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Q. No.A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle:
| 1. | does not deflect. |
| 2. | deflects for a very short time and then comes back to the original position. |
| 3. | deflects and remains deflected as long as the battery is connected. |
| 4. | deflects and gradually comes to the original position in a time which is large compared to the time constant. |
Subtopic: Â Displacement Current |
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Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor:
| (a) | increases |
| (b) | decreases |
| (c) | does not change |
| (d) | is zero |
Choose the correct option:
1. (a), (b)
2. (b), (c)
3. only (c)
4. (a), (d)
Subtopic: Â Displacement Current |
 71%
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A capacitor of capacitance \(C\) is connected across an AC source of voltage \(V\), given by;
\(V=V_0 \sin \omega t\)
The displacement current between the plates of the capacitor would then be given by:
1. \( I_d=\dfrac{V_0}{\omega C} \sin \omega t \)
2. \( I_d=V_0 \omega C \sin \omega t \)
3. \( I_d=V_0 \omega C \cos \omega t \)
4. \( I_d=\dfrac{V_0}{\omega C} \cos \omega t\)
Subtopic: Â Displacement Current |
 57%
From NCERT
NEET - 2021
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To produce an instantaneous displacement current of \(2~\text{mA}\) in the space between the parallel plates of a capacitor of capacitance \(4~\mu\text{F}\), the rate of change of applied variable potential difference \(\left(\frac{dV}{dt}\right)\) must be:
1. \( 800~ \text{V} / \text{s} \)
2. \( 500~ \text{V} / \text{s} \)
3. \( 200~ \text{V} / \text{s} \)
4. \( 400 ~\text{V} / \text{s}\)
1. \( 800~ \text{V} / \text{s} \)
2. \( 500~ \text{V} / \text{s} \)
3. \( 200~ \text{V} / \text{s} \)
4. \( 400 ~\text{V} / \text{s}\)
Subtopic: Â Displacement Current |
 78%
From NCERT
NEET - 2023
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Which of the following situation(s), can cause the radiation of an electromagnetic wave?
Choose the correct options:
| (A) | a varying sinusoidal current flowing through a capacitor |
| (B) | an electric dipole, whose size (and magnitude) is oscillating with time |
| (C) | a steady current flowing through a toroid |
Choose the correct options:
| 1. | only (A) |
| 2. | only (B) |
| 3. | only (A) & (B) |
| 4. | (A), (B), (C) |
Subtopic: Â Generation of EM Waves |
 65%
From NCERT
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Consider the following statements:
| Statement I: | Charged particles which undergo acceleration or deceleration radiate their energy away. |
| Statement II: | Therefore, charged particles moving in circular paths in a uniform magnetic field should also radiate their energy. |
| 1. | Statement I is true, Statement II is true and Statement I implies Statement II. |
| 2. | Statement I is true, Statement II is true and Statement I does not imply Statement II. |
| 3. | Statement I is true, Statement II is false. |
| 4. | Statement I is false, Statement II is true. |
Subtopic: Â Generation of EM Waves |
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Out of the following options which one can be used to produce a propagating electromagnetic wave?
| 1. | a stationary charge. |
| 2. | a chargeless particle. |
| 3. | an accelerating charge. |
| 4. | a charge moving at constant velocity. |
Subtopic: Â Generation of EM Waves |
 89%
From NCERT
NEET - 2016
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Assume a bulb of efficiency \(2.5\%\) as a point source. The peak values of the electric field and magnetic field produced by the radiation coming from a \(100~\text{W}\) bulb at a distance of \(3~\text{m}\) are respectively:
| 1. | \( 2.5 ~\text{V/m}, ~2.2 \times 10^{-8} ~\text{T} \) |
| 2. | \( 3.6 ~\text{V/m}, ~ 3.6 ~\text{T} \) |
| 3. | \( 4.07~\text{V/m},~ 1.4 \times 10^{-8} ~\text{T}\) |
| 4. | \( 4.2 ~\text{V/m}, ~3.4 \times 10^{-6}~\text{T}\) |
Subtopic: Â Properties of EM Waves |
 53%
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The magnetic field in a plane electromagnetic wave is given by \({B}=\left(2 \times 10^{-7}\right)\sin \left(0.5 \times 10^3 {x}+1.5 \times 10^{11} {t}\right )~\text{T}\). The wavelength and frequency of the wave are respectively:
| 1. | \( 2.16~\text{cm}, 24.1~\text{GHz} \) | 2. | \( 0.29~\text{cm}, 13.7~\text{GHz} \) |
| 3. | \( 3.23 ~\text{cm}, 20.0~\text{GHz} \) | 4. | \( 1.26~\text{cm}, 23.9~\text{GHz}\) |
Subtopic: Â Properties of EM Waves |
 73%
From NCERT
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The electric field of an electromagnetic wave is given by \(\overrightarrow E = E_0 \hat j \cos (\omega t - kx)+ E_0\hat i \sin (\omega t -kx)\).
The maximum value of the electric field in the wave is:
1. \(\dfrac {E_0} {\sqrt 2}\)
2. \(E_0\)
3. \(\sqrt 2 E_0\)
4. \(\sqrt 3 E_0\)
The maximum value of the electric field in the wave is:
1. \(\dfrac {E_0} {\sqrt 2}\)
2. \(E_0\)
3. \(\sqrt 2 E_0\)
4. \(\sqrt 3 E_0\)
Subtopic: Â Properties of EM Waves |
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