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Q.

The magnitude of magnetic induction at mid point O due to current arrangement as shown in figure will be

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a

$\frac{μ_{0} I}{4 πa}$

b

0

c

$\frac{μ_{0} I}{πa}$

d

$\frac{μ_{0} I}{2 πa}$

answer is A.

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Detailed Solution

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$\vec{B}$ due to wire segments AB and ED is zero at O,
$\begin{array}{l} {(B_{net})}_{at O} = \frac{μ_{0} I}{4 π \times \frac{a}{2}} [\sin 90^{\circ} + \sin 0^{\circ}] \times 2 \\ = \frac{μ_{0} I}{πa} \end{array}$

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