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Q. No.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 |
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Level 3: 35%-60%
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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 |
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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 |
51%
Level 3: 35%-60%
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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 |
Level 3: 35%-60%
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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 |
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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 |
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Level 2: 60%+
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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 |
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Level 3: 35%-60%
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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 |
Level 3: 35%-60%
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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 |
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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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