A thin metallic rod XY of length L is pivoted at a point M on the rod which is at a distance $\frac{L}{4}$ from end X. If the rod is rotated at a constant angular velocity $ω$ about an axis passing through M and perpendicular to the rod in the presence of a uniform magnetic field of magnitude B directed perpendicular to the plane of rotation of the rod, then:

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The emf induced between M and Y is $\frac{9}{32} {BωL}^{2}$

The emf induced between M and Y is $\frac{9}{16} {BωL}^{2}$

The emf induced between X and Y is $\frac{1}{8} {BωL}^{2}$

The emf induced between X and Y is $\frac{1}{4} {BωL}^{2}$

answer is A, D.

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Detailed Solution

We know that the magnitude of the emf induced across two points on the rod that lie at distances $L_{1}$ and $L_{2}$ (such that $L_{1} > L_{2}$ ) from the axis of rotation is $ε = \frac{1}{2} B ω (L_{1}^{2} - L_{2}^{2}) .$ Therefore, the emf induced between M and Y is $ε_{M T} = \frac{1}{2} B ω ({(\frac{3 L}{4})}^{2} - 0) = \frac{9}{32} B ω L^{2} .$ And, the emf induced between X and Y is $ε_{X T} = \frac{1}{2} B ω ({(\frac{3 L}{4})}^{2} - {(\frac{L}{4})}^{2}) = \frac{1}{4} B ω L^{2}$