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The figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normally to the plane of the loop away from the reader. Then:

1. | for the rectangular loop abcd, the induced current is clockwise. |

2. | for the triangular loop abc, the induced current is clockwise. |

3. | for the irregularly shaped loop abcd, the induced current is anti-clockwise. |

4. | none of these. |

Subtopic: Â Faraday's Law & Lenz Law |

Â 66%

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A wheel with \(10\) metallic spokes each \(0.5\) m long is rotated with a speed of \(120\) rev/min in a plane normal to the horizontal component of earth’s magnetic field H_{E} at a place. If \(H_E=0.4\) G at the place, what is the induced emf between the axle and the rim of the wheel? (\(1\) G=\(10^{-4}\) T)

1. \(5.12\times10^{-5}\) T

2. \(0\)

3. \(3.33\times10^{-5}\)

4. \(6.28\times10^{-5}\)

Subtopic: Â Motional emf |

Â 64%

From NCERT

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Two concentric circular coils, one of small radius \(\mathrm{r_1}\)_{} and the other of large radius \(\mathrm{r_2},\) such that \(\mathrm{r_1<<r_2},\) are placed co-axially with centres coinciding. The mutual inductance of the arrangement is:

1. \(\frac{\mu_0\pi r_1^2}{3r_2}\)

2. \(\frac{2\mu_0\pi r_1^2}{r_2}\)

3. \(\frac{\mu_0\pi r_1^2}{r_2}\)

4. \(\frac{\mu_0\pi r_1^2}{2r_2}\)

Subtopic: Â Mutual Inductance |

Â 67%

From NCERT

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The expression for the magnetic energy stored in a solenoid in terms of magnetic field \(B\), area \(A\) and length \(l\) of the solenoid is:

1. \(
\frac{1}{\mu_0}B^2Al\)

2. \(
\frac{1}{2\mu_0}B^2Al\)

3. \(
\frac{2}{\mu_0}B^2Al\)

4. \(
\frac{3}{2\mu_0}B^2Al\)

Subtopic: Â Self - Inductance |

Â 82%

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In a coil of resistance \(10\) \(\Omega\), the induced current developed by changing magnetic flux through it is shown in the figure as a function of time. The magnitude of change in flux through the coil in Weber is:

1. \(2\)

2. \(6\)

3. \(4\)

4. \(8\)

Subtopic: Â Magnetic Flux |

Â 67%

From NCERT

AIPMT - 2012

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A rod of length \(l\) rotates with a uniform angular velocity \(\omega\) about its perpendicular bisector. A uniform magnetic field \(B\) exists parallel to the axis of rotation. The potential difference between the two ends of the rod is:

1. zero

2. \(\frac{1}{2}Bl\omega ^{2}\)

3. \(Bl\omega ^{2}\)

4. \(2Bl\omega ^{2}\)

Subtopic: Â Motional emf |

Â 60%

From NCERT

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A conducting rod is moved with a constant velocity \(v\) in a magnetic field. A potential difference appears across the two ends,

a. | \(\overrightarrow v \|\overrightarrow l\) | if

b. | if \(\overrightarrow v \|\overrightarrow B\) |

c. | \(\overrightarrow l \|\overrightarrow B\) | if

d. | none of these |

Choose the correct option:

1. | (a), (b) |

2. | (b), (c) |

3. | (d) only |

4. | (a), (d) |

Subtopic: Â Motional emf |

Â 78%

From NCERT

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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. \(vBl \over 2\)

3. \({\sqrt 3 \over 2}vBl\)

4. \({1 \over \sqrt 3}vBl\)

1. \(vBl\)

2. \(vBl \over 2\)

3. \({\sqrt 3 \over 2}vBl\)

4. \({1 \over \sqrt 3}vBl\)

Subtopic: Â Motional emf |

Â 73%

From NCERT

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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: Â Magnetic Flux |

From NCERT

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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. \(\angle \text { PQR }=\angle S T U=90^{\circ}\)). 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:

\(1.~ (\cos \alpha+\sin \alpha) \frac{d B}{d t}\\ 2.~ (\cos \alpha-\sin \alpha) \frac{d B}{d t}\\ 3.~ (\tan \alpha+\cot \alpha) \frac{d B}{d t}\\ 4.~ (\tan \alpha-\cot \alpha) \frac{dB}{d t}\)

\(1.~ (\cos \alpha+\sin \alpha) \frac{d B}{d t}\\ 2.~ (\cos \alpha-\sin \alpha) \frac{d B}{d t}\\ 3.~ (\tan \alpha+\cot \alpha) \frac{d B}{d t}\\ 4.~ (\tan \alpha-\cot \alpha) \frac{dB}{d t}\)

Subtopic: Â Faraday's Law & Lenz Law |

From NCERT

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