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Q. No.A rod \(XY\) of length \(l\) is placed in a uniform magnetic field \(B\), as shown in the diagram. The rod moves with a velocity \(v\), making an angle of \(60^\circ\) with its length. The emf induced in the rod is:
1.
\(vBl\)
2.
\(\dfrac{vBl}{2}\)
3.
\(\dfrac{\sqrt 3}{2}vBl\)
4.
\(\dfrac{1}{\sqrt 3}vBl\)
Subtopic: Motional emf |
73%
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An inductor \((L)\) and a capacitor \((C)\) are connected in a circuit, with the capacitor initially charged to a maximum voltage \(V_{0}\). The switch is now closed. The maximum current in the circuit is:
| 1. | \( \dfrac{V_{0}}{\sqrt{L C}}\) | 2. | \(V_{0}\sqrt{LC}\) |
| 3. | \(V_{0} \sqrt{\dfrac{L}{C}}\) | 4. | \(V_{0} \sqrt{\dfrac{C}{L}}\) |
Subtopic: Self - Inductance |
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An inductor \((L)\) and a resistor \((R)\) are connected in series across a battery of emf \(E,\) and the circuit is switched on. The current rises steadily. The rate of increase of the current \(\left(\text{i.e.,}\dfrac {di} {dt}\right),\) when the voltage drops across the resistor is \(\dfrac{E}{2}\), is given by: \(\dfrac {di} {dt}\) =
| 1. | \(\dfrac{E}{L}\) | 2. | \(\dfrac{E}{2L}\) |
| 3. | \(\dfrac{2E}{L}\) | 4. | \(\dfrac{E}{L}e^{-1}\) |
Subtopic: LR circuit |
84%
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A horizontal square loop of area \(A\) has \(n\) turns of wire. It is immersed in a uniform, rotating magnetic field \(B\) which is initially perpendicular to the plane of the loop. The field rotates with an angular speed \(\omega\) about a diagonal of the loop. The EMF induced across the loop is:
| 1. | constant, of magnitude \(n\omega BA\). |
| 2. | increasing with time \(t\), of magnitude \(n\omega^2BAt\). |
| 3. | decreasing with time \(t\), of magnitude \(\dfrac{nBA}{t}\). |
| 4. | sinusoidal with time \(t\), of amplitude \(n\omega BA\). |
Subtopic: Motional emf |
58%
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The mutual inductance between the two circuits can be determined by simply letting a current \(i\) flow through one circuit and finding the flux of the magnetic field through the second circuit: \(\phi_{2}=M_{12} i_{1}\), where \(M_{12}\) is the mutual inductance. Using this method, or otherwise determine the mutual inductance \((M)\) between a long straight wire, and a small coplanar loop of the area \(A\), located at a distance \(l\) from the wire. The value of \(M\) is:
| 1. | \( \dfrac{\mu_{0} l}{2 \pi}\) | 2. | \(\dfrac{\mu_{0} A}{2 \pi l}\) |
| 3. | \(\dfrac{\mu_{0} l^{3}}{4 \pi A}\) | 4. | \(\dfrac{\mu_{0} A^{2}}{2 \pi l^{3}}\) |
Subtopic: Mutual Inductance |
80%
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A metallic rod of length \(3~\text{m}\) rotates with an angular speed of \(4~\text{rad/s}\) in a uniform magnetic field. The field makes an angle of \(30^{\circ}\) with the plane of rotation. The emf induced across the rod is \(72~\text{mV}\). The magnitude of the field is:
1. \(4 \times 10^{-3}~\text{T}\)
2. \(8 \times 10^{-3}~\text{T}\)
3. \(16 \times 10^{-3}~\text{T}\)
4. \(48 \times 10^{-3}~\text{T}\)
1. \(4 \times 10^{-3}~\text{T}\)
2. \(8 \times 10^{-3}~\text{T}\)
3. \(16 \times 10^{-3}~\text{T}\)
4. \(48 \times 10^{-3}~\text{T}\)
Subtopic: Motional emf |
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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 |
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Given below are two statements:
A rectangular loop of resistance \(R\) is placed in a region where there is a magnetic field \(B\), passing perpendicularly through the plane of the loop, as shown in the figure. The loop is pulled with a constant velocity \(v\) so that it is partially within the field.
A rectangular loop of resistance \(R\) is placed in a region where there is a magnetic field \(B\), passing perpendicularly through the plane of the loop, as shown in the figure. The loop is pulled with a constant velocity \(v\) so that it is partially within the field.
| Assertion (A): | An external force \(F\) is needed to be applied in the direction of the velocity \(v\) so that the loop can move with constant velocity \(v\). |
| Reason (R): | As the loop moves towards the right, the magnetic flux decreases inducing an emf and a corresponding current. This current causes a retarding force to be exerted on the wire. |
| 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: Motional emf |
65%
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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%
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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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