The emf induced in a straight line conductor of length l, moving perpendicular to the magnetic field B_o with a velocity v at angle of 45° to the normal to the rod is

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$B_{0} ℓ \sqrt{2}$

$\sqrt{\frac{2}{3}} B_{0} ℓ$

$2 \sqrt{2} B_{0} ℓv$

$B_{0} ℓv \frac{1}{\sqrt{2}}$

answer is D.

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Detailed Solution

$E = (\bar{v} \times \bar{B}) \cdot \bar{ℓ}$ $From fig. δ^{'} = δ$
$\bar{ℓ} = \bar{AB}$ $E = vBℓcos δ = vBℓsin θ$
$vBℓ \cdot \frac{1}{\sqrt{2}}$ [Note End of rod is +ve]

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