Q. No.During the propagation of electromagnetic waves in a medium:
| 1. | Electric energy density is half of the magnetic energy density |
| 2. | Electric energy density is equal to the magnetic energy density |
| 3. | Both electric and magnetic energy densities are zero |
| 4. | Electric energy density is double of the magnetic energy density |
Subtopic: Properties of EM Waves |
From NCERT
JEE
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A red LED emits light at \(0.1\) watt uniformly around it. The amplitude of the electric field of the light at a distance of \(1~\text{m}\) from the diode is:
1. \(1.73~\text{V/m}\)
2. \(2.45~\text{V/m}\)
3. \(5.48~\text{V/m}\)
4. \(7.75~\text{V/m}\)
Subtopic: Properties of EM Waves |
From NCERT
JEE
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An electromagnetic wave traveling in the \(x\text-\)direction has the frequency of \(2\times 10^{14}~\text{Hz}\) and electric field amplitude of \(27~\text{Vm}^{-1}.\) From the options given below, which one describes the magnetic field for this wave?
| 1. | \(\begin{aligned} & \vec{B}(x, t)=\left(9 \times 10^{-8}~\text T\right) \hat{j} \sin \left[1.5 \times 10^{-6} x-2 \times 10^{14} t\right] \end{aligned}\) |
| 2. | \(\begin{aligned} & \vec{B}(x, t)=\left(9 \times 10^{-8}~\text{T}\right) \hat{i} \sin \left[2 \pi\left(1.5 \times 10^{-8} x-2 \times 10^{14} t\right)\right] \end{aligned}\) |
| 3. | \(\begin{aligned} & \vec{B}(x, t)=\left(9 \times 10^{-8}~\text{T}\right) \hat{{k}} \sin \left[2 \pi\left(1.5 \times 10^{-6}{x}-2 \times 10^{14}{t}\right)\right] \end{aligned}\) |
| 4. | \(\begin{aligned} & \vec{B}(x, t)=\left(3 \times 10^{-8}~\text{T}\right) \hat{j} \sin \left[2 \pi\left(1.5 \times 10^{-8} x-2 \times 10^{14} t\right)\right] \end{aligned}\) |
Subtopic: Properties of EM Waves |
From NCERT
JEE
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For plane electromagnetic waves propagating in the \(z\) direction, which one of the following combinations gives the correct possible direction for \({\vec{E}}\) and \({\vec{B}}\) field respectively?
1. \({(2 \hat{i}+3 \hat{j})}\) and \({(\hat{i}+2 \hat{j})}\)
2. \({(-2 \hat{i}-3 \hat{j})}\) and \({(3\hat{i}-2 \hat{j})}\)
3. \({(3 \hat{i}+4 \hat{j})}\) and \({(4\hat{i}-3 \hat{j})}\)
4. \({(\hat{i}+2 \hat{j})}\) and \({(2\hat{i}-\hat{j})}\)
1. \({(2 \hat{i}+3 \hat{j})}\) and \({(\hat{i}+2 \hat{j})}\)
2. \({(-2 \hat{i}-3 \hat{j})}\) and \({(3\hat{i}-2 \hat{j})}\)
3. \({(3 \hat{i}+4 \hat{j})}\) and \({(4\hat{i}-3 \hat{j})}\)
4. \({(\hat{i}+2 \hat{j})}\) and \({(2\hat{i}-\hat{j})}\)
Subtopic: Properties of EM Waves |
From NCERT
JEE
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The magnetic field in a plane electromagnetic wave is given by \(\vec B=B_0\sin(K x+\omega t)\hat j~\text T.\) The expression for the corresponding electric field will be:
(where \(c\) is the speed of light)
1. \(\vec E=\frac{B_0}{c}\sin (kx+\omega t)\hat k~\text{V/m}\)
2. \(\vec E=-B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
3. \(\vec E=B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
4. \(\vec E=B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
(where \(c\) is the speed of light)
1. \(\vec E=\frac{B_0}{c}\sin (kx+\omega t)\hat k~\text{V/m}\)
2. \(\vec E=-B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
3. \(\vec E=B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
4. \(\vec E=B_0c\sin (kx+\omega t)\hat k~\text{V/m}\)
Subtopic: Properties of EM Waves |
From NCERT
JEE
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The electric field component of a monochromatic radiation is given by \(\vec E=2~E_0\hat i\cos{kz}\cos\Omega t.\)
Its magnetic field \(\vec B\) is given by:
1. \(\dfrac{2~E_0}{c}~\hat j\cos kz\cos \omega t\)
2. \(\dfrac{2~E_0}{c}~\hat j\sin kz\cos \omega t\)
3. \(\dfrac{2~E_0}{c}~\hat j\sin kz\sin \omega t\)
4. \(-\dfrac{2~E_0}{c}~\hat j\sin kz \sin \omega t\)
Its magnetic field \(\vec B\) is given by:
1. \(\dfrac{2~E_0}{c}~\hat j\cos kz\cos \omega t\)
2. \(\dfrac{2~E_0}{c}~\hat j\sin kz\cos \omega t\)
3. \(\dfrac{2~E_0}{c}~\hat j\sin kz\sin \omega t\)
4. \(-\dfrac{2~E_0}{c}~\hat j\sin kz \sin \omega t\)
Subtopic: Properties of EM Waves |
From NCERT
JEE
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An EM wave from air enters a medium. The electric fields are \(\overrightarrow{{E}}_1={E}_{01} \hat{{x}} \cos \left[2 \pi {\nu}\left(\frac{{z}}{{c}}-{t}\right)\right] \) in air and \(\overrightarrow{{E}}_2={E}_{02} \hat{{x}} \cos {k}(2 {z}-{ct})]\) in medium, where the wave number \(k\) and frequency \(\nu\) refer to their values in air. The medium is non-magnetic. If \(\epsilon_{r_1}\) and \(\epsilon_{r_2}\) refer to relative permittivities of air and medium respectively, which of the following options is correct?
1. \( \frac{\epsilon_{r_1}}{\epsilon_{r_2}}=4 \)
2. \( \frac{\epsilon_{r_1}}{\epsilon_{r_2}}=2 \)
3. \( \frac{\epsilon_{r_1}}{\epsilon_{r_2}}=\frac{1}{4} \)
4. \( \frac{\epsilon_{r_1}}{\epsilon_{r_2}}=\frac{1}{2}\)
Subtopic: Properties of EM Waves |
From NCERT
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A monochromatic beam of light has a frequency \(v=\frac{3}{2\pi}\times10 ^{12}~\text{Hz}\) and is propagating along the direction \(\frac{i+j}{\sqrt{2}}\) It is polarized along the \(\hat{k}\) direction. The acceptable form of the magnetic field is:
| 1. | \(\frac{E_0}{C}\bigg(\frac{\hat{i}-\hat{j}}{\sqrt2}\bigg) \cos\bigg[ 10^4 \frac{(\hat{i}-\hat{j})}{\sqrt2}.\overrightarrow{r} - (3 \times10 ^{12})t \bigg]\) |
| 2. | \(\frac{E_0}{C}\hat{k}\cos\bigg[ 10^4 \frac{(\hat{i}+\hat{j})}{\sqrt2}.\overrightarrow{r} +(3 \times10 ^{12})t \bigg]\) |
| 3. | \(\frac{E_0}{C}\bigg(\frac{\hat{i}-\hat{j}}{\sqrt2}\bigg)\cos\bigg[ 10^4 \frac{(\hat{i}+\hat{j})}{\sqrt2}.\overrightarrow{r} + (3 \times10 ^{12})t \bigg]\) |
| 4. | \(\frac{E_0}{C}\bigg(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt3}\bigg)\cos\bigg[ 10^4 \frac{(\hat{i}+\hat{j})}{\sqrt2}.\overrightarrow{r} + (3 \times10 ^{12})t \bigg]\) |
Subtopic: Properties of EM Waves |
From NCERT
JEE
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A plane electromagnetic wave of wavelength \(\lambda\) has an intensity \(I.\) It is propagating along the positive \(y\text{-direction}.\) The allowed expressions for the electric and magnetic fields are given by:
| 1. | \(\vec{E} = \sqrt{\frac{2I}{\varepsilon_0c}}\cos\bigg[ \frac{2\pi}{\lambda}(y-ct)\bigg]\hat{k;}~~~ \vec{B} = +\frac{1}{c}E\hat{i}\) |
| 2. | \(\vec{E} = \sqrt{\frac{I}{\varepsilon_0c}}\cos\bigg[ \frac{2\pi}{\lambda}(y-ct)\bigg]\hat{k;}~~~\vec{B} = +\frac{1}{c}E\hat{i}\) |
| 3. | \(\vec{E} = \sqrt{\frac{2I}{\varepsilon_0c}}\cos\bigg[ \frac{2\pi}{\lambda}(y+ct)\bigg]\hat{k;}~~~\vec{B} = \frac{1}{c}E\hat{i}\) |
| 4. | \(\vec{E} = \sqrt{\frac{1}{\varepsilon_0c}}\cos\bigg[ \frac{2\pi}{\lambda}(y-ct)\bigg]\hat{i;}~~~\vec{B} = +\frac{1}{c}E\hat{i}\) |
Subtopic: Properties of EM Waves |
From NCERT
JEE
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The magnetic field of an electromagnetic wave is given by:
\(\vec{B}=1.6 \times 10^{-6} ~\text{cos} \left(2 \times 10^{-7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j})~ {\text{Wb/m}^2}\)
The associated electric field will be:
| 1. | \( \vec{E}=4.8 \times 10^2 ~\text{cos} \left(2 \times 10^7 z+6 \times 10^{15} t\right)(-\hat{i}+2 \hat{j})~ {\text{V/m}}\) |
| 2. | \(\vec{E}=4.8 \times 10^2 ~\text{cos} \left(2 \times 10^7 z-6 \times 10^{15} t\right)(2 \hat{i}+2 \hat{j})~{\text{V/m}}\) |
| 3. | \( \vec{E}=4.8 \times 10^2 ~\text{cos} \left(2 \times 10^7 z-6 \times 10^{15} t\right)(\hat{i}-2 \hat{j}) ~{\text{V/m}}\) |
| 4. | \( \vec{E}=4.8 \times 10^2 ~\text{cos} \left(2 \times 10^7 z-6 \times 10^{15} t\right)(-2 \hat{i}+\hat{j}) ~{\text{V/m}}\) |
Subtopic: Properties of EM Waves |
From NCERT
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