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A conducting rod $P Q$ of length $8 l$ is rotated about a point $O$ at distance $3 l$ from $P$ , in a uniform magnetic field $\overset{⇀}{B}$ directed into the paper. Then

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$V_{P} - V_{O} = \frac{B ω l^{2}}{2}$

$V_{O} - V_{Q} = \frac{25 B ω l^{2}}{2}$

$V_{P} - V_{Q} = 8 B ω l^{2}$

$V_{O} - V_{P} = \frac{9}{2} B ω l^{2}$

answer is B, C, D.

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Detailed Solution

Using Fleming’s Right Hand Rule,

$ξ_{1} = V_{O} - V_{P} = \frac{1}{2} B ω {(3 l)}^{2} = \frac{9}{2} B ω l^{2}$

$ξ_{2} = V_{O} - V_{Q} = \frac{1}{2} B ω {(5 l)}^{2} = \frac{25}{2} B ω l^{2}$

$\Rightarrow ξ_{2} - ξ_{1} = V_{O} - V_{Q} - V_{O} + V_{P} \Rightarrow ξ_{2} - ξ_{1} = - V_{Q} + V_{P} = 8 B ω l^{2}$

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