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A thin rod AB of length 2m is moving with velocity $v = 3 m / s$ in a uniform magnetic field of induction B = 1.5 Tesla. Direction of velocity vector makes an angle of $θ = 30^{0}$ with the length of the rod. Then potential difference between the ends is given by ______

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$V_{A} - V_{B} = 4.5 v o l t$

$V_{B} - V_{A} = 4.5 v o l t$

$V_{A} - V_{B} = 4 v o l t$

$V_{B} - V_{A} = 3 v o l t$

answer is A.

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Detailed Solution

End A will be at higher potential.

$\begin{array}{l} ∴ V_{A} - V_{B} = B l V \sin θ = 1.5 \times 2 \times 3 \times \sin 30^{0} v o l t \\ \Rightarrow V_{A} - V_{B} = 4.5 v o l t \end{array}$

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