A conducting rod AB of length l is rotated about point O which lies between point A and B in a uniform magnetic field B directed into the paper. If length AO = l/3, then (VA - VB) is ω = angular velocity (a) (5Bl²ω)/(6) (b) (6Bl²ω)/(5) (c) (Bl²ω)/(6) (d) (Bl²ω)/(3)

Answer: (c). Emf between pivot O and a point at distance r: 1/2Bω r²; both ends are at lower potential than O (same sense). AO = l/3: VO - VA = 1/2B²/9 = (Bω l²)/(18) OB = 2l/3: VO - VB = 1/2Bω4l²/9 = (4Bω l²)/(18) VA - VB = (4Bω l²)/(18) - (Bω l²)/(18) = (Bω l²)/(6) So the answer is (c) (Bl²ω)/(6).

A conducting rod AB of length l is rotated about point O which lies… | MHT CET 2026 (12 May, Shift 1)

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Rotating Conductors & AC Generator

A conducting rod AB of length $l$ is rotated about point O which lies between point A and B in a uniform magnetic field $\vec{B}$ directed into the paper. If length $AO = \frac{l}{3}$ , then $(V_A - V_B)$ is $\{\omega =$ angular velocity $\}$

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Answer: (c)

Emf between pivot O and a point at distance $r$ : $\frac{1}{2}B\omega r^2$ ; both ends are at lower potential than O (same sense).
$AO = \frac{l}{3}$ : $V_O - V_A = \frac{1}{2}B\omega\frac{l^2}{9} = \frac{B\omega l^2}{18}$
$OB = \frac{2l}{3}$ : $V_O - V_B = \frac{1}{2}B\omega\frac{4l^2}{9} = \frac{4B\omega l^2}{18}$
$V_A - V_B = \frac{4B\omega l^2}{18} - \frac{B\omega l^2}{18} = \frac{B\omega l^2}{6}$
So the answer is (c) $\frac{Bl^2\omega}{6}$ .

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