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प्रश्न
A conducting disc of radius r rotates with a small but constant angular velocity ω about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery of the disc.
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उत्तर
The angular velocity of the disc is ω. Also, the magnetic field of magnitude B is perpendicular to the disc.
Let us take a circular element of thickness da at a distance a from the centre.
Linear speed of the element at a from the centre, v = ωa
Now,
\[de = Blv\]
\[de = B \times da \times a\omega\]
\[ \Rightarrow e = \int_0^r \left( B\omega a \right)da\]
\[ e = \frac{1}{2}B\omega r^2\]
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