Question: Q. 6. A metallic rod of length ’ $L$ ’ is rotated with angular frequency of ’ $\omega$ ’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $R$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field $B$ parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. $U$ [Delhi I, II, III 2012]
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Solution:
Ans.
The magnitude of the emf, generated across a length $d r$ of the rod, as it moves at right angles to the magnetic field, is given by
$$ d \varepsilon=B v d r $$
Therefore,
$\varepsilon=\oint d \varepsilon=\int_{0}^{R} B v d r=\int_{0}^{R} B \omega r d r=\frac{B \omega R^{2}}{2}$
[CBSE Marking Scheme 2012]
