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Q. No.A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
| 1. | non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates |
| 2. | zero between the plates and non-zero outside |
| 3. | zero at all places |
| 4. | constant between the plates and zero outside the plates |
Subtopic: Displacement Current |
Level 3: 35%-60%
NEET - 2025
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A parallel plate capacitor is charged by connecting it to a battery through a resistor. If \(i\) is the current in the circuit, then in the gap between the plates:
| 1. | A displacement current of magnitude equal to \(i\) flows in the same direction as \(i.\) |
| 2. | A displacement current of magnitude equal to \(i\) flows in the opposite direction to \(i.\) |
| 3. | A displacement current of magnitude greater than \(i\) flows but it can be in any direction. |
| 4. | There is no current. |
Subtopic: Displacement Current |
63%
Level 2: 60%+
NEET - 2024
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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: Displacement Current |
80%
Level 1: 80%+
NEET - 2023
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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:
\(V=V_0 \sin \omega t\)
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: Displacement Current |
59%
Level 3: 35%-60%
NEET - 2021
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A parallel plate capacitor of capacitance \(20~\mu\text{F}\) is being charged by a voltage source whose potential is changing at the rate of \(3~\text{V/s}.\) The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:
| 1. | zero, zero | 2. | zero, \(60~\mu\text{A}\) |
| 3. | \(60~\mu\text{A},\) \(60~\mu\text{A}\) | 4. | \(60~\mu\text{A},\) zero |
Subtopic: Displacement Current |
71%
Level 2: 60%+
NEET - 2019
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A \(100~\Omega\) resistance and a capacitor of \(100~\Omega\) reactance are connected in series across a \(220~\text{V}\) source. When the capacitor is \(50\%\) charged, the peak value of the displacement current is:
1. \(2.2~\text{A}\)
2. \(11~\text{A}\)
3. \(4.4~\text{A}\)
4. \(11\sqrt{2}~\text{A}\)
Subtopic: Displacement Current |
62%
Level 2: 60%+
NEET - 2016
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Match List-I with List-II (the symbols carry their usual meaning).
Choose the correct answer from the options given below:
| List-I | List-II | ||
| A. | \( \oint \vec{E} \cdot d \vec{A}=\dfrac{Q}{\varepsilon_0}\) | I. | Ampere-Maxwell's law |
| B. | \( \oint \vec{B} \cdot d \vec{A}=0 \) | II. | Faraday's law |
| C. | \( \oint \vec{E} \cdot d\vec{ l}=\dfrac{-d(\phi)}{d t} \) | III. | Gauss's law of electrostatics |
| D. | \( \oint \vec{B} \cdot d\vec{l}=\mu_0 i_c+ \mu_0 \varepsilon_0 \dfrac{d\left(\phi_E\right)}{d t}\) | IV. | Gauss's law of magnetism |
| 1. | A-III, B-IV, C-II, D-I | 2. | A-IV, B-III, C-II, D-I |
| 3. | A-III, B-II, C-IV, D-I | 4. | A-IV, B-I, C-III, D-II |
Subtopic: Maxwell's Equations |
83%
Level 1: 80%+
NEET - 2024
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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: Generation of EM Waves |
90%
Level 1: 80%+
NEET - 2016
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The electric field in a plane electromagnetic wave is given by \(E_z=60\cos(5x+1.5\times10^9t)~\text{V/m}.\) Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
| 1. | \(B_z=60\cos(5x+1.5\times10^9t)~\text T\) |
| 2. | \(B_y=60\sin(5x+1.5\times10^9t)~\text T\) |
| 3. | \(B_y=2\times10^{-7}\cos(5x+1.5\times10^9t)~\text T\) |
| 4. | \(B_x=2\times10^{-7}\cos(5x+1.5\times10^9t)~\text T\) |
Subtopic: Properties of EM Waves |
55%
Level 3: 35%-60%
NEET - 2025
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If the ratio of relative permeability and relative permittivity of a uniform medium is \(1 : 4.\) The ratio of the magnitudes of electric field intensity \((E)\) to the magnetic field intensity \((H)\) of an EM wave propagating in that medium is:\(\left(\text{Given that}\sqrt{\frac{\mu_0}{\varepsilon_0}}=120\pi\right)\)
| 1. | \(30\pi:1\) | 2. | \(1:120\pi\) |
| 3. | \(60\pi:1\) | 4. | \(120\pi:1\) |
Subtopic: Properties of EM Waves |
Level 3: 35%-60%
NEET - 2024
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