SATHEE: electromagnetic-induction Question 12

इलेक्ट्रोमैग्नेटिक इंडक्शन

Question: Q. 6. A metallic rod of length ’ $L$ ’ is rotated with angular frequency of ’ $ω$ ’ 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

$d ε = B v d r$

Therefore,

$ε = \oint d ε = \int_{0}^{R} B v d r = \int_{0}^{R} B ω r d r = \frac{B ω R^{2}}{2}$

[CBSE Marking Scheme 2012]

Short Answer Type Questions-II

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