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Copper disc of radius 0.1 m is rotated about its centre with 20 rev/s in a uniform magnetic field. EMF INDUCED across disc is pi/20 V, pi/10V, 20 pi mV, 10 pi mV

Copper disc of radius 0.1 m is rotated about its centre with 20 rev/s in a uniform magnetic field. EMF INDUCED across disc is pi/20 V, pi/10V, 20 pi mV, 10 pi mV

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2 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
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The emf in this case would be given by:
\varepsilon =\frac{\omega Br^{2}}{2}=\frac{20\times 2\pi\times B\times 0.1^{2}}{2}=0.2\pi B volts
Regards
Sumit
ankit singh
askIITians Faculty 614 Points
3 years ago
Answer : option (a) \frac{\pi}{10}V
explanation : use formula, \xi_{in}=\frac{1}{2}B\omega R^2
where \xi_{in} denotes induced emf, B is magnetic field, \omega denotes angular frequency of disc and R denotes the radius of disc.
here, B = 0.1 T , R = 0.1 m and \omega=10rps=20π rad/s
hence, \xi_{in} = 1/2 × 0.1 × 20π × (0.1)²
= 1/2 × 20π × 0.01
= π/10 V

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