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Q. No.In the circuit shown in the adjoining figure, the switch was kept at the position \('1'\) for a long time. The switch \(K\) is suddenly (and smoothly) shifted to position \('2'.\)
The current through the cell, just after the shift, is:
The current through the cell, just after the shift, is:
| 1. | \(\dfrac{V_0}{2R}\) | 2. | \(\dfrac{V_0}{R}\) |
| 3. | \(\dfrac{3V_0}{4R}\) | 4. | zero |
Subtopic: LR circuit |
Level 4: Below 35%
Hints
Faraday's law of electromagnetic induction is used in the operation of:
1. metal detectors
2. jet engines
3. electromagnets
4. LEDs
1. metal detectors
2. jet engines
3. electromagnets
4. LEDs
Subtopic: Faraday's Law & Lenz Law |
Level 4: Below 35%
Hints
A circular wire of radius \(R\) is placed in a uniform magnetic field \(B,\) which acts into the plane as shown. The wire is given a half-turn about a diameter. The resistance per unit length of the wire is \(\lambda.\) The total charge flowing through the wire is:
| 1. | \(\dfrac{2BR}{\lambda}\) | 2. | \(\dfrac{BR}{\lambda}\) |
| 3. | \(\dfrac{BR}{2\lambda}\) | 4. | zero |
Subtopic: Motional emf |
Level 4: Below 35%
Hints
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 |
Level 4: Below 35%
Hints
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 |
Level 3: 35%-60%
Hints
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 |
Level 3: 35%-60%
Hints
Given below are two statements:
| Assertion (A): | Faraday's law of electromagnetic induction is a consequence of Biot-Savart's law. |
| Reason (R): | Currents cause magnetic fields and interact with magnetic flux. |
| 1. | (A) is True but (R) is False. |
| 2. | (A) is False but (R) is True. |
| 3. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 4. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
Subtopic: Faraday's Law & Lenz Law |
Level 3: 35%-60%
Hints
A rod \(PQ\) of length \(L\) moves in a uniform magnetic field \(B\) with a velocity \(v,\) perpendicular to its own length. The magnetic field \(B\) acts in the plane of motion, making an angle \(\theta\) with the rod. The motional emf, \(V_P-V_Q\) is:
1. zero
2. \(BLv\cos\theta\)
3. \(BLv\sin\theta\)
4. \(BLv\)
1. zero
2. \(BLv\cos\theta\)
3. \(BLv\sin\theta\)
4. \(BLv\)
Subtopic: Motional emf |
Level 3: 35%-60%
Hints
A square wire loop of resistance \(0.5\) \(\Omega\)/m, having a side \(10\) cm and made of \(100\) turns is suddenly flipped in a magnetic field \(B,\) which is perpendicular to the plane of the loop. A charge of \(2\times10^{-4}
\) C passes through the loop. The magnetic field \(B\) has the magnitude of:
1. \(2\times10^{-6} \) T
2. \(4\times10^{-6} \) T
3. \(2\times10^{-3} \) T
4. \(4\times10^{-3} \) T
1. \(2\times10^{-6} \) T
2. \(4\times10^{-6} \) T
3. \(2\times10^{-3} \) T
4. \(4\times10^{-3} \) T
Subtopic: Magnetic Flux |
Level 3: 35%-60%
Hints
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 |
Level 3: 35%-60%
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