Q. No.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.
\(\dfrac{2 \mu_{0} A K}{\pi l}\)
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| 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. | \(\dfrac{1}{2}\dfrac{\mu_0 i}{2\pi R}\) |
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| 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. |
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| 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. | \( (\tan \alpha-\cot \alpha) \dfrac{dB}{d t}\) |
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| 1. | increases continuously. |
| 2. | decreases continuously. |
| 3. | first increases and then decreases. |
| 4. | remains constant throughout. |
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| 1. | zero | 2. | \(-\dfrac{I_{0} R}{L}\) |
| 3. | \(\dfrac{I_{0} R}{L}\) | 4. | \(\dfrac{I_{0} R}{2L}\) |
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| 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. | \(\omega Ba^2(\cos\omega t-\sin\omega t)\) |
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| 1. | \(Bv^2t\) | 2. | \(2Bv^2t\) |
| 3. | \(\dfrac{\sqrt3}{2}Bv^2t\) | 4. | \(\dfrac{2}{\sqrt3}Bv^2t\) |
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The current through the circuit is \(i\). Then:
1. \(i= CBlg\)
2. \(i> CBlg\)
3. \(i < CBlg\)
4. \(i= 0\)
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Then:
| 1. | \(a=g\) |
| 2. | \(a>g\) |
| 3. | \(a<g\) |
| 4. | \(a\) is initially less than \(g\), but later it is greater than \(g\). |
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