A conducting rod AC of length 4 ℓ is rotated about a point O in a uniform magnetic field B directed into the paper. AO = ℓ and OC = 3 ℓ . Then :

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For a rod rotating about an interior point O, each arm acts as a source of emf $\frac{1}{2} B ω r^{2}$ ; for the rotation shown, both ends are at lower potential than O.

With AO= $ℓ$ and OC= $3 ℓ$ : $V_{A} - V_{O} = - \frac{1}{2} B ω ℓ^{2}$ and $V_{C} - V_{O} = - \frac{9}{2} B ω ℓ^{2}$ .

$V_{A} - V_{C} = (- \frac{1}{2} + \frac{9}{2}) B ω ℓ^{2} = 4 B ω ℓ^{2}$ . Answer: (C).

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