Q. No.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.
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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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1. \(\varepsilon_1=0,\varepsilon_2\neq0\)
2. \(\varepsilon_1\neq0,\varepsilon_2=0\)
3. \(\varepsilon_1=0,\varepsilon_2=0\)
4. \(\varepsilon_1\neq0,\varepsilon_2\neq0\)
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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 |
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| 1. | falls with uniform velocity. |
| 2. | accelerates down with acceleration less than \(g\). |
| 3. | accelerates down with acceleration equal to \(g\). |
| 4. | moves down and eventually comes to rest. |
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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. | \(\dfrac{U_L}{P_R}\) | 2. | \(\dfrac{U_L~\mathrm {ln}2}{P_R}\) |
| 3. | \(\dfrac{2U_L~\mathrm{ln 2}}{P_R}\) | 4. | \(\dfrac{2U_L}{P_R}\) |
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| 1. | \(\dfrac{2BR}{\lambda}\) | 2. | \(\dfrac{BR}{\lambda}\) |
| 3. | \(\dfrac{BR}{2\lambda}\) | 4. | zero |
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