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Electromagnetic Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Waves4 Marks
Question
A stationary charge produces only an electrostatic field while a charge in uniform motion produces a magnetic field, that does not change with time. An oscillating charge is an example of accelerating charge. It produces an oscillating magnetic field, which in turn produces an oscillating electric fields and so on. The oscillating electric and magnetic fields regenerate each other as a wave which propagates through space.

Magnetic field in a plane electromagnetic wave is given by $\vec{\text{B}}=\text{B}_0\sin(\text{kx}+\omega\text{t}) \hat{\text{j}}\text{T}.$
  1. Expression for corresponding electric field will be $($Where $c$ is speed of light$)$.
  1. $\vec{\text{E}}=-\text{B}_0\text{c}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
  2. $\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{kx}-\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
  3. $\vec{\text{E}}=\frac{\text{B}_0}{\text{c}}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
  4. $\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
  1. The electric field component ofa monochromatic radiation is given by $\vec{\text{E}} = 2\epsilon_0\hat{\text{i}}\cos\text{kz}\cos\omega\text{t}.$ Its magnetic field $\vec{\text{B}}$ is then given by:
  1. $\frac{2\epsilon_0}{\text{c}}\hat{\text{j}}\cos\text{kz}\cos\omega\text{t}$
  2. $\frac{2\epsilon_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\cos\omega\text{t}$
  3. $\frac{2\epsilon_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$
  4. $-\frac{2\epsilon_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$
  1. A plane em wave of frequency $25MHz$ travels in a free space along $x-$ direction. At a particular point in space and time, $\text{E}=(6.3\ \hat{\text{j}})\frac{\text{V}}{\text{m}}.$ What is magnetic field at that time?
  1. $0.095\mu\text{T}$
  2. $0.124\mu\text{T}$
  3. $0.089\mu\text{T}$
  4. $0.021\mu\text{T}$
  1. A plane electromagnetic wave travelling along the $x-$ direction has a wavelength of $3\ mm$. The variation in the electric field occurs in the $y-$ direction with an amplitude $66Vm^1$. The equations for the electric and magnetic fields as a function of $x$ and tare respectively.
  1. $\text{E}_\text{y}=33\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
    $\text{B}_\text{z}=1.1\times10^{-7}\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
  2. $\text{E}_\text{y}=11\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
    $\text{B}_\text{y}=11\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
  3. $\text{E}_\text{x}=33\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
    $\text{B}_\text{x}=11\times10^{-7}\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
  4. $\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
    $\text{B}_\text{z}=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
  1. A plane electromagnetic wave travels in free space along $x-$ axis. At a particular point in space, the electric field along $y-$ axis is $9.3Vm^{-1.}$ The magnetic induction $(B)$ along $z-$ axis is:
  1. $3.1 \times 10^{-8}T$
  2. $3 \times 10^{-5}T$
  3. $3 \times 10^{-6}T$
  4. $9.3 \times 10^{-6}T$
✓

Answer

  1. $(d)\ \vec{\text{E}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
Given : $\vec{\text{B}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
The relation between electric and magnetic field is,
$\text{c}=\frac{\text{E}}{\text{B}}$ or $E = cB$
The electric $6$ eld component is perpendicular to the direction of propagation and the direction of magnetic field.
Therefore, the electric field component along $z-$ axis is obtained as $\vec{\text{E}}=\text{cB}_0\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}.$
  1. $(c)\ \frac{2\epsilon_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$
$\frac{\text{dE}}{\text{dz}}=-\frac{\text{dB}}{\text{dt}}$
$\frac{\text{dE}}{\text{dz}}=-2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}=-\frac{\text{dB}}{\text{dt}}$
${\text{dB}}=+2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}{\text{dt}}$
${\text{B}}=+2\text{E}_0\text{k}\sin\text{kz}\int\cos\omega\text{t}{\text{dt}}$
$=+2\text{E}_0\frac{\text{k}}{\omega}\sin\text{kz}\sin\omega\text{t}$
$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$
$\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}$
$\therefore\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}\hat{\text{j}}$
$E$ is along $y-$ direction and the wave propagates along $x-$ axis.
$\therefore B$ should be in a direction perpendicular to both $x$ and $y-$ axis.
  1. $(d)\ 0.021\mu\text{T}$
Here, $\vec{\text{E}}=6.3\hat{\text{j}};\text{c}=3\times10^8\frac{\text{m}}{\text{s}}$
The magnitude of $B$ is:
${\text{Bz}}=\frac{\text{E}}{\text{c}}=\frac{6.3}{3\times10^8}$
$=2.1\times10^8\ \text{T}=0.021\mu\text{T}$
  1. $(d) \ \text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\text{B}_\text{z}=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
Here : $\text{E}_0=66\text{Vm}^{-1},\text{E}_\text{y}=66\cos\omega\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
$\lambda=3\text{ mm}=3\times10^{-3}\text{m},\text{k}=\frac{2\pi}{\lambda}$
$\frac{\omega}{\text{k}}=\text{c}$
$\Rightarrow\omega=\text{ck}=3\times10^8\times\frac{2\pi}{3\times10^{-3}}$
or $\omega=2\pi\times10^{11}$
$\therefore\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
${\text{Bz}}=\frac{\text{E}_\text{y}}{\text{c}}=\Big(\frac{66}{3\times10^8}\Big)\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
  1. $(a)\ 3.1 \times 10^{-8}T$
At a particular point $ , E = 9.3Vm^{-1}$​​​​​​​
$\therefore$ Magnetic field at the same point $=\frac{9.3}{3\times10^8}$
$= 3.1 \times 10^{-8}T$

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