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Q. No.A parallel plate capacitor is being charged by means of a constant current \(i.\) The plates are circular (of radius \(R)\) and are separated by a distance \(d.\) The magnetic field between the plates, at a distance \(\dfrac{R}{2}\) from the central axis is:
1.
zero
2.
\(\dfrac{\mu_{0} i}{2 \pi \left(\dfrac{R}{2}\right)}\)
3.
\(\dfrac{1}{4}\dfrac{\mu_{0} i}{2 \pi R}\)
4.
\(\dfrac{1}{2}\dfrac{\mu_0 i}{2\pi R}\)
Subtopic: Magnetic Flux |
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A thin metallic plate is allowed to fall through the space between two magnetic poles creating a horizontal magnetic field. The plate is vertical, and its face is perpendicular to the field lines as it falls. While it is entering the region of the magnetic field,
| 1. | The acceleration of the plate is equal to \(g.\) |
| 2. | The acceleration of the plate is greater than \(g.\) |
| 3. | The acceleration of the plate is less than \(g.\) |
| 4. | The plate comes to a stop and rebounds upward. |
Subtopic: Eddy Current |
71%
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A rectangular loop of conducting wire is bent symmetrically so that its two plane halves are inclined at right angles with respect to each other (i.e. \(\angle { PQR }=\angle S T U=90^{\circ}\)). Every segment has a length '\(a\)' \((PQ=QR=RS=...=UP=a)\). A uniform time-dependent magnetic field \(B(t)\) acts on the loop, making an angle '\(\alpha\)' with the lower half of the loop and '\(90^o - \alpha \)' with the upper half. The EMF induced in the loop is proportional to:
| 1. | \((\cos \alpha+\sin \alpha) \dfrac{d B}{d t}\) |
| 2. | \( (\cos \alpha-\sin \alpha) \dfrac{d B}{d t}\) |
| 3. | \((\tan \alpha+\cot \alpha) \dfrac{d B}{d t}\) |
| 4. | \( (\tan \alpha-\cot \alpha) \dfrac{dB}{d t}\) |
Subtopic: Faraday's Law & Lenz Law |
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A circular loop of radius \(R\), enters a region of uniform magnetic field \(B\) as shown in the diagram. The field \((B)\) is perpendicular to the plane of the loop while the velocity of the loop, \(v\), is along its plane. The induced EMF:
| 1. | increases continuously. |
| 2. | decreases continuously. |
| 3. | first increases and then decreases. |
| 4. | remains constant throughout. |
Subtopic: Faraday's Law & Lenz Law |
70%
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The current through the inductor in the figure is initially zero. The initial rate of change of the current \(i\) through the inductor (i.e. \(\dfrac{di}{dt}\)) is:
| 1. | zero | 2. | \(-\dfrac{I_{0} R}{L}\) |
| 3. | \(\dfrac{I_{0} R}{L}\) | 4. | \(\dfrac{I_{0} R}{2L}\) |
Subtopic: LR circuit |
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A rectangular conducting wire-frame having dimensions of \(a × 2a\) is bent symmetrically so that its two halves are at right-angle with respect to each other. A uniform, constant magnetic field \(B\) acts parallel to one of the bent sides, initially. The wire frame begins to rotate with a uniform angular speed \(\omega\) about the bend-line, \(PQ\). The emf induced in the loop will have the form:
| 1. | \(2\omega Ba^2\sin\omega t\) |
| 2. | \(2\omega Ba^2\cos\omega t\) |
| 3. | \(\omega Ba^2(\cos\omega t+\sin\omega t)\) |
| 4. | \(\omega Ba^2(\cos\omega t-\sin\omega t)\) |
Subtopic: Faraday's Law & Lenz Law |
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A triangular wire frame, in the form of an equilateral triangle \(PQR\) moves with a uniform velocity into a region where there is a uniform magnetic field \(B\). The edge \(PQ\) is parallel to the boundary of the region and the velocity \(v\) is perpendicular to it. The emf(\(E\)) induced within the frame is plotted as a function of time \(t,\) starting from when the frame enters the magnetic field. \(E\) is given by:
| 1. | \(Bv^2t\) | 2. | \(2Bv^2t\) |
| 3. | \(\dfrac{\sqrt3}{2}Bv^2t\) | 4. | \(\dfrac{2}{\sqrt3}Bv^2t\) |
Subtopic: Motional emf |
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The two long, parallel wires shown in the diagram carry equal and opposite currents \(i\). The currents change linearly with time: \(\dfrac{di} {dt}\) = a constant = \(K\). The small circuit is situated midway between the wires and has an area \(A\). The emf induced in the small circuit is:
| 1. | zero | 2. | \(\dfrac{\mu_{0} A K}{2 \pi l}\) |
| 3. | \(\dfrac{\mu_{0} A K}{ \pi l}\) | 4. | \(\dfrac{2 \mu_{0} A K}{\pi l}\) |
Subtopic: Magnetic Flux |
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In the system shown in the figure the horizontal rod falls vertically down under its own weight while retaining electrical contact with parallel rails. There is no resistance in the circuit, and there is a uniform horizontal magnetic field into the plane.
The current through the circuit is \(i\). Then:
1. \(i= CBlg\)
2. \(i> CBlg\)
3. \(i < CBlg\)
4. \(i= 0\)
The current through the circuit is \(i\). Then:
1. \(i= CBlg\)
2. \(i> CBlg\)
3. \(i < CBlg\)
4. \(i= 0\)
Subtopic: Motional emf |
52%
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In the system shown in the figure, the horizontal rod falls vertically down under its own weight while retaining electrical contact with parallel rails. There is no resistance in the circuit, and there is a uniform horizontal magnetic field into the plane. The acceleration of the rod \(PQ\), as it falls down is '\(a\)'.
Then:
Then:
| 1. | \(a=g\) |
| 2. | \(a>g\) |
| 3. | \(a<g\) |
| 4. | \(a\) is initially less than \(g\), but later it is greater than \(g\). |
Subtopic: Motional emf |
70%
From NCERT
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