AB and CD are long straight conductors, distance d apart, carrying a current I. The magnetic field at the midpoint of BC is

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$\frac{{ - {\mu _0}I}}{{\pi d}}\hat k$

$\frac{{ - {\mu _0}I}}{{8\pi d}}\hat k$

$\frac{{ - {\mu _0}I}}{{2\pi d}}\hat k$

$\frac{{ - {\mu _0}I}}{{4\pi d}}\hat k$

answer is B.

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Detailed Solution

The field at the midpoint of BC due to AB is $\left( { - \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{{d/2}}\hat k} \right)$ and the same is due to CD. Therefore, the total field is $\left[ { - \left( {\frac{{{\mu _0}i}}{{\pi d}}} \right)\,\hat k} \right]$

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