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A conducting rod PQ is rotated in about an axis passing through O and perpendicular to the plane of rotation of the rod with constant angular speed $ω$ . A magnetic field of induction B is applied perpendicular to the plane of rotation. The one end of the rod is at a distance " $a$ " and other end is at a distance " $b$ " from O. Find induced emf across the ends of the rod.

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$\frac{B ω (b^{2} - a^{2})}{4}$

$\frac{B ω (b^{2} - a^{2})}{2}$

$\frac{B ω (b^{2} + a^{2})}{2}$

$\frac{B ω (b^{2} + a^{2})}{3}$

answer is B.

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Detailed Solution

The induced emf across the element of length dx is

$d e = B v_{x} d x = B (ω x) d x .$

The emf across the whole rod $e = B ω \int_{a}^{b} x d x = \frac{B ω (b^{2} - a^{2})}{2}$

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