Electromagnetic Waves

$$\begin{gathered} C=\frac{E}{B} \\ B=\frac{E}{C} \\ B=\frac{9.3}{3\times {10}^8} \\ B=3.1\times {10}^{-8}T \end{gathered}$$

Infrared is the least energetic thus having biggest wavelength ($\mathrm{\lambda }$) & gamma rays are most energetic thus having smallest wavelength ($\mathrm{\lambda }$).

The electromagnetic waves used in radar systems are microwaves, which have a frequency range of approximately $1$Sv6Kpe[] GHz to $300$Sv6Kpe[] GHz

Displacement current density,

$J_d = {\epsilon }_0\frac{dE}{dt}$

Region I : $$\begin{gathered} E_x = E_0\sin \left(kz-\omega t\right) \\ \frac{d{E}_x}{dt} =-\omega E_0\cos (kz-\omega t) \\ {J}_d = -{\epsilon }_0\omega E_0\cos (kz-\omega t) \end{gathered}$$

Region II : $$\begin{gathered} E_x =E_0 \Rightarrow \frac{d{E}_x}{dt} =0 \\ {J}_d =0 \end{gathered}$$

Region III : $$\begin{gathered} E_x =E_0\sin \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

Region IV : $$\begin{gathered} E_x =E_0\cos \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

Displacement current density,

$J_d = {\epsilon }_0\frac{dE}{dt}$

Region I : $$\begin{gathered} E_x = E_0\sin \left(kz-\omega t\right) \\ \frac{d{E}_x}{dt} =-\omega E_0\cos (kz-\omega t) \\ {J}_d = -{\epsilon }_0\omega E_0\cos (kz-\omega t) \end{gathered}$$

Region II : $$\begin{gathered} E_x =E_0 \Rightarrow \frac{d{E}_x}{dt} =0 \\ {J}_d =0 \end{gathered}$$

Region III : $$\begin{gathered} E_x =E_0\sin \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

Region IV : $$\begin{gathered} E_x =E_0\cos \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

The decreasing order of wavelength is

Radio waves > Microwaves > Infrared waves > Visible rays > Ultra violet rays > X rays > Gamma rays

From electromagnetic spectrum we know that X-ray has smallest and microwave has large wavelength among these.

Infrared is the least energetic thus having biggest wavelength ($\mathrm{\lambda }$) & gamma rays are most energetic thus having smallest wavelength ($\mathrm{\lambda }$).

Both the electric and magnetic field vectors oscillate in phase, meaning their maxima and minima occur at the same points in time and space.

the phase difference is zero.

Both the electric and magnetic field vectors oscillate in phase, meaning their maxima and minima occur at the same points in time and space.

the phase difference is zero.

Name of scientist is J.C. Bose

Displacement current density,

$J_d = {\epsilon }_0\frac{dE}{dt}$

Region I : $$\begin{gathered} E_x = E_0\sin \left(kz-\omega t\right) \\ \frac{d{E}_x}{dt} =-\omega E_0\cos (kz-\omega t) \\ {J}_d = -{\epsilon }_0\omega E_0\cos (kz-\omega t) \end{gathered}$$

Region II : $$\begin{gathered} E_x =E_0 \Rightarrow \frac{d{E}_x}{dt} =0 \\ {J}_d =0 \end{gathered}$$

Region III : $$\begin{gathered} E_x =E_0\sin \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

Region IV : $$\begin{gathered} E_x =E_0\cos \left(kz\right) \\ \frac{dE}{dt}= 0 \Rightarrow J_d =0 \end{gathered}$$

$$\begin{gathered} \text{ Average energy density }=\frac{1}{2}{\epsilon }_0{E}_0^2 \\ \text{ Average energy }\left(E_{AV}\right)=\frac{1}{2}{\epsilon }_0{E}_0^{2}AL \\ =\frac{1}{2}{\epsilon }_0{E}_0^2\pi {\left(\frac{D}{2}\right)}^{2}L \\ =\frac{\pi {\epsilon }_0{E}_0^2{D}^{2}L}{8} \\ As, E_1=E_2 \\ \frac{\pi {\epsilon }_0{E}_{01}^2{D}^{2}L}{8}=\frac{\pi {\epsilon }_0{E}_{02}^2{\left(\frac{D}{2}\right)}^{2}L}{8} \\ \Rightarrow E_{01}^2=\frac{E_{02}^2}{4} \\ E_{02}=2E_{01}=200 \end{gathered}$$

From electromagnetic spectrum we know that X-ray has smallest and microwave has large wavelength among these.

In free space, the electric and magnetic fields $\overrightarrow{E}$ and $\overrightarrow{B}$ are related by the following equation:

$$\begin{gathered} \overrightarrow{B}=\frac{1}{c}\hat{K}\times \overrightarrow{E} \\ \overrightarrow{K}=3\hat{i}+4\hat{j} \\ \hat{K}=\frac{3\hat{i}+4\hat{j}}{5} \\ \hat{E}=\frac{4\hat{i}-3\hat{j}}{5} \\ \hat{B}=\hat{K}\times \hat{E} \\ \hat{B}=-\hat{Z} \\ B_0=\frac{E_0}{C}=\frac{57}{3\times {10}^8} \end{gathered}$$

Gamma rays has highest frequency

The relationship between the electric field ($E_0$) and magnetic field ($B_0$) in electromagnetic waves is given by:

$$E_0 = c \cdot B_0$$​

Rearranging for $B_0$​:

$$B_0 = \frac{E_0}{c}$$​​

Substitute the given values:

$$B_0 = \frac{9.3}{3 \times 10^8}$$​

Simplify:

$$B_0 = 3.1 \times 10^{-8} \, \text{T}$$

Rounding $B_0$​ is approximately 3 (depending on units being normalized for $B_0$​).

Correct answer: (d).

Gamma rays has highest frequency

To understand the change in wave characteristics from microwaves to X-rays in the electromagnetic spectrum, we need to consider the general properties of electromagnetic waves:

• Speed: Electromagnetic waves travel at the same speed in a vacuum, which is approximately c = 3 × 10⁸ m/s. Therefore, the speed remains constant across the spectrum.
• Wavelength and Frequency: The relationship between speed (c), frequency (f), and wavelength (λ) is given by the equation c = λf. When the speed is constant, any change in wavelength results in an inverse change in frequency.

From microwaves to X-rays:

• Wavelength: Generally decreases. Microwaves have longer wavelengths compared to X-rays.
• Frequency: Increases. Since their wavelengths decrease, their frequencies must increase to maintain the constant speed, according to the formula.

Thus, the correct change in wave characteristics from microwaves to X-rays is: "Speed remains same, Wavelength decreases, Frequency increases."

$$\begin{gathered} \text{ Average energy density }=\frac{1}{2}{\epsilon }_0{E}_0^2 \\ \text{ Average energy }\left(E_{AV}\right)=\frac{1}{2}{\epsilon }_0{E}_0^{2}AL \\ =\frac{1}{2}{\epsilon }_0{E}_0^2\pi {\left(\frac{D}{2}\right)}^{2}L \\ =\frac{\pi {\epsilon }_0{E}_0^2{D}^{2}L}{8} \\ As, E_1=E_2 \\ \frac{\pi {\epsilon }_0{E}_{01}^2{D}^{2}L}{8}=\frac{\pi {\epsilon }_0{E}_{02}^2{\left(\frac{D}{2}\right)}^{2}L}{8} \\ \Rightarrow E_{01}^2=\frac{E_{02}^2}{4} \\ E_{02}=2E_{01}=200 \end{gathered}$$

Name of scientist is J.C. Bose

Name of scientist is J.C. Bose

The displacement current in any region of the capacitor is proportional to the current in the external circuit and depends on the ratio of the areas involved.

The displacement current $I_d$​ in the smaller region is given by:

where:

• $I = 6 \, \text{A}$ is the total current in the circuit,
• $A_{\text{small}} = 3.2 \, \text{cm}^2$ is the area of the smaller region,
• $A_{\text{total}} = 16 \, \text{cm}^2$is the total area of the capacitor plates.

Simplify:

The displacement current in the smaller region is:

To understand the change in wave characteristics from microwaves to X-rays in the electromagnetic spectrum, we need to consider the general properties of electromagnetic waves:

• Speed: Electromagnetic waves travel at the same speed in a vacuum, which is approximately c = 3 × 10⁸ m/s. Therefore, the speed remains constant across the spectrum.
• Wavelength and Frequency: The relationship between speed (c), frequency (f), and wavelength (λ) is given by the equation c = λf. When the speed is constant, any change in wavelength results in an inverse change in frequency.

From microwaves to X-rays:

• Wavelength: Generally decreases. Microwaves have longer wavelengths compared to X-rays.
• Frequency: Increases. Since their wavelengths decrease, their frequencies must increase to maintain the constant speed, according to the formula.

Thus, the correct change in wave characteristics from microwaves to X-rays is: "Speed remains same, Wavelength decreases, Frequency increases."

We are given the magnetic field of an electromagnetic wave as:

$$\mathbf{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin (\omega t - kz)$$

We need to determine the corresponding electric field.

The electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ are related as:

$$\mathbf{E} = c \mathbf{B}$$

where $c$ is the speed of light in a vacuum:

$$c = 3 \times 10^8 \text{ m/s}$$

The magnitude of $B_0$​ from the given equation is:

$$B_0 = 30$$

Thus, the corresponding electric field magnitude is:

$$E_0 = c B_0 = (3 \times 10^8) \times 30 = 9 \times 10^9$$

Since the electromagnetic wave propagates in the z-direction, the electric and magnetic fields are perpendicular to each other. The direction of the electric field $\mathbf{E}$ can be found using the right-hand rule:

$\mathbf{E}\propto B\times K$

For the given magnetic field components, performing the cross-product operation gives:

$$\mathbf{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 9 \times 10^9 \sin (\omega t - kz)$$

From the given options, the correct answer matches:

$$\boxed{ \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 9 \times 10^9 \sin (\omega t - kz) }$$​

This corresponds to Option B.

Gamma rays has highest frequency

Both the electric and magnetic field vectors oscillate in phase, meaning their maxima and minima occur at the same points in time and space.

the phase difference is zero.

1. Wave Propagation Direction: The wave propagates along the $z$-axis ($\hat{k} = \hat{z}$).

2. Given Electric Field:

   $$\vec{E} = E_0 (4 \hat{i} - 3 \hat{j}) \cos(\omega t - kz).$$.

3. Find $\hat{k} \times \vec{E}$: Using the cross-product rules:

   • $\hat{z} \times \hat{i} = \hat{j}$,
   • $\hat{z} \times \hat{j} = -\hat{i}$,

   Compute the cross-product:

   $$\hat{z} \times (4 \hat{i} - 3 \hat{j}) = (4 \hat{j}) - (-3 \hat{i}),$$ $$\hat{z} \times (4 \hat{i} - 3 \hat{j}) = 4 \hat{j} + 3 \hat{i}.$$

4. Write Magnetic Field: The magnetic field is:

   $$\vec{B} = \frac{E_0}{c} (3 \hat{i} + 4 \hat{j}) \cos(\omega t - kz).$$

(A)

$\mathrm{I}=\frac{1}{2}{\epsilon }_0{\mathrm{E}}_0^2\mathrm{c}$
$\mathrm{I}=\frac{1}{2}\times 9\times {10}^{-12}\times (600{)}^2\times 3\times {10}^8$
$I=\frac{9\times 3}{2}\times 36\times {10}^4\times {10}^{-12}\times {10}^8$
$I=27\times 18$
$I=486\frac{\text{ watt }}{{\mathrm{m}}^2}$

The given equation \(E_y = E_0 \sin(kx - \omega t)\) represents a wave propagating in the positive x-direction. The argument of the sine function, \((kx - \omega t)\), indicates the direction of propagation. A negative sign between \(kx\) and \(\omega t\) signifies propagation in the positive x direction.

Electromagnetic waves are transverse waves, meaning the electric and magnetic fields are perpendicular to both each other and the direction of propagation. The equation shows the electric field oscillating in the y-direction (\(E_y\)).

The general form of a wave equation is given by: \(A \sin(kx - \omega t + \phi)\), where \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(t\) is time, \(x\) is the position, and \(\phi\) is the phase constant. In this case, the wave is propagating along the x-axis because the x-coordinate is present in the argument of the sine function with a negative sign in front of the \(\omega t\) term.

The refractive index of the medium is \(n = \frac{c}{v} = \frac{3 \times 10^{8}}{1.5 \times 10^{8}} = 2\).

Using the relation \(n = \sqrt{\mu_r \epsilon_r}\), we get:

\(2 = \sqrt{2.0 \times \epsilon_r}\) \(\Longrightarrow\) \(4 = 2.0 \times \epsilon_r\) \(\Longrightarrow\) \(\epsilon_r = 2\).

The given equation \(E_y = E_0 \sin(kx - \omega t)\) represents a wave propagating in the positive x-direction. The argument of the sine function, \((kx - \omega t)\), indicates the direction of propagation. A negative sign between \(kx\) and \(\omega t\) signifies propagation in the positive x direction.

Electromagnetic waves are transverse waves, meaning the electric and magnetic fields are perpendicular to both each other and the direction of propagation. The equation shows the electric field oscillating in the y-direction (\(E_y\)).

The general form of a wave equation is given by: \(A \sin(kx - \omega t + \phi)\), where \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(t\) is time, \(x\) is the position, and \(\phi\) is the phase constant. In this case, the wave is propagating along the x-axis because the x-coordinate is present in the argument of the sine function with a negative sign in front of the \(\omega t\) term.

Gamma rays has highest frequency

Infrared is the least energetic thus having biggest wavelength ($\mathrm{\lambda }$) & gamma rays are most energetic thus having smallest wavelength ($\mathrm{\lambda }$).

$\mathrm{I}=\frac{1}{2}{\epsilon }_0{\mathrm{E}}_0^2\mathrm{c}$
$\mathrm{I}=\frac{1}{2}\times 9\times {10}^{-12}\times (600{)}^2\times 3\times {10}^8$
$I=\frac{9\times 3}{2}\times 36\times {10}^4\times {10}^{-12}\times {10}^8$
$I=27\times 18$
$I=486\frac{\text{ watt }}{{\mathrm{m}}^2}$

Em waves have momentum so assertion is false

• An electromagnetic wave (EM wave) consists of:

  • An electric field ($E$)
  • A magnetic field ($B$)
  • A wave propagation direction

• These three quantities are mutually perpendicular following the right-hand rule:

  • If the wave propagates along +X-direction, then:
    • The electric field ($E$) is along the Y-axis.
    • The magnetic field ($B$) is along the Z-axis.

• Given that the wave propagates in the +X direction:
  • $E$ is along Y-axis.
  • $B$ is along Z-axis.

Correct answer:

$$\boxed{B} \quad (Y, Z)$$

The given equation \(E_y = E_0 \sin(kx - \omega t)\) represents a wave propagating in the positive x-direction. The argument of the sine function, \((kx - \omega t)\), indicates the direction of propagation. A negative sign between \(kx\) and \(\omega t\) signifies propagation in the positive x direction.

Electromagnetic waves are transverse waves, meaning the electric and magnetic fields are perpendicular to both each other and the direction of propagation. The equation shows the electric field oscillating in the y-direction (\(E_y\)).

The general form of a wave equation is given by: \(A \sin(kx - \omega t + \phi)\), where \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(t\) is time, \(x\) is the position, and \(\phi\) is the phase constant. In this case, the wave is propagating along the x-axis because the x-coordinate is present in the argument of the sine function with a negative sign in front of the \(\omega t\) term.

• Original EM wave equation:

  $$E = 100 \sin(kx - \omega t)$$

• Energy stored in a cylindrical region:

  $$U_0 = \frac{1}{2} \varepsilon_0 E_0^2 \pi R^2 L$$

  where:

  • $E_0 = 100$ (initial amplitude),
  • $R$ is the cylinder's radius,
  • $L$ is the length.

• New cylinder conditions:

  • Same length $L$
  • Half the diameter, so the radius becomes $R' = \frac{R}{2}$
  • New energy $U_0$​ remains the same.

For the new wave, let the amplitude be $E_0'$​. The new energy equation:

$$U_0 = \frac{1}{2} \varepsilon_0 E_0'^2 \pi \left(\frac{R}{2}\right)^2 L$$

$$U_0 = \frac{1}{2} \varepsilon_0 E_0'^2 \pi \frac{R^2}{4} L$$

Dividing by the original equation:

$$\frac{E_0'^2}{4} = \frac{E_0^2}{1}$$

$$E_0'^2 = 4 E_0^2$$

​ $$E_0' = 2 E_0 = 200$$

Since the form remains the same, the new equation is:

$$E = 200 \sin(kx - \omega t)$$

• Original EM wave equation:

  $$E = 100 \sin(kx - \omega t)$$

• Energy stored in a cylindrical region:

  $$U_0 = \frac{1}{2} \varepsilon_0 E_0^2 \pi R^2 L$$

  where:

  • $E_0 = 100$ (initial amplitude),
  • $R$ is the cylinder's radius,
  • $L$ is the length.

• New cylinder conditions:

  • Same length $L$
  • Half the diameter, so the radius becomes $R' = \frac{R}{2}$
  • New energy $U_0$​ remains the same.

For the new wave, let the amplitude be $E_0'$​. The new energy equation:

$$U_0 = \frac{1}{2} \varepsilon_0 E_0'^2 \pi \left(\frac{R}{2}\right)^2 L$$

$$U_0 = \frac{1}{2} \varepsilon_0 E_0'^2 \pi \frac{R^2}{4} L$$

Dividing by the original equation:

$$\frac{E_0'^2}{4} = \frac{E_0^2}{1}$$

$$E_0'^2 = 4 E_0^2$$

​ $$E_0' = 2 E_0 = 200$$

Since the form remains the same, the new equation is:

$$E = 200 \sin(kx - \omega t)$$

To understand the change in wave characteristics from microwaves to X-rays in the electromagnetic spectrum, we need to consider the general properties of electromagnetic waves:

• Speed: Electromagnetic waves travel at the same speed in a vacuum, which is approximately c = 3 × 10⁸ m/s. Therefore, the speed remains constant across the spectrum.
• Wavelength and Frequency: The relationship between speed (c), frequency (f), and wavelength (λ) is given by the equation c = λf. When the speed is constant, any change in wavelength results in an inverse change in frequency.

From microwaves to X-rays:

• Wavelength: Generally decreases. Microwaves have longer wavelengths compared to X-rays.
• Frequency: Increases. Since their wavelengths decrease, their frequencies must increase to maintain the constant speed, according to the formula.

Thus, the correct change in wave characteristics from microwaves to X-rays is: "Speed remains same, Wavelength decreases, Frequency increases."

Infrared is the least energetic thus having biggest wavelength ($\mathrm{\lambda }$) & gamma rays are most energetic thus having smallest wavelength ($\mathrm{\lambda }$).

The given equation \(E_y = E_0 \sin(kx - \omega t)\) represents a wave propagating in the positive x-direction. The argument of the sine function, \((kx - \omega t)\), indicates the direction of propagation. A negative sign between \(kx\) and \(\omega t\) signifies propagation in the positive x direction.

Electromagnetic waves are transverse waves, meaning the electric and magnetic fields are perpendicular to both each other and the direction of propagation. The equation shows the electric field oscillating in the y-direction (\(E_y\)).

The general form of a wave equation is given by: \(A \sin(kx - \omega t + \phi)\), where \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(t\) is time, \(x\) is the position, and \(\phi\) is the phase constant. In this case, the wave is propagating along the x-axis because the x-coordinate is present in the argument of the sine function with a negative sign in front of the \(\omega t\) term.

• An electromagnetic wave (EM wave) consists of:

  • An electric field ($E$)
  • A magnetic field ($B$)
  • A wave propagation direction

• These three quantities are mutually perpendicular following the right-hand rule:

  • If the wave propagates along +X-direction, then:
    • The electric field ($E$) is along the Y-axis.
    • The magnetic field ($B$) is along the Z-axis.

• Given that the wave propagates in the +X direction:
  • $E$ is along Y-axis.
  • $B$ is along Z-axis.

Correct answer:

$$\boxed{B} \quad (Y, Z)$$