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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.
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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