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AB and DE are very long straight non-coplanar conductors at right angles. A conductor BCD in the form of 3/4th of a circle of radius r connects the two straight conductors. If the circular coil is in the plane of AB (i.e.,xy - plane) and normal to DE, then the magnetic induction at the centre O, due to current I is

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$\frac{μ_{0} I}{8 πr} [(3 π + 2) (- \hat{k}) + 2 (\hat{j})]$

$\frac{μ_{0} I}{8 πr} [3 π + 4] (- \hat{k})$

$\frac{μ_{0} I}{8 πr} [(4 + 3 π) (\hat{k}) + 2 \hat{j}]$

$\frac{μ_{0} I}{8 πr} [3 π + 4] (\hat{k})$

answer is B.

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