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A conducting rod AC of length 3L is rotated with angular velocity $ω$ as shown in diagram about a point O in a uniform magnetic field B directed into the paper. AO = L and OC = 2L. Then

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$V_{A} - V_{C} = \frac{5}{2} B ω L^{2}$

$V_{O} - V_{C} = 2 B ω L^{2}$

$V_{A} - V_{C} = \frac{3}{2} B ω L^{2}$

Both 2 and 3

answer is D.

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Detailed Solution

$V_{0} = 0, V_{A} = \frac{- B ω L^{2}}{2}; V_{B} = \frac{B ω {(2 L)}^{2}}{2}$

$V_{A} - V_{C} = \frac{3 B ω L^{2}}{2}; V_{O} - V_{C} = 2 B ω L^{2}$

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