Q. No.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.
\(\dfrac{2}{\sqrt3}Bv^2t\)
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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. | \(\dfrac{U_L}{P_R}\) | 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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| 1. | \(\dfrac{\mu_0A}{L}\cdot N\) | 2. | \(\dfrac{\mu_0A}{L}\cdot N^2\) |
| 3. | \(\dfrac{\mu_0L^3}{A}\cdot N\) | 4. | \(\dfrac{\mu_0L^3}{A}\cdot N^2\) |
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| 1. | \(\dfrac{1}{2\pi}\sqrt{\dfrac{L}{C}}\) | 2. | \(\sqrt{\dfrac{L}{C}}\) |
| 3. | \(2\pi\sqrt{\dfrac{L}{C}}\) | 4. | \(2\sqrt{\dfrac{L}{C}}\) |
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1. \(2\) A/s
2. \(2\times10^{-3}\) A/s
3. \(2\times10^{3}\) A/s
4. \(2\times10^{-6}\) A/s
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1. \(x\)
2. \(\sqrt{r^2-x^2}\)
3. \(r\)
4. \(x\sqrt{r^2-x^2}\)
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