A conducting rod $P Q$ of length $l = 2 m$ is moving at a speed of $2 m$ $s^{- 1}$ making an angle of $30^{o}$ with its length. A uniform magnetic field $B = 2 T$ exists in a direction perpendicular to the plane of motion. Then

V_{P} - V_{Q} = 8 V

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V_{P} - V_{Q} = 4 V

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V_{Q} - V_{P} = 8 V

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V_{Q} - V_{P} = 4 V

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Solution

The correct option is B $V_{P} - V_{Q} = 4 V$
$E = B v L s i n (θ)$
$E = 2 \times 2 \times 2 s i n (30^{o}) = 4 V$
By right hand rule:
$V_{P} > V_{Q}$
$V_{P} - V_{Q} = 4 V$

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