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Q. No.A magnetic field vector in an electromagnetic wave is represented by \(\vec{B}=B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{j} .\) Its associated electric field vector is:
1. \(\vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{k}\)
2. \(\vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{i}\)
3. \(\vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{k}\)
4. \(\vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{i}\)
1. \(\vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{k}\)
2. \(\vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{i}\)
3. \(\vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{k}\)
4. \(\vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\dfrac{2 \pi x}{\lambda}\right) \hat{i}\)
Subtopic: Properties of EM Waves |
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Level 2: 60%+
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An electromagnetic wave travels in free space along the \(x\text-\)direction. At a particular point in space and time, \(\vec{B}=2 \times 10^{-7} \hat{j} ~\text{T}\) is associated with this wave. The value of corresponding electric field \(\vec{E}\) at this point is: (in \(\text{V/m}.\))
1. \(60 \hat{k}~ \)
2. \(-60 \hat{k} \)
3. \( 30 \hat{k}\)
4. \(-600 \hat{k}\)
1. \(60 \hat{k}~ \)
2. \(-60 \hat{k} \)
3. \( 30 \hat{k}\)
4. \(-600 \hat{k}\)
Subtopic: Properties of EM Waves |
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Level 1: 80%+
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For an electromagnetic wave propagating through vacuum, \(\vec{k},\vec{E}\) and \(\omega\) represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by:
1. \(\dfrac{\vec{E} \times \vec{k}}{\omega}\)
2. \(\dfrac{\vec{k} \times \vec{E}}{\omega}\)
3. \(\omega(\vec{E} \times \vec{k})\)
4. \(\omega(\vec{k} \times \vec{E})\)
1. \(\dfrac{\vec{E} \times \vec{k}}{\omega}\)
2. \(\dfrac{\vec{k} \times \vec{E}}{\omega}\)
3. \(\omega(\vec{E} \times \vec{k})\)
4. \(\omega(\vec{k} \times \vec{E})\)
Subtopic: Properties of EM Waves |
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Level 3: 35%-60%
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A point light source emits E.M. waves in free space. A detector, placed at a distance of \(L~\text{m},\) \(45^\circ\)
1. \(\dfrac{I_0}{4}\)
2. \( I_0\)
3. \(\dfrac{I_0}{\sqrt{2}} \)
4. \(\dfrac{I_0}{2}\)
1. \(\dfrac{I_0}{4}\)
2. \( I_0\)
3. \(\dfrac{I_0}{\sqrt{2}} \)
4. \(\dfrac{I_0}{2}\)
Subtopic: Properties of EM Waves |
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Level 2: 60%+
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An electromagnetic wave travelling in \(x\text-\)direction is described by field equation \({E}_y=300 \sin \omega\left(t-\dfrac{x}{{c}}\right)\). If the electron is restricted to move in \(y\text-\)direction only with speed of \(1.5\times 10^6~\text{m/s}\) then ratio of maximum electric and magnetic forces on the electron is:
1. \(200\)
2. \(150\)
3. \(400\)
4. \(300\)
1. \(200\)
2. \(150\)
3. \(400\)
4. \(300\)
Subtopic: Properties of EM Waves |
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Level 3: 35%-60%
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A displacement current of \(4.0~\text A\) can be set up in the space between two parallel plates of \(6~\mu \text{F}\) capacitor. The rate of change of potential difference across the plates of the capacitor is nearly \(\alpha \times 10^6 ~\text{V/s}\). The value of \(\alpha\) is:
1. \(0.58\)
2. \(0.67\)
3. \(0.82\)
4. \(0.75\)
1. \(0.58\)
2. \(0.67\)
3. \(0.82\)
4. \(0.75\)
Subtopic: Displacement Current |
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Level 1: 80%+
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Given below are two statements:
In the light of the above statements, choose the correct answer from the options given below:
| Assertion (A): | The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. |
| Reason (R): | There is no mass associated with the electromagnetic waves. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
Subtopic: Properties of EM Waves |
73%
Level 2: 60%+
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A monochromatic source of light operating at \(15~\text{kW}\) emits \(2.5 \times 10^{22}\) /s
(Take\(h=6.6 \times 10^{-34} ~\text{J/s}\) \(c=3 \times 10^8 ~\text{m/s}\)
1. Microwave
2. Infrared
3. Visible
4. Ultraviolet
(Take
1. Microwave
2. Infrared
3. Visible
4. Ultraviolet
Subtopic: Electromagnetic Spectrum |
Level 4: Below 35%
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The electric field a plane electromagnetic wave is given by :
\({E}_{y}=69 \sin \left[0.6 \times 10^3 {x}-1.8 \times 10^{11} {t}\right] ~\text{V/m}.\)
The expression for magnetic field associated with this electromagnetic wave is: (in T)
\({E}_{y}=69 \sin \left[0.6 \times 10^3 {x}-1.8 \times 10^{11} {t}\right] ~\text{V/m}.\)
The expression for magnetic field associated with this electromagnetic wave is: (in T)
| 1. | \(B_z=2.3 \times 10^{-7} \sin \left[0.6 \times 10^3 x-1.8 \times 10^{11} t\right]\) |
| 2. | \(B_z=2.3 \times 10^{-7} \sin \left[0.6 \times 10^3 x+1.8 \times 10^{11} t\right]\) |
| 3. | \(B_y=69 \sin \left[0.6 \times 10^3 x+1.8 \times 10^{11} t\right]\) |
| 4. | \( B_y=2.3 \times 10^{-7} \sin \left[0.6 \times 10^3 x-1.8 \times 10^{11} t\right] \) |
Subtopic: Properties of EM Waves |
78%
Level 2: 60%+
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An electromagnetic wave of frequency \(100~\text{MHz}\) propagates through a medium of conductivity, \(\sigma= 10~\text{mho/m}\). The ratio of maximum conducting current density to maximum displacement current density is:
\(\left[\text {Take}~ \dfrac{1}{4 \pi \varepsilon_0}=9 \times 10^9 ~\text{Nm}^2 / \text{C}^2\right] \)
1. \(1200\)
2. \(1400\)
3. \(1800\)
4. \(2000\)
\(\left[\text {Take}~ \dfrac{1}{4 \pi \varepsilon_0}=9 \times 10^9 ~\text{Nm}^2 / \text{C}^2\right] \)
1. \(1200\)
2. \(1400\)
3. \(1800\)
4. \(2000\)
Subtopic: Displacement Current |
Level 3: 35%-60%
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