

For relation B

This equation $\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}=\mu_0\left(\mathrm{I}+\epsilon_0 \frac{\mathrm{~d} \Phi_{\mathrm{E}}}{\mathrm{dt}}\right)$ relates the magnetic field line integral to both the conduction current (I) and the displacement current $\left(\epsilon_0 \frac{\mathrm{~d} \Phi_{\mathrm{E}}}{\mathrm{dt}}\right)$. This modification to Ampere's Law was introduced by Maxwell to account for changing electric fields. This is the Ampere-Maxwell Law, i.e., $\mathrm{B} \rightarrow$ III

For relation C

In relation $C, \oint \vec{E} \cdot d \vec{a}=\frac{1}{\epsilon_0} \int_V \rho d V$ the total charge $Q_{\text {enclosed }}$ is represented as $\oint \vec{E} \cdot d \vec{a}$. The equation states that the total electric flux through a closed surface is equal to the enclosed charge divided by $\epsilon_0$. This is Gauss's Law of Electrostatics, i.e., $\mathrm{C} \rightarrow$ IV

For relation D

This relation $\oint \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{l}}=\mu_0 \mathrm{I}$ states that the line integral of the magnetic field around a closed loop is equal to $\mu_0$ times the current enclosed by the loop. This is the original form of Ampere's Circuital Law which is valid for steady currents, i.e., $\mathrm{D} \rightarrow \mathrm{I}$

So, the correct match is $\mathrm{A} \rightarrow$ II, $\mathrm{B} \rightarrow$ III, $\mathrm{C} \rightarrow$ IV, $\mathrm{D} \rightarrow$ I. Hence, the correct option is (B).

The intensity I of an electromagnetic wave is related to the maximum electric field ( $\mathrm{E}_0$ ) and the maximum magnetic field $\left(\mathrm{B}_0\right)$. The formula for the intensity to the magnetic field amplitude is :

$ I=\frac{c B_0^2}{2 \mu_0} $

Where :

• I is the intensity.

• c is the speed of light in a vacuum.

• $\mathrm{B}_0$ is the amplitude of the magnetic field.

• $\mu_0$ is the permeability of free space.

The relationship between the speed of light (c), permeability ( $\mu_0$ ), and permittivity ( $\epsilon_0$ ) of free space is,

$ c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \Rightarrow \mu_0=\frac{1}{\epsilon_0 c^2} $

By substituting $\mu_0$ into the intensity equation, we get a formula directly relating the given variables :

$ I=\frac{c B_0^2}{2\left(\frac{1}{\epsilon_0 c^2}\right)}=\frac{1}{2} c^3 \epsilon_0 B_0^2 $

The intensity of the laser beam is, $\mathrm{I}=4.0 \times 10^{14} \mathrm{~W} / \mathrm{m}^2$

The permittivity of free space, $\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{N} \cdot \mathrm{m}^2$

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

From the intensity formula, $B_0=\sqrt{\frac{2 I}{c^3 \epsilon_0}}$

Putting the values,

$ B_0=\sqrt{\frac{2 \times 4.0 \times 10^{14}}{\left(3 \times 10^8\right)^3 \times 8.85 \times 10^{-12}}}=\sqrt{\frac{8.0 \times 10^{14}}{27 \times 10^{24} \times 8.85 \times 10^{-12}}} $

$ \mathrm{B}_0=\sqrt{\frac{8.0}{238.95} \times 10^2} \approx 1.83 \mathrm{~T} $

Therefore, the amplitude of the magnetic field associated with beam is approximately 1.83 T.

Hence, the correct option is (A).

In a vacuum or air, the amplitudes of the electric field ( $\mathrm{E}_0$ ) and magnetic field ( $\mathrm{B}_0$ ) are related by the speed of light (c) :

$ c=\frac{E_0}{B_0} \Rightarrow B_0=\frac{E_0}{c} $

From the given equation $\mathrm{E}_{\mathrm{y}}=69 \sin \left[0.6 \times 10^3 \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right], \mathrm{E}_0=69 \mathrm{~V} / \mathrm{m}, \omega$ (angular frequency) $= 1.8 \times 10^{11} \mathrm{rad} / \mathrm{s}$ and k (wave number) $=0.6 \times 10^3 \mathrm{rad} / \mathrm{m}$

So, the amplitude of the magnetic field is,

$ B_0=\frac{69}{3 \times 10^8}=23 \times 10^{-8}=2.3 \times 10^{-7} T $

In electromagnetic waves which is a transverse waves consists of the electric field ( $\overrightarrow{\mathrm{E}}$ ), magnetic field ( $\overrightarrow{\mathrm{B}}$ ), and the direction of propagation ( $\overrightarrow{\mathrm{s}}$ ) are all mutually perpendicular.

The wave is traveling along the +x direction and the electric field is oscillating along the y -axis.

Using the direction vector $\hat{\mathrm{s}}=\widehat{\mathrm{E}} \times \widehat{\mathrm{B}}$ :

$ \hat{\imath}=\hat{\jmath} \times \widehat{B} $

For this to be true, the magnetic field must be along the $\mathrm{z}-\operatorname{axis}(\hat{\mathrm{k}})$, because $\hat{\mathrm{j}} \times \hat{\mathrm{k}}=\hat{\mathrm{i}}$.

In an electromagnetic wave in a vacuum, the electric and magnetic fields are in phase.

So, the phase of magnetic field vector is, $\left(0.6 \times 10^3 \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right)$

Therefore, the magnetic field vector is $\mathrm{B}_{\mathrm{z}}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^3 \mathrm{x}-1.8 \times 10^{11} \mathrm{t}\right]$ T.

Hence, the correct option is (A).

In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and the propagation direction $\vec{k}$ are mutually perpendicular to each other. The wave travels in the direction of the cross product $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$.

Also the magnetic field vector is in the direction of $\widehat{\mathrm{k}} \times \widehat{\mathrm{E}}$

The magnitude of magnetic field is given as $\mathrm{B}=\frac{\mathrm{E}}{\mathrm{c}}$, where c is the speed of light.

Maxwell's equation (Faraday's Law) in differential form is :

$ \nabla \times \overrightarrow{\mathrm{E}}=-\frac{\partial \overrightarrow{\mathrm{B}}}{\partial \mathrm{t}} $

For a plane monochromatic wave, the fields can be represented asb :

$ \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}}_0 \mathrm{e}^{\mathrm{i}(\overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}}-\omega \mathrm{t})} $

$ \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_0 \mathrm{e}^{\mathrm{i}(\overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}}-\omega \mathrm{t})} $

Applying the curl operator ( $\nabla \times$ ) to the electric field expression :

$ \nabla \times \overrightarrow{\mathrm{E}}=\mathrm{i} \overrightarrow{\mathrm{k}} \times \overrightarrow{\mathrm{E}} $

Applying the time derivative to the magnetic field expression :

$ \frac{\partial \overrightarrow{\mathrm{B}}}{\partial \mathrm{t}}=-\mathrm{i} \omega \overrightarrow{\mathrm{~B}} $

Substituting into Faraday's Law :

$ \mathrm{i} \overrightarrow{\mathrm{k}} \times \overrightarrow{\mathrm{E}}=-(-\mathrm{i} \omega \overrightarrow{\mathrm{~B}}) $

$\Rightarrow $ $\mathrm{i} \overrightarrow{\mathrm{k}} \times \overrightarrow{\mathrm{E}}=\mathrm{i} \omega \overrightarrow{\mathrm{B}}$

$\Rightarrow $ $ \overrightarrow{\mathrm{B}}=\frac{\overrightarrow{\mathrm{k}} \times \overrightarrow{\mathrm{E}}}{\omega} $

Therefore, the correct option is (B)

The intensity (I) of electromagnetic waves from a point source in free space follows the inverse square law.

$ I=\frac{P}{4 \pi r^2} $

Where P is the power of the source and r is the distance from the source.

Initial distance is L and the intensity recorded is $\mathrm{I}_0$.

It is given that the detector is shifted to another location on the same spherical surface.

On a spherical surface is all points are at a constant distance r from the center (the source).

Since the detector remains on the same spherical surface, its distance from the point source remains L .

The angle ( $45^{\circ}$ ) represents a change in direction, but not a change in distance r .

As r is constant $\mathrm{r}_1=\mathrm{r}_2=\mathrm{L}$ :

$ I_{\text {new }}=\frac{P}{4 \pi L^2}=I_0 $

Therefore, the measured intensity at new location will be $\mathrm{I}_0$. Hence, the correct option is (B).

Assertion (A) is true.

Electromagnetic waves carry momentum as well as energy. When they fall on a surface, they transfer momentum to that surface. Due to this transfer of momentum, the surface experiences a pressure, called radiation pressure.

So, (A) is true.

Reason (R) is also true.

Electromagnetic waves do not have rest mass associated with them.

But this is not the explanation of why electromagnetic waves exert pressure. They exert pressure because they carry momentum, not because they have no mass.

Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).

Hence, the correct answer is:

$\boxed{\text{Option B}}$

For a capacitor, the displacement current is given by

$ I_d = C \frac{dV}{dt} $

Given:

$ I_d = 4.0 \, \text{A} $

$ C = 6\,\mu\text{F} = 6 \times 10^{-6}\,\text{F} $

We need to find:

$ \frac{dV}{dt} $

Using the formula,

$ \frac{dV}{dt} = \frac{I_d}{C} $

Substitute the values:

$ \frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} $

$ = \frac{2}{3} \times 10^6 $

$ = 0.67 \times 10^6 \,\text{V/s} $

So,

$ \alpha = 0.67 $

Hence, the correct option is B.

Given,

$ \vec{B}=B_0 \sin \left(2 \pi \nu t-\frac{2 \pi x}{\lambda}\right)\hat{j} $

We compare it with the standard wave form

$ \sin(\omega t-kx) $

This represents a wave travelling in the $+x$ direction.

In an electromagnetic wave:

• $\vec{E}$, $\vec{B}$ and direction of propagation are mutually perpendicular.

• Direction of propagation is given by $\vec{E} \times \vec{B}$.

Here,

• propagation is along $+\hat{i}$

• magnetic field is along $+\hat{j}$

So electric field must be along $\hat{k}$ or $-\hat{k}$.

Now use

$ \vec{E} \times \vec{B} = +\hat{i} $

If $\vec{B}$ is along $\hat{j}$, then for

$ \hat{k} \times \hat{j} = -\hat{i} $

which is wrong.

But

$ (-\hat{k}) \times \hat{j} = \hat{i} $

So $\vec{E}$ must be along $-\hat{k}$.

Also, for electromagnetic waves,

$ E_0 = c B_0 $

and since

$ c=\nu \lambda $

we get

$ E_0=\nu \lambda B_0 $

Hence,

$ \vec{E}=-\nu \lambda B_0 \sin \left(2 \pi \nu t-\frac{2 \pi x}{\lambda}\right)\hat{k} $

So the correct answer is:

$ \boxed{\vec{E}=-\nu \lambda B_0 \sin \left(2 \pi \nu t-\frac{2 \pi x}{\lambda}\right)\hat{k}} $

Option A.

The unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ can be determined by exploring the relationship between intensity ($I$) and the electric field ($E_0$) in an electromagnetic wave.

The intensity of an electromagnetic wave is given by:

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

From this equation, we can solve for the electric field $E_0$:

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

The symbol $E_0$ represents the electric field, which is measured in units of Newtons per Coulomb (N/C). Therefore, the unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is also $N/C$.

Thus, the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ has units of $N/C$.

We know, for a plane electromagnetic wave,

$\widehat E \times \widehat B = \widehat C$

Given, $\widehat C = \widehat i$ (+ x direction)

So, $\widehat E \times \widehat B = \widehat i$

i.e. $\widehat j \times \widehat k = \widehat i$

So, $\overrightarrow E = {E_y}$

$\overrightarrow B = {B_2}$

As, $\widehat i,\,\widehat j$ and $\widehat k$ are unit vectors in the directions of x, y and z axes respectively.

Hence, option 3 is correct.

Assertion is false because em waves have momentum.

$\begin{aligned} & \vec{B}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \\ & \vec{E}=\vec{B} \times \vec{c} \text { and } E=B_0 c \\ & \text { Here } \vec{E}\left(\frac{\sqrt{3}}{2}(-\hat{j})+\frac{1}{2} \hat{i}\right) \\ & E_0=30 c \\ & \vec{E}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 c \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \end{aligned}$

We know,

Energy density $ = {1 \over 2}{\varepsilon _0}{E^2}C$

$ \Rightarrow {\text{Energy} \over \text{Volume}} = {1 \over 2}{\varepsilon _0}{E^2}C$

$ \Rightarrow \text{Energy}= {1 \over 2}{\varepsilon _0}E_c^2x\,vol$

Given that, ${\left( {\text{Energy}} \right)_1} = {\left( {\text{Energy}} \right)_2}$

$ \Rightarrow {1 \over 2}{\varepsilon _0}E_1^2cx\pi R_1^2{L_1} = {1 \over 2}{\varepsilon _0}E_2^2cx\pi R_2^2{L_2}$

$ \Rightarrow E_1^2R_1^2{L_1} = E_2^2R_2^2{L_1}\,(as\,{L_1} = {L_2})$

$ \Rightarrow E_1^2R_1^2 = E_2^2R_2^2$

$ \Rightarrow {E_1}{R_1} = {E_2}{R_2}$

$ \Rightarrow {E_1}{R_1} = {E_2}\left( {{{{R_1}} \over 2}} \right)\,\left( {As\,{R_2} = {{{R_1}} \over 2}} \right)$

$ \Rightarrow {E_2} = 2{E_1} = 2 \times 100$

$ \Rightarrow {E_2} = 200\,N/C$

Hence, option (3) is correct.

For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, $c \approx 3 \times 10^8 \, \mathrm{m/s},$ through the equation

$ E = cB. $

Here's how to determine the magnetic field:

We are given the electric field magnitude: $E=9.3 \, \mathrm{V/m}.$

Use the relationship to find the magnetic field magnitude:

$ B = \frac{E}{c} = \frac{9.3}{3 \times 10^8} \, \mathrm{T}. $

Calculating the value gives:

$ B \approx 3.1 \times 10^{-8} \, \mathrm{T}. $

Since the electric field is along the $y$ direction and the wave propagates along the $+x$ direction, by the right-hand rule, the magnetic field is oriented along the $z$ direction.

Thus, the magnetic field vector is

$ \mathrm{B}_z = 3.1 \times 10^{-8} \, \mathrm{T}, $

which corresponds to Option C.

We start with the given electric field:

$ \vec{E} = 57 \cos\Bigl[7.5\times10^6\, t - 5\times10^{-3}(3x+4y)\Bigr](4\,\hat{i} - 3\,\hat{j}) \quad \text{(N/C)} $

Since we are dealing with an electromagnetic wave in free space, the electric field, magnetic field, and the propagation direction are all mutually perpendicular. In free space the magnetic field is related to the electric field by:

$ \vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E}, $

where:

• $c = 3\times10^8 \, \text{m/s}$ is the speed of light, and

• $\hat{n}$ is the unit vector in the direction of wave propagation.

Step 1. Identify the propagation direction:

The phase of the wave is written as

$ 7.5\times10^6\,t - 5\times10^{-3}(3x+4y). $

This tells us that the wave vector, up to a multiplicative constant, is

$ \vec{k} = 5\times10^{-3}(3\,\hat{i} + 4\,\hat{j}). $

The unit vector in the direction of propagation is then

$ \hat{n} = \frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}+4\hat{j}}{5}. $

Step 2. Compute the cross product:

We have

$ \vec{E} \propto (4\,\hat{i}-3\,\hat{j}), $

so we need to calculate

$ \hat{n} \times \vec{E} = \frac{1}{5}(3\hat{i}+4\hat{j}) \times \Bigl[57 \cos(\ldots)(4\hat{i}-3\hat{j})\Bigr]. $

Taking the constant factors out of the cross product gives

$ \hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,\Bigl[(3\hat{i}+4\hat{j}) \times (4\hat{i}-3\hat{j})\Bigr]. $

Recall that the cross product in the plane (where both vectors have zero $\hat{k}$ components) will point in the $\hat{k}$ (or $-\hat{k}$) direction. To compute the cross product of vectors in the $xy$-plane, we use the formula

$ \vec{a} \times \vec{b} = (a_xb_y - a_yb_x)\,\hat{k}. $

Let

• $\vec{a} = 3\hat{i}+4\hat{j},$

• $\vec{b} = 4\hat{i}-3\hat{j}.$

Then

$ \vec{a} \times \vec{b} = \Bigl(3\cdot (-3) - 4\cdot 4\Bigr)\,\hat{k} = (-9 - 16)\,\hat{k} = -25\,\hat{k}. $

Thus,

$ \hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,(-25\,\hat{k}) = -\frac{57 \times 25}{5}\, \cos(\ldots) \,\hat{k}. $

Simplify the constant:

$ \frac{25}{5} = 5, $

so we get

$ \hat{n} \times \vec{E} = -\frac{57 \times 5}{1} \cos(\ldots)\,\hat{k} = -285 \cos(\ldots) \,\hat{k}. $

Step 3. Apply the relation for the magnetic field:

Now, substituting back into the formula for $\vec{B}$,

$ \vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E} = -\frac{285}{3\times10^8} \cos(\ldots)\,\hat{k}. $

Notice that

$ \frac{285}{3\times10^8} = \frac{57 \times 5}{3\times10^8}, $

so we have

$ \vec{B} = -\frac{57}{3\times10^8} \cos\Bigl[7.5\times10^6\,t - 5\times10^{-3}(3x+4y)\Bigr](5\,\hat{k})\quad \text{(T)}. $

This expression exactly matches Option D.

Final Answer: Option D

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

$ \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} $

$ \begin{aligned} & I=\frac{P}{A} \\ & =\frac{200 \times 0.11}{4 \pi r^2}=\frac{22}{4 \pi\left(1^2\right)}=1.75 \mathrm{~W} / \mathrm{m}^2 \end{aligned} $

Given:

• Power of bulb $ P = 10\ \text{W} $

• Distance $ r = 0.5\ \text{m} $

• The bulb emits uniformly in all directions (i.e., isotropically).

Step 1: Formula for intensity

$ I = \frac{P}{A} = \frac{P}{4\pi r^2} $

Step 2: Substitute values

$ I = \frac{10}{4\pi (0.5)^2} $

$ I = \frac{10}{4\pi \times 0.25} = \frac{10}{\pi} $

Step 3: Simplify numerically

$ I \approx \frac{10}{3.14} \approx 3.18\ \text{W m}^{-2} $

✅ Final Answer:

$ \boxed{I \approx 3.18\ \text{W m}^{-2}} $

Correct Option: A.

Question:

The layer of the atmosphere that reflects low frequency (LF) electromagnetic waves during daytime only is —

Step 1: Recall the ionospheric layers

The ionosphere has several layers:

• D layer — lowest layer, present only during the daytime

• E layer

• F₁ and F₂ layers — present at higher altitudes

Step 2: Function of the D layer

• The D layer is ionized by solar radiation and disappears at night (it recombines quickly when sunlight is gone).

• It primarily affects low-frequency (LF) and medium-frequency (MF) radio waves.

• It does not strongly reflect HF waves—it mostly causes absorption at higher frequencies.

Step 3: Reflection characteristic

• LF waves (30–300 kHz) are reflected by the D layer during daytime only.

• At night, the D layer disappears, and these LF waves can instead travel farther by ground or sky wave propagation.

✅ Final Answer:

Option A — D layer

Answer:

$ \boxed{\text{A. D layer}} $

In an electromagnetic wave, oscillating electric and magnetic fields increases and decreases simultaneously i.e., they are always in same phase. Also $\mathbf{E}$ and $\mathbf{B}$ are always perpendicular to each other.

The electromagnetic spectrum (in order of increasing wavelength) is:

Gamma rays → X-rays → Ultraviolet (UV) → Visible light → Infrared → Microwaves → Radio waves

So, radio waves have the maximum wavelength among the given options.

✅ Correct answer: Option B — Radio waves

The electric field ($E_0$) and magnetic field ($B_0$) in a plane electromagnetic wave have a special relationship.

We know that $\frac{E_0}{B_0} = c$ where $c$ is the speed of light, which is $3 \times 10^8$ m/s.

If we compare $E_0$ to $10^8$ times $B_0$, we get: $\frac{E_0}{10^8 B_0} = \frac{E_0}{B_0} \times \frac{1}{10^8}$

Now substitute $c$ for $\frac{E_0}{B_0}$: $\frac{E_0}{10^8 B_0} = \frac{3 \times 10^8}{10^8} = 3$

So, the answer is: $\frac{E_0}{10^8 B_0} = 3$

$ \begin{aligned} &\text { Power of the source }\\ &\begin{aligned} P & =I \times 4 \pi r^2 \\ & =\frac{1}{2} C \varepsilon_0 E_{\mathrm{rms}}^2 \times 4 \pi r^2 \\ & =\frac{1}{2} \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times 3^2 \times 4 \pi \times 3^2\\ & =2.7 \mathrm{~W} \end{aligned} \end{aligned} $

Step 1: Relationship between $ E $ and $ B $ in an electromagnetic wave

For a plane electromagnetic wave in free space,

$ E = c B $

where $ c $ is the speed of light in free space.

Thus,

$ B = \frac{E}{c}. $

Step 2: Express $ c $ in terms of $ \mu_0 $ and $ \varepsilon_0 $

The speed of light is given by

$ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}. $

Step 3: Substitute into expression for $ B $

$ B = \frac{E}{c} = E \sqrt{\mu_0 \varepsilon_0}. $

✅ Final Answer:

$ \boxed{B = E \sqrt{\mu_0 \varepsilon_0}} $

Correct Option: C

$ \text { } \begin{aligned} B & =\frac{F}{c}=\frac{6.3}{3 \times 10^8} \\ & =2.1 \times 10^{-8} \mathrm{~T} \end{aligned} $

Magnetic energy stored per unit volume

$ \begin{aligned} & =\frac{B^2}{2 \mu_0}=\frac{B^2}{2 \times \frac{1}{\varepsilon_0 c^2}} \quad\left[c^2=\frac{1}{\mu_0 \varepsilon_0}\right] \\ & =\frac{\varepsilon_0 B^2 c^2}{2} \end{aligned} $

$\mathbf{B}=3 \times 10^{-7} \sin \left(100 \pi x+10^{12} t\right) \mathrm{T}$

Here, $K=100 \pi \mathrm{~m}^{-1}$

$ \therefore \quad \lambda=\frac{2 \pi}{K}=\frac{2 \pi}{100 \pi}=0.02 \mathrm{~m} $

For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($E$) and the magnetic field ($B$) can be described using the equation:

$E = cB$

where

• $c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \text{m/s}$.
• $B$ is the magnitude of the magnetic field.

Given the magnetic field $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}$, we can calculate the corresponding electric field magnitude using the formula above:

$E = (3.0 \times 10^8) \times (3.5 \times 10^{-7})$

$= 105 \, \text{Vm}^{-1}$

Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, \text{Vm}^{-1}$. The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation. Given $B_y$, this means $E$ will have components in the $x-z$ plane. Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$-direction, the corresponding electric field component must lie in a plane perpendicular to the $y$-axis, which could be either the $x$ or the $z$ direction.

However, knowing electromagnetic wave properties, if the wave is propagating along the $x$-axis and the magnetic field ($B_y$) is along the $y$-axis, then by right-hand rule, the electric field ($E$) must be along the $z$-axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.

Therefore, the correct option is:

$\text{Option A: } E_z = 105 \sin \left(1.5 \times 10^3 x + 0.5 \times 10^{11} t\right) \, \text{Vm}^{-1}$

To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.

Firstly, the wavelength ($ \lambda $) of the wave is given as $4 \mathrm{~mm} = 4 \times 10^{-3} \mathrm{~m}$. The speed of light (and all electromagnetic waves in vacuum) is $ c = 3 \times 10^8 \mathrm{~m/s} $. Using these values, we can find the frequency ($ f $) of the wave using the relationship:

$ c = \lambda f $

$ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-3}} = 75 \times 10^9 \mathrm{~Hz} $

The angular frequency ($ \omega $) which appears in wave equations is related to the frequency by $ \omega = 2\pi f $. However, in this context, what we need is the wave vector ($ k $), which defines how the phase of the wave changes with space. The wave vector $ k = \frac{2\pi}{\lambda} $. So, for this wave, $ k = \frac{2\pi}{4 \times 10^{-3}} = \frac{\pi}{2} \times 10^3 \, \mathrm{m}^{-1} $.

Knowing that the electric field ($ \mathbf{E} $) and magnetic field ($ \mathbf{B} $) are related as $ E = cB $ in a vacuum, where $ E $ and $ B $ are the magnitudes of the electric and magnetic fields, respectively, we can calculate the magnitude of the magnetic field using the provided maximum electric field magnitude ($ E = 60 \, \mathrm{Vm}^{-1} $).

$ B = \frac{E}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \mathrm{~T} $

Therefore, the wave equation for the magnetic field, considering it propagates in the $ x $-direction and oscillates in a direction perpendicular to both the $ x $-direction and the direction of the electric field (thus, in the $ z $-direction if $ \mathbf{E} $ is in the $ y $-direction), is:

$ \mathbf{B} = B \sin(kx - \omega t) \hat{\mathbf{k}} $

$ \mathbf{B} = 2 \times 10^{-7} \sin \left(\frac{\pi}{2} \times 10^3(x - 3 \times 10^8t)\right) \hat{\mathbf{k}} \mathrm{T} $

This is represented by Option B:

$ \mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T} $

Therefore, the correct answer is Option B.

To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface. The pressure exerted by the light on a non-reflecting surface is given by the formula:

$ P = \frac{F}{A} $

where $P$ is the pressure, $F$ is the force, and $A$ is the area. The pressure due to the light can also be related to the energy flux $I$ by the relationship:

$ P = \frac{I}{c} $

where $I$ is the light energy flux and $c$ is the speed of light in a vacuum ($3 \times 10^8 \mathrm{~m/s}$).

Given that the average force $F$ is $2.4 \times 10^{-4} \mathrm{~N}$ and the light energy flux $I$ is $360 \mathrm{~W/cm}^2$, we first convert the flux to $\mathrm{W/m}^2$:

$ 360 \mathrm{~W/cm}^2 = 360 \times 10^4 \mathrm{~W/m}^2 $

Now we can use the relation between pressure and energy flux:

$ P = \frac{I}{c} = \frac{360 \times 10^4}{3 \times 10^8} = 1.2 \mathrm{~N/m}^2 $

Next, we use the pressure formula to find the area of the surface:

$ P = \frac{F}{A} \implies A = \frac{F}{P} $

Substituting the values we have:

$ A = \frac{2.4 \times 10^{-4} \mathrm{~N}}{1.2 \mathrm{~N/m}^2} = 2 \times 10^{-4} \mathrm{~m}^2 $

The area of the surface is:

$ A = 2 \times 10^{-4} \mathrm{~m}^2 = 0.02 \mathrm{~m}^2 $

Hence, the correct option is:

Option D $0.02 \mathrm{~m}^2$

To find the intensity of the given electromagnetic wave, we need to use the formula for the intensity of an electromagnetic wave:

$ I = \frac{1}{2} \epsilon_0 c E_0^2 $

where:

• $I$ is the intensity in $\mathrm{W} / \mathrm{m}^2$
• $\epsilon_0$ is the permittivity of free space, given as $9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
• $c$ is the speed of light in vacuum, approximately $3 \times 10^8 \mathrm{~m/s}$
• $E_0$ is the peak electric field, given as $600 \mathrm{Vm}^{-1}$

Now, substitute these values into the formula:

$ I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times (600)^2 $

Simplify the expression step-by-step:

$ I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times 360000 $

First, calculate $9 \times 3 \times 360000$:

$ I = \frac{1}{2} \times 9.72 \times 10^{-4} \times 360000 $

Combine 9 and 3 into 27, giving you:

$ I = 13.5 \times 10^{-4} \times 360000 $

Then, calculate the multiplication:

$ I = 13.5 \times 36 $

Finally, multiply the remaining values:

$ I = 486 \times 10^{-4} $

The final value is:

The intensity, $I$, is 486 $\mathrm{W} / \mathrm{m}^2$. Therefore, the correct option is:

Option A: 486

$\begin{aligned} & v=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \cdot \varepsilon_r}} \\ & \Rightarrow 1.5 \times 10^8=\frac{3 \times 10^8}{\sqrt{2 \cdot \varepsilon_r}} \Rightarrow \varepsilon_r=2 \end{aligned}$

To match the given List I (EM-Wave) with List II (Wavelength Range), we need to understand the typical wavelength ranges associated with each type of electromagnetic (EM) wave mentioned in List I.

Here's the matching based on known wavelength ranges:

• Infra-red: Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. The wavelength range for infrared is typically from about 700 nm (nanometers) to 1 mm (millimeter), which means it matches with option (III) - 1 mm to 700 nm.


• Ultraviolet: Ultraviolet (UV) light sits between visible light and X-rays along the electromagnetic spectrum, with wavelengths shorter than visible light but longer than X-rays. The typical wavelength range for UV light is from 400 nm down to 10 nm, so it matches with option (II) - 400 nm to 1 nm (even though the strict lower bound should be 10 nm, this is the closest option provided).


• X-rays: X-rays have a wavelength range shorter than UV rays and longer than gamma rays, typically ranging from 10 nm down to 0.01 nm (or $10^{-2}$ nm). Thus, X-rays match with option (IV) - 1 nm to $10^{-3}$ nm, as this is the closest range provided that fits the typical scope of X-ray wavelengths.


• Gamma rays: Gamma rays have the shortest wavelength on the electromagnetic spectrum, with wavelength less than 10 picometers (pm), which equates to less than $10^{-3}$ nm. This corresponds to option (I) - $<10^{-3}$ nm.

So, based on the above analysis, the correct option for the matching would be:

• (A)-(III)


• (B)-(II)


• (C)-(IV)


• (D)-(I)

Therefore, the correct answer is Option B: (A)-(III), (B)-(II), (C)-(IV), (D)-(I).

Wavelengths are as

Gamma rays $<1 \mathrm{~nm}$

x-ray < $(1-10)$ nm

Infrared $<(700-10^5) \mathrm{~nm}$

Microwave $<(10^5-10^8) \mathrm{~nm}$

To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $\overrightarrow{\mathrm{E}}$ and the magnetic field $\overrightarrow{\mathrm{B}}$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:

$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}$

where:

• $\overrightarrow{\mathrm{E}}$ is the electric field.
• $\hat{\mathrm{k}}$ is the unit vector in the direction of propagation of the wave.
• $\mathrm{c}$ is the speed of light in a vacuum.

Given the electric field:

$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$

The wave is propagating in the $z$-direction, so $\hat{\mathrm{k}} = \hat{z}$. The unit vector $\hat{\mathrm{i}}$ represents the $x$-direction.

The magnetic field induction is given by:

$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}$

The cross product $\hat{\mathrm{i}} \times \hat{\mathrm{z}}$ yields $\hat{\mathrm{j}}$ (the unit vector in the $y$-direction):

$\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}$

Therefore, the magnetic field induction is:

$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})$

The correct answer is:

Option A:

$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as $ c $. The speed of light $ c $ is $ 3 \times 10^8 $ meters per second. The relationship between the speed of light $ c $, the frequency $ f $, and the wavelength $ \lambda $ of an electromagnetic wave is given by the equation:

$ c = f \lambda $

Here, we are given that the frequency $ f $ of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by $ 10^6 $ (because 1 MHz = $ 10^6 $ Hz).

$ f = 60 \times 10^6 \text{ Hz} $

To find the wavelength $ \lambda $ in meters, we can rearrange the wave equation to solve for $ \lambda $:

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

Substitute the given values of $ c $ and $ f $ into the equation to find the wavelength:

$ \lambda = \frac{3 \times 10^8 \text{ m/s}}{60 \times 10^6 \text{ Hz}} $

Now, simplify the equation by dividing the numbers:

$ \lambda = \frac{3}{60} \times 10^{8-6} \text{ m} $

$ \lambda = \frac{1}{20} \times 10^2 \text{ m} $

$ \lambda = 5 \text{ m} $

Therefore, the wavelength of the wave is 5 meters, corresponding to option B. Additionally, in an electromagnetic wave, the electric and magnetic field vectors are indeed mutually perpendicular to each other and to the direction of propagation, which in this case is the $ z $ direction.

$\begin{aligned} & \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{B}^2}{2 \mu_0} \\ & \because \mathrm{E}=\mathrm{CB} \text { and } \mathrm{C}=\frac{1}{\mu_0 \varepsilon_0} \end{aligned}$

The average energy density of an electromagnetic wave is given by the formula:

$U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$

For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:

$U_E = \frac{1}{2} \varepsilon_0 E^2$

Given that the electric field amplitude ($E$) is $50 \, \mathrm{Vm}^{-1}$ and the permittivity of free space ($\varepsilon_0$) is $8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$, we can calculate the energy density due to the electric field as follows:

$U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 = 1.10625 \times 10^{-8} \, \mathrm{J/m^3}$

Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.

$\begin{aligned} & \text { Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\ & \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\ & \hat{C}=\hat{E} \times \hat{B} \end{aligned}$