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:  Magnetic Flux |
 67%
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AIPMT - 2012
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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:  Faraday's Law & Lenz Law |
 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 HE 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:  Motional emf |
 64%
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Two concentric circular coils, one of small radius \({r_1}\) and the other of large radius \({r_2},\) such that \({r_1<<r_2},\)  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:  Mutual Inductance |
 67%
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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:  Self - Inductance |
 82%
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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:  Motional emf |
 60%
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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.  if \(\overrightarrow v \|\overrightarrow l\)
b. if  \(\overrightarrow v \|\overrightarrow B\)
c. if  \(\overrightarrow l \|\overrightarrow B\)
d. none of these

Choose the correct option:

1. (a), (b)
2. (b), (c)
3. (d) only
4. (a), (d)

Subtopic:  Motional emf |
 78%
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A rod \(\mathrm{XY}\) 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. \({1 \over \sqrt 3}vBl\)
Subtopic:  Motional emf |
 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
Subtopic:  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:
                 
\(1.~ (\cos \alpha+\sin \alpha) \frac{d B}{d t}\\ 2.~ (\cos \alpha-\sin \alpha) \frac{d B}{d t}\\ 3.~ (\tan \alpha+\cot \alpha) \frac{d B}{d t}\\ 4.~ (\tan \alpha-\cot \alpha) \frac{dB}{d t}\)
Subtopic:  Faraday's Law & Lenz Law |
From NCERT
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